Re: [PHP-DB] Loading extensions

2001-03-20 Thread Yasuo Ohgaki

I've posted notes on PHP Manual

http://www.php.net/manual/en/install-windows.php

It might help.

--
Yasuo Ohgaki


""Shahmat Dahlan"" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I have IIS 4 with PHP 4pl1 running on Win NT 4. I can't seem to load the
> (*.dll) extensions. There seem to be something wrong with the extension
> path. The browsercap.ini path is pointing to the right path. Can anyone
> help?
>
> Thanks in advance.
> Shahmat
>
>
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[PHP-DB] A NETWORK MARKETER?

2001-03-20 Thread timmyralley8627




ATTENTION:  ANY SERIOUS NETWORK MARKETER
TAKE A LOOK AT THE FOLLOWING OPPORTUNITY I CAME ACROSS!!
IF YOU PLUG INTO THEIR PROVEN SYSTEM, THEY GUARANTEE YOU SUCCESS, 
AND PROVIDE YOU WITH FREE LEADS TO BUILD YOUR GROUP!!


DO NOT LET THIS SLIP AWAY!!  CLICK BELOW AND FIND OUT AS I DID,
WHY THESE GUYS ARE THE LEADERS IN THE INDUSTRY!!



http://www.geocities.com/businessupdates2000




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[PHP-DB] Loading extensions

2001-03-20 Thread Shahmat Dahlan

I have IIS 4 with PHP 4pl1 running on Win NT 4. I can't seem to load the
(*.dll) extensions. There seem to be something wrong with the extension
path. The browsercap.ini path is pointing to the right path. Can anyone
help?

Thanks in advance.
Shahmat


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Re: [PHP-DB] Select where

2001-03-20 Thread Beau Lebens


>___
>
> > $query = "SELECT * FROM members where status='deceased'"
> >
> > Then just call the $query in your script
>___
>
>Thanks, I only gave the mySQL but the php scripting is
>
>
>$deceased = mysql_query(SELECT * FROM members where
>status=\'deceased\'");
>  and later
>while ($myrow = mysql_fetch_row($deceased))
>
>MySQL now says in relation to the *while* line
>Warning: Supplied argument is not a valid MySQL result resource
>
>Tim Morris

in that case, you don't need to escape the single ' since you are using 
double " to enclose your statement, however, you should have got a parse 
error if that is your code exactly, because you are missing a " at the 
start of the statement =)

also, make sure that the values in the db are always lower-case to get the 
match (i *think* it's case-sensitive off the top of my head)

HTH
Beau





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-- 
Beau Lebens - Web Master
Insurance My Way, Your Friendly Online Broker
(08) 9226 5888
http://www.insurancemyway.com.au/


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[PHP-DB] Sessions and MySQL

2001-03-20 Thread Rob Wheeldon

I am trying to implement session support with mysql. I am using
session.set_save_handler but don't really have a great set of functions does
anyone have some good ones?



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Re: [PHP-DB] class method

2001-03-20 Thread bryan

Yeah, the var $query is defined to the class.

I did mistype the result, thanks.

I was able to make this bad boy work with placing

$field = $row['row_name'];  // whatever the row is called.

But, if that is the case, why would I need a class?
I am just trying to make it so that I can use the
loop for any query that needs it, and can show any row
when I need it.

I appreciate your help!  Thanks a lot
bryan



- Original Message -
From: "JJeffman" <[EMAIL PROTECTED]>
To: "db" <[EMAIL PROTECTED]>
Sent: Tuesday, March 20, 2001 1:46 PM
Subject: Re: [PHP-DB] class method


> I need a little more information to help you.
> Is the $query variable a global one ? Or is it a property of the class (
var
> $query; )  ?
> Why are you assigning the "mysql_fetch_array" function to a variable if
call
> it is easier ?
>
> You have misplaced the "if $result" statement, the right position, I think
,
> is :
>
> function getdata() {
> $this->sql = $query;
> $result = mysql_query($query);
> if( $result ) { // Check if $result is valid instead of NOT valid
>// $fetch = "mysql_fetch_array";
>while ($row = mysql_fetch_array($result)) {
>  $this->field = $row['$this->field'];
> }
> }
> else  $this->error = mysql_error();
> return;
> }
>
> HTH
>
> Jayme.
>
> www.conex.com.br/jjeffman/antispam.html
>
> -Mensagem Original-
> De: bryan <[EMAIL PROTECTED]>
> Para: db <[EMAIL PROTECTED]>
> Enviada em: terça-feira, 20 de março de 2001 16:21
> Assunto: [PHP-DB] class method
>
>
> I am really new working with classes.  Some of the ideas I am getting,
some
> I am not.
> I wrote this simple little class, with a connection method, and a method
to
> loop through the
> data.  I am not even sure if my code is write (for the loop), but I am
> having trouble
> displaying the data.  My connection works fine, but my loop is jacked.
> Could anyone
> suggest or maybe correct a mistake (if you see any)?  Just looking to see
if
> I am
> started in the right direction, or if I have this all bass ackwards.
> I have no clue if this is even possible.  I am looking to create a while
> loop for whatever
> query I run, and make each row equal to whatever field I put in.  I have
all
> variables
> defined within the scope of my class.
>
> Thanks for your help.
>
> Code :
> 
> function getdata() {
>
> $this->sql = $query;
>
> $result = mysql_query($query);
>
> $fetch = "mysql_fetch_array";
>
>  while ($row = $fetch($result)) {
>
>  $this->field = $row['$this->field'];
>
>  }
>
>  if(!$result) {
>
>   $this->error = mysql_error();
>
>  return;
>
> }
>
> }
>
> **
> End
>
> thanks!!!
>
> [ bryan fitch . programmer . [EMAIL PROTECTED] ]
>
>
>
>
>
>
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Re: [PHP-DB] No. of Users online right now?

2001-03-20 Thread JJeffman

Have a look on Sessions. You can also write to a database every time a new
visitor runs into your site and check if the other are still connected.

You have to create a script that tracks every movement of the visitor. It is
not a simple script. As far as I know the web server doesn't give you this
information .

HTH.

Jayme.

-Mensagem Original-
De: Dan Eskildsen <[EMAIL PROTECTED]>
Para: <[EMAIL PROTECTED]>
Enviada em: terça-feira, 20 de março de 2001 07:46
Assunto: [PHP-DB] No. of Users online right now?


> Hi there,
>
> This is a NewBie Question:
> While out surfing I have often seen a message like:  "There are 17 users
on
> this site at the moment."
>
> Can I do this with php?  Just to show how many people are on (browsing)
the
> site at that moment?  I run a community website.
>
> Thanks
>
> ==
> Regards from Denmark, Europe
> Please cc your reply to [EMAIL PROTECTED]
> ==
> I haven't lost my mind, I've got it backed up on TAPE somewhere...
>
>
>
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Re: [PHP-DB] Dynamically build a form from mySQL table

2001-03-20 Thread JJeffman

If you are using MySQL you can use the "mysql_list_fields" to get the fields
names from a table.

HTH

Jayme.

-Mensagem Original-
De: Martin E. Koss <[EMAIL PROTECTED]>
Para: <[EMAIL PROTECTED]>
Enviada em: terça-feira, 20 de março de 2001 12:02
Assunto: [PHP-DB] Dynamically build a form from mySQL table


> Hi,
> I've been trying to figure this out by looking through the code in
> phpMyAdmin but I'm not getting very far.
>
> What I want to do is dynamically generate a form where I get a table row
> containing the field name in one cell and the input field in the next
cell,
> for example:
>
> $DatabaseName = "FocusDynamics";
> $TableName = "Items";
>
> // open the table and get all field names, sizes, etc.,
>
> // for each Field name, print something like:
>
> PRINT "\n";
> PRINT "$FieldName\n";
> PRINT " maxlength=\"FieldSize\">\n";
> PRINT "";
>
> How do I get all this info from the table?
>
> Thanks for any help.
>
>
> Martin E. Koss
> M: 07946-706459
> E: [EMAIL PROTECTED]
>
>
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Re: [PHP-DB] class method

2001-03-20 Thread JJeffman

I need a little more information to help you.
Is the $query variable a global one ? Or is it a property of the class ( var
$query; )  ?
Why are you assigning the "mysql_fetch_array" function to a variable if call
it is easier ?

You have misplaced the "if $result" statement, the right position, I think ,
is :

function getdata() {
$this->sql = $query;
$result = mysql_query($query);
if( $result ) { // Check if $result is valid instead of NOT valid
   // $fetch = "mysql_fetch_array";
   while ($row = mysql_fetch_array($result)) {
 $this->field = $row['$this->field'];
}
}
else  $this->error = mysql_error();
return;
}

HTH

Jayme.

www.conex.com.br/jjeffman/antispam.html

-Mensagem Original-
De: bryan <[EMAIL PROTECTED]>
Para: db <[EMAIL PROTECTED]>
Enviada em: terça-feira, 20 de março de 2001 16:21
Assunto: [PHP-DB] class method


I am really new working with classes.  Some of the ideas I am getting, some
I am not.
I wrote this simple little class, with a connection method, and a method to
loop through the
data.  I am not even sure if my code is write (for the loop), but I am
having trouble
displaying the data.  My connection works fine, but my loop is jacked.
Could anyone
suggest or maybe correct a mistake (if you see any)?  Just looking to see if
I am
started in the right direction, or if I have this all bass ackwards.
I have no clue if this is even possible.  I am looking to create a while
loop for whatever
query I run, and make each row equal to whatever field I put in.  I have all
variables
defined within the scope of my class.

Thanks for your help.

Code :

function getdata() {

$this->sql = $query;

$result = mysql_query($query);

$fetch = "mysql_fetch_array";

 while ($row = $fetch($result)) {

 $this->field = $row['$this->field'];

 }

 if(!$result) {

  $this->error = mysql_error();

 return;

}

}

**
End

thanks!!!

[ bryan fitch . programmer . [EMAIL PROTECTED] ]






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[PHP-DB] Transmitting Arrays

2001-03-20 Thread Mark Collin

I have a form which dynamically builds a table by taking data out of a mysql
database

http://www.fyrespray.net/screenshots/admin/modifyscreenshot.php4?pageid=1

but i'm having problems whn i try to modify the data i'm trying to put all
the data into arrays serialize it, urlencode it then send it on to the next
page to put it into database and then send back to the original page when
its done, i'm having a problem actually sendig my arrays around though, the
second page doesn't seem to get any data after i have unserialized it and
urlunecoded it.

anyone have any ideas, i've been fiddleing about a load with the code and
there is a good chance i have totally messed it up now heres the state its
in atm though if you would like to look

http://www.fyrespray.net/screenshots/admin/modifyscreenshot.txt
http://www.fyrespray.net/screenshots/admin/modify.txt

(prolly need to save the first page if your browser is picking up the html
bits and making half of a web page up like mine does)


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Re: [PHP-DB] problems configuring php and postgres - FIXED

2001-03-20 Thread Gary Huntress

My thanks to everyone who provided me info and other hints/tips.

The vast majority of the problem was human error of course, sprinkling in a
few bad assumptions and some configuration problems...its amazing I got it
working at allbut it is :)

Thanks again to everyone.

Gary
"Rasmus Lerdorf" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> But did you have /usr/local/pgsql/include and /usr/local/pgsql/lib
> directories with the correct files in them?
>
> On Tue, 20 Mar 2001, Steve Brett wrote:
>
> >   --with-pgsql[=DIR]  Include PostgreSQL support.  DIR is the
PostgreSQL
> >   base install directory, defaults to
> > /usr/local/pgsql.
> >   Set DIR to shared to build as a dl, or
shared,DIR
> >   to build as a dl and still specify DIR.
> >
> > this did not work when i compiled.
> >
> > pointing at the source did.
> >
> > i tried pointing php at both dirs to no avail.
> >
> >

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[PHP-DB] class method

2001-03-20 Thread bryan

I am really new working with classes.  Some of the ideas I am getting, some I am not.
I wrote this simple little class, with a connection method, and a method to loop 
through the
data.  I am not even sure if my code is write (for the loop), but I am having trouble
displaying the data.  My connection works fine, but my loop is jacked.  Could anyone
suggest or maybe correct a mistake (if you see any)?  Just looking to see if I am 
started in the right direction, or if I have this all bass ackwards.  
I have no clue if this is even possible.  I am looking to create a while loop for 
whatever
query I run, and make each row equal to whatever field I put in.  I have all variables
defined within the scope of my class.

Thanks for your help.
 
Code :

function getdata() {

$this->sql = $query;

$result = mysql_query($query);

$fetch = "mysql_fetch_array";

 while ($row = $fetch($result)) {
 
 $this->field = $row['$this->field'];

 }

 if(!$result) {
 
  $this->error = mysql_error();

 return;
 
}

}

**
End

thanks!!!

[ bryan fitch . programmer . [EMAIL PROTECTED] ]






Re: [PHP-DB] How to display HTML code?

2001-03-20 Thread Phillip Bow

Now that is cool.  I wish I had noticed that one before.  Would have saved a
lot of odd keystrokes.
--
phill

""Joe Brown"" <[EMAIL PROTECTED]> wrote in message
996a64$k3a$[EMAIL PROTECTED]">news:996a64$k3a$[EMAIL PROTECTED]...
> try:
>  echo htmlentities("Home");
> ?>
> ""Osman Omar"" <[EMAIL PROTECTED]> wrote in message
> 00f901c0b0cf$c99e7da0$da604a2b@sony">news:00f901c0b0cf$c99e7da0$da604a2b@sony...
> > Hi,
> >
> > Suppose I want to display HTML code eg "Home"
> >
> > How can I do that in php?
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, e-mail: [EMAIL PROTECTED]
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> >
>
>
>
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RE: [PHP-DB] Select where

2001-03-20 Thread Chris Andrew

Tim,

Your missing a double quotation prior to your SELECT.

Further, are you connected to a database? :
$result=mysql_query($sql,$db);

Chris

> -Original Message-
> From: boclair [mailto:[EMAIL PROTECTED]]
> Sent: 20 March 2001 14:54
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Select where
> 
> 
> 
> - Original Message -
> From: boclair <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Wednesday, March 21, 2001 12:02 AM
> Subject: [PHP-DB] Select where
> 
> 
> > This is simple but  I cannot see where I am going wrong
> >
> > I have a table members with one of the fields
> > status, varchar(10)
> >
> > The values may be active or retired or deceased or null
> >
> > If I run the select
> >
> > SELECT * FROM members WHERE status = 'deceased';
> >
> > I getMySQL said: You have an error in your SQL syntax near
> > '\'deceased\';' at line 1
> >
> > Will somebody show me the correct syntax
> 
> ___
> 
> > $query = "SELECT * FROM members where status='deceased'"
> >
> > Then just call the $query in your script
> ___
> 
> Thanks, I only gave the mySQL but the php scripting is
> 
> 
> $deceased = mysql_query(SELECT * FROM members where
> status=\'deceased\'");
>  and later
> while ($myrow = mysql_fetch_row($deceased))
> 
> MySQL now says in relation to the *while* line
> Warning: Supplied argument is not a valid MySQL result resource
> 
> Tim Morris
> 
> 
> 
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RE: [PHP-DB] problems configuring php and postgres - "call to und efined function"

2001-03-20 Thread Steve Brett


from phpinfo()

'./configure' '--with-mysql=no' '--with-apache=/usr/src/apache_1.3.14/'
'--with-pgsql=/usr/src/postgresql-7.0.3/' '--with-xml' '--enable-track-vars'
'--enable-trans-sid' '--with-gd'

Steve

> -Original Message-
> From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> Sent: 20 March 2001 16:09
> To: Steve Brett
> Subject: RE: [PHP-DB] problems configuring php and postgres - "call to
> und efined function"
> 
> 
> Well, you just didn't point PHP at the right place then.
> 
> On Tue, 20 Mar 2001, Steve Brett wrote:
> 
> > it wouldn't compile without it.
> >
> > really.
> >
> >
> >
> > > -Original Message-
> > > From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> > > Sent: 20 March 2001 15:46
> > > To: Steve Brett
> > > Subject: RE: [PHP-DB] problems configuring php and 
> postgres - "call to
> > > und efined function"
> > >
> > >
> > > Well, that's not true.  If you install from source it 
> will install the
> > > libpq.a and the postgres.h files.  You should never point PHP
> > > at the raw
> > > source directory.
> > >
> > > -Rasmus
> > >
> > > On Tue, 20 Mar 2001, Steve Brett wrote:
> > >
> > > > just an add-on.
> > > >
> > > > if you compile form source the --with-pgsql= needs to point
> > > to the source
> > > > directory in order to find the files it needs.
> > > >
> > > > it does say in the php configure help option that it should
> > > point to the
> > > > install dir i think whihc is a bit confusing as i learnt to
> > > my peril (and
> > > > boredom)
> > > >
> > > > Steve
> > > >
> > > > > -Original Message-
> > > > > From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> > > > > Sent: 20 March 2001 08:03
> > > > > To: Gary Huntress
> > > > > Cc: [EMAIL PROTECTED]
> > > > > Subject: Re: [PHP-DB] problems configuring php and
> > > postgres - "call to
> > > > > undefined function"
> > > > >
> > > > >
> > > > > What kind of binary install did you do?  An rpm?  If so,
> > > simply also
> > > > > install the rpm-devel package and it will compile nicely.
> > > > >
> > > > > -Rasmus
> > > > >
> > > > > On Tue, 20 Mar 2001, Gary Huntress wrote:
> > > > >
> > > > > > I am trying to get postgresql (pg) support for php working
> > > > > on my RH7 box.  I
> > > > > > had a working binary installation of postgres 7.03.  I
> > > > > thought I had read
> > > > > > that php4 had default support for pg, but my call to
> > > > > pg_connect() gave me a
> > > > > > "Call to undefined function" error.
> > > > > >
> > > > > > So, next I tried to re ./configure and add
> > > > > --with-pgsql=/usr/local/pgsql.
> > > > > > That would not compile, bombing when it looked for
> > > > > "postgres.h".   I did not
> > > > > > have the pg source (since this was a binary installation)
> > > > > so I installed
> > > > > > that next (in /usr/src, if that matters)
> > > > > >
> > > > > > I deleted my existing binary installation of pg, and ran
> > > > > ./configure; gmake;
> > > > > > gmake install with no errors, and I added an entry in
> > > > > /etc/ld.so.conf as
> > > > > > stated in the pg INSTALL.   Pg is running fine with several
> > > > > db and I can
> > > > > > remotely connect, so I think thats ok.
> > > > > >
> > > > > > After finishing with the source installation of pg, I re-
> > > > > ./configure php,
> > > > > > make and get the same error looking for "postgres.h"
> > > > > >
> > > > > >  I then manually edited the ./configure file and
> > > > > added the include
> > > > > > path to /usr/src/postgres/install  After this, I
> > > > > could successfully
> > > > > > make php, then make install.
> > > > > >
> > > > > > I restarted apache, but I am still getting "Call to
> > > > > undefined function".
> > > > > >
> > > > > > Any thoughts?   I'm pretty much out of ideas right 
> now.
> > > > > >
> > > > > > Regards,
> > > > > >
> > > > > > Gary Huntress
> > > > > >
> > > > > >
> > > > > >
> > > > > > --
> > > > > > PHP Database Mailing List (http://www.php.net/)
> > > > > > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > > > > > For additional commands, e-mail: [EMAIL PROTECTED]
> > > > > > To contact the list administrators, e-mail:
> > > > > [EMAIL PROTECTED]
> > > > > >
> > > > >
> > > > >
> > > > > --
> > > > > PHP Database Mailing List (http://www.php.net/)
> > > > > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > > > > For additional commands, e-mail: [EMAIL PROTECTED]
> > > > > To contact the list administrators, e-mail:
> > > > > [EMAIL PROTECTED]
> > > > >
> > > >
> > >
> >
> 

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RE: [PHP-DB] problems configuring php and postgres - "call to und efined function"

2001-03-20 Thread Rasmus Lerdorf

But did you have /usr/local/pgsql/include and /usr/local/pgsql/lib
directories with the correct files in them?

On Tue, 20 Mar 2001, Steve Brett wrote:

>   --with-pgsql[=DIR]  Include PostgreSQL support.  DIR is the PostgreSQL
>   base install directory, defaults to
> /usr/local/pgsql.
>   Set DIR to shared to build as a dl, or shared,DIR
>   to build as a dl and still specify DIR.
>
> this did not work when i compiled.
>
> pointing at the source did.
>
> i tried pointing php at both dirs to no avail.
>
>
> > -Original Message-
> > From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> > Sent: 20 March 2001 16:09
> > To: Steve Brett
> > Subject: RE: [PHP-DB] problems configuring php and postgres - "call to
> > und efined function"
> >
> >
> > Well, you just didn't point PHP at the right place then.
> >
> > On Tue, 20 Mar 2001, Steve Brett wrote:
> >
> > > it wouldn't compile without it.
> > >
> > > really.
> > >
> > >
> > >
> > > > -Original Message-
> > > > From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> > > > Sent: 20 March 2001 15:46
> > > > To: Steve Brett
> > > > Subject: RE: [PHP-DB] problems configuring php and
> > postgres - "call to
> > > > und efined function"
> > > >
> > > >
> > > > Well, that's not true.  If you install from source it
> > will install the
> > > > libpq.a and the postgres.h files.  You should never point PHP
> > > > at the raw
> > > > source directory.
> > > >
> > > > -Rasmus
> > > >
> > > > On Tue, 20 Mar 2001, Steve Brett wrote:
> > > >
> > > > > just an add-on.
> > > > >
> > > > > if you compile form source the --with-pgsql= needs to point
> > > > to the source
> > > > > directory in order to find the files it needs.
> > > > >
> > > > > it does say in the php configure help option that it should
> > > > point to the
> > > > > install dir i think whihc is a bit confusing as i learnt to
> > > > my peril (and
> > > > > boredom)
> > > > >
> > > > > Steve
> > > > >
> > > > > > -Original Message-
> > > > > > From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> > > > > > Sent: 20 March 2001 08:03
> > > > > > To: Gary Huntress
> > > > > > Cc: [EMAIL PROTECTED]
> > > > > > Subject: Re: [PHP-DB] problems configuring php and
> > > > postgres - "call to
> > > > > > undefined function"
> > > > > >
> > > > > >
> > > > > > What kind of binary install did you do?  An rpm?  If so,
> > > > simply also
> > > > > > install the rpm-devel package and it will compile nicely.
> > > > > >
> > > > > > -Rasmus
> > > > > >
> > > > > > On Tue, 20 Mar 2001, Gary Huntress wrote:
> > > > > >
> > > > > > > I am trying to get postgresql (pg) support for php working
> > > > > > on my RH7 box.  I
> > > > > > > had a working binary installation of postgres 7.03.  I
> > > > > > thought I had read
> > > > > > > that php4 had default support for pg, but my call to
> > > > > > pg_connect() gave me a
> > > > > > > "Call to undefined function" error.
> > > > > > >
> > > > > > > So, next I tried to re ./configure and add
> > > > > > --with-pgsql=/usr/local/pgsql.
> > > > > > > That would not compile, bombing when it looked for
> > > > > > "postgres.h".   I did not
> > > > > > > have the pg source (since this was a binary installation)
> > > > > > so I installed
> > > > > > > that next (in /usr/src, if that matters)
> > > > > > >
> > > > > > > I deleted my existing binary installation of pg, and ran
> > > > > > ./configure; gmake;
> > > > > > > gmake install with no errors, and I added an entry in
> > > > > > /etc/ld.so.conf as
> > > > > > > stated in the pg INSTALL.   Pg is running fine with several
> > > > > > db and I can
> > > > > > > remotely connect, so I think thats ok.
> > > > > > >
> > > > > > > After finishing with the source installation of pg, I re-
> > > > > > ./configure php,
> > > > > > > make and get the same error looking for "postgres.h"
> > > > > > >
> > > > > > >  I then manually edited the ./configure file and
> > > > > > added the include
> > > > > > > path to /usr/src/postgres/install  After this, I
> > > > > > could successfully
> > > > > > > make php, then make install.
> > > > > > >
> > > > > > > I restarted apache, but I am still getting "Call to
> > > > > > undefined function".
> > > > > > >
> > > > > > > Any thoughts?   I'm pretty much out of ideas right
> > now.
> > > > > > >
> > > > > > > Regards,
> > > > > > >
> > > > > > > Gary Huntress
> > > > > > >
> > > > > > >
> > > > > > >
> > > > > > > --
> > > > > > > PHP Database Mailing List (http://www.php.net/)
> > > > > > > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > > > > > > For additional commands, e-mail: [EMAIL PROTECTED]
> > > > > > > To contact the list administrators, e-mail:
> > > > > > [EMAIL PROTECTED]
> > > > > > >
> > > > > >
> > > > > >
> > > > > > --
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> > > > > > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > > > > > For additional commands, e-mail: [EMAIL PROTECTED]

[PHP-DB] getmxrr() timed out

2001-03-20 Thread Vlad

Hi all
Such problem - I cant get MX records from my php script
When I use getmxrr() function it times out. But such functions as gethostbyname() work.
Anybody knows the problem?

Thanks, Vlad





RE: [PHP-DB] problems configuring php and postgres - "call to und efined function"

2001-03-20 Thread Steve Brett

  --with-pgsql[=DIR]  Include PostgreSQL support.  DIR is the PostgreSQL
  base install directory, defaults to
/usr/local/pgsql.
  Set DIR to shared to build as a dl, or shared,DIR 
  to build as a dl and still specify DIR.

this did not work when i compiled.

pointing at the source did.

i tried pointing php at both dirs to no avail.


> -Original Message-
> From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> Sent: 20 March 2001 16:09
> To: Steve Brett
> Subject: RE: [PHP-DB] problems configuring php and postgres - "call to
> und efined function"
> 
> 
> Well, you just didn't point PHP at the right place then.
> 
> On Tue, 20 Mar 2001, Steve Brett wrote:
> 
> > it wouldn't compile without it.
> >
> > really.
> >
> >
> >
> > > -Original Message-
> > > From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> > > Sent: 20 March 2001 15:46
> > > To: Steve Brett
> > > Subject: RE: [PHP-DB] problems configuring php and 
> postgres - "call to
> > > und efined function"
> > >
> > >
> > > Well, that's not true.  If you install from source it 
> will install the
> > > libpq.a and the postgres.h files.  You should never point PHP
> > > at the raw
> > > source directory.
> > >
> > > -Rasmus
> > >
> > > On Tue, 20 Mar 2001, Steve Brett wrote:
> > >
> > > > just an add-on.
> > > >
> > > > if you compile form source the --with-pgsql= needs to point
> > > to the source
> > > > directory in order to find the files it needs.
> > > >
> > > > it does say in the php configure help option that it should
> > > point to the
> > > > install dir i think whihc is a bit confusing as i learnt to
> > > my peril (and
> > > > boredom)
> > > >
> > > > Steve
> > > >
> > > > > -Original Message-
> > > > > From: Rasmus Lerdorf [mailto:[EMAIL PROTECTED]]
> > > > > Sent: 20 March 2001 08:03
> > > > > To: Gary Huntress
> > > > > Cc: [EMAIL PROTECTED]
> > > > > Subject: Re: [PHP-DB] problems configuring php and
> > > postgres - "call to
> > > > > undefined function"
> > > > >
> > > > >
> > > > > What kind of binary install did you do?  An rpm?  If so,
> > > simply also
> > > > > install the rpm-devel package and it will compile nicely.
> > > > >
> > > > > -Rasmus
> > > > >
> > > > > On Tue, 20 Mar 2001, Gary Huntress wrote:
> > > > >
> > > > > > I am trying to get postgresql (pg) support for php working
> > > > > on my RH7 box.  I
> > > > > > had a working binary installation of postgres 7.03.  I
> > > > > thought I had read
> > > > > > that php4 had default support for pg, but my call to
> > > > > pg_connect() gave me a
> > > > > > "Call to undefined function" error.
> > > > > >
> > > > > > So, next I tried to re ./configure and add
> > > > > --with-pgsql=/usr/local/pgsql.
> > > > > > That would not compile, bombing when it looked for
> > > > > "postgres.h".   I did not
> > > > > > have the pg source (since this was a binary installation)
> > > > > so I installed
> > > > > > that next (in /usr/src, if that matters)
> > > > > >
> > > > > > I deleted my existing binary installation of pg, and ran
> > > > > ./configure; gmake;
> > > > > > gmake install with no errors, and I added an entry in
> > > > > /etc/ld.so.conf as
> > > > > > stated in the pg INSTALL.   Pg is running fine with several
> > > > > db and I can
> > > > > > remotely connect, so I think thats ok.
> > > > > >
> > > > > > After finishing with the source installation of pg, I re-
> > > > > ./configure php,
> > > > > > make and get the same error looking for "postgres.h"
> > > > > >
> > > > > >  I then manually edited the ./configure file and
> > > > > added the include
> > > > > > path to /usr/src/postgres/install  After this, I
> > > > > could successfully
> > > > > > make php, then make install.
> > > > > >
> > > > > > I restarted apache, but I am still getting "Call to
> > > > > undefined function".
> > > > > >
> > > > > > Any thoughts?   I'm pretty much out of ideas right 
> now.
> > > > > >
> > > > > > Regards,
> > > > > >
> > > > > > Gary Huntress
> > > > > >
> > > > > >
> > > > > >
> > > > > > --
> > > > > > PHP Database Mailing List (http://www.php.net/)
> > > > > > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > > > > > For additional commands, e-mail: [EMAIL PROTECTED]
> > > > > > To contact the list administrators, e-mail:
> > > > > [EMAIL PROTECTED]
> > > > > >
> > > > >
> > > > >
> > > > > --
> > > > > PHP Database Mailing List (http://www.php.net/)
> > > > > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > > > > For additional commands, e-mail: [EMAIL PROTECTED]
> > > > > To contact the list administrators, e-mail:
> > > > > [EMAIL PROTECTED]
> > > > >
> > > >
> > >
> >
> 

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Re: [PHP-DB] More on strings

2001-03-20 Thread CC Zona

In article <007901c0b153$ae861a80$660083ca@oemcomputer>,
 [EMAIL PROTECTED] ("Mick Lloyd") wrote:

> Thanks for the reply. I tried the variable route but again failed. I think
> from your suggested line of code you may have misunderstood the problem. The
> LIKE option works fine without having to put it into a variable.
> 
> $sql = "select Primaryid from primarycodes where Code =
> '$row[Primaryexpertise]'";
> print $sql."\n";
> 
> \\Prints: select Primaryid from primarycodes where Code = 'Biology' \\which
> seems OK
> 
> $resultp = mysql_query($sql) or die ("Failed"); \\This seems to work
> $pcodeid = mysql_fetch_array($resultp) or die("No pcodeid");
> 
> \\Prints: No pcodeid \\ie failed

mysql_query() returns true as long as the query syntax appears valid; it 
does not indicate whether the query actually found rows or not.  So just 
because the code didn't die on the mysql_query() line, you cannot assume 
the query "works" as intended.  The fact that the mysql_fetch_array() line 
dies would suggest that the query above likely is failing for some reason.  
One thing you can do is check mysql_num_rows() first to determine whether 
the query found any records.  Another thing you can do is echo $sql, then 
copy/paste that query at the command line to determine whether it's the SQL 
itself that's the problem or the PHP code.  Finally, instead of displaying 
your own ambiguous error messages on die(), let MySQL & PHP report to you 
what they know about the problem.  Example:

$resultp = mysql_query($sql) or die('Query failed.  MySQL says: ' . 
mysql_error() . '. PHP says: ' . $errormsg . '.');

-- 
CC

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Re: [PHP-DB] Select where

2001-03-20 Thread boclair


- Original Message -
From: Darryl Friesen <[EMAIL PROTECTED]>
To: boclair <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Wednesday, March 21, 2001 1:22 AM
Subject: Re: [PHP-DB] Select where


> > Thanks, I only gave the mySQL but the php scripting is
> >
> > $deceased = mysql_query(SELECT * FROM members where
> > status=\'deceased\'");
> >  and later
> > while ($myrow = mysql_fetch_row($deceased))
> >
> > MySQL now says in relation to the *while* line
> > Warning: Supplied argument is not a valid MySQL result resource
>
> Yikes!  I hope those are typos, and that you didn't cut and paste
that query
> from your script.  If you did, then it needs some work.  Try:
>
> $deceased = mysql_query("SELECT * FROM members where
status='deceased'");
>
> You need the leading double quote before SELECT, and you don't need
to
> escape the single quotes.

 It's always the simple things.  I had the leading " but had escaped
the ' around deceased.
No excuse but too many lines of printf for a table with styling.
Started to escape everything.

Many, many thanks

Tim Morris



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Re: [PHP-DB] More on strings

2001-03-20 Thread Mick Lloyd

Darryl

Thanks for the reply. I tried the variable route but again failed. I think
from your suggested line of code you may have misunderstood the problem. The
LIKE option works fine without having to put it into a variable.

$sql = "select Primaryid from primarycodes where Code =
'$row[Primaryexpertise]'";
print $sql."\n";

\\Prints: select Primaryid from primarycodes where Code = 'Biology' \\which
seems OK

$resultp = mysql_query($sql) or die ("Failed"); \\This seems to work
$pcodeid = mysql_fetch_array($resultp) or die("No pcodeid");

\\Prints: No pcodeid \\ie failed

Regards

Mick Lloyd
[EMAIL PROTECTED]
Tel: +44 (0)1684 560224
- Original Message -
From: Darryl Friesen <[EMAIL PROTECTED]>
To: Database Help PHP <[EMAIL PROTECTED]>
Sent: Tuesday, March 20, 2001 1:41 PM
Subject: Re: [PHP-DB] More on strings


> >  $resultp = mysql_query("select Primaryid from primarycodes where Code =
> > '".$row['Primaryexpertise']."'") or die (mysql_error());
> >
> >  $resultp = mysql_query("select Primaryid from primarycodes where Code
> like
> > '%$row[Primaryexpertise]%'") or die (mysql_error());
> >
> >  $pcodeid = mysql_fetch_array($resultp) or die("No pcodeid");
>
> Two suggestions.:
>
> First, stick the query into a variable instead of using it directly; that
> way you can do some simple debugging (i.e. print the query to the page
> instead of posting this email).  Second, take the array variable out of
the
> double quotes, like you did in the first line.
>
> $sql = "select Primaryid from primarycodes where Code like'%" .
> $row[Primaryexpertise] . "%'";
> print $sql . "\n";
> $resultp = mysql_query($sql) or die (mysql_error());
>
>
> - Darryl
>
>  --
>   Darryl Friesen, B.Sc., Programmer/Analyst[EMAIL PROTECTED]
>   Education & Research Technology Services, http://gollum.usask.ca/
>   Department of Computing Services,
>   University of Saskatchewan
>  --
>   "Go not to the Elves for counsel, for they will say both no and yes"
>
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
> To contact the list administrators, e-mail: [EMAIL PROTECTED]
>



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[PHP-DB] Dynamically build a form from mySQL table

2001-03-20 Thread Martin E. Koss

Hi,
I've been trying to figure this out by looking through the code in
phpMyAdmin but I'm not getting very far.

What I want to do is dynamically generate a form where I get a table row
containing the field name in one cell and the input field in the next cell,
for example:

$DatabaseName = "FocusDynamics";
$TableName = "Items";

// open the table and get all field names, sizes, etc.,

// for each Field name, print something like:

PRINT "\n";
PRINT "$FieldName\n";
PRINT "\n";
PRINT "";

How do I get all this info from the table?

Thanks for any help.


Martin E. Koss
M: 07946-706459
E: [EMAIL PROTECTED]


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Re: [PHP-DB] Select where

2001-03-20 Thread Darryl Friesen

> Thanks, I only gave the mySQL but the php scripting is
>
> $deceased = mysql_query(SELECT * FROM members where
> status=\'deceased\'");
>  and later
> while ($myrow = mysql_fetch_row($deceased))
>
> MySQL now says in relation to the *while* line
> Warning: Supplied argument is not a valid MySQL result resource

Yikes!  I hope those are typos, and that you didn't cut and paste that query
from your script.  If you did, then it needs some work.  Try:

$deceased = mysql_query("SELECT * FROM members where status='deceased'");

You need the leading double quote before SELECT, and you don't need to
escape the single quotes.


- Darryl

 --
  Darryl Friesen, B.Sc., Programmer/Analyst[EMAIL PROTECTED]
  Education & Research Technology Services, http://gollum.usask.ca/
  Department of Computing Services,
  University of Saskatchewan
 --
  "Go not to the Elves for counsel, for they will say both no and yes"



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Re: [PHP-DB] mysql_result

2001-03-20 Thread Shahmat Dahlan

That didn't quite work, I tried using single quote enclose within double
quotes and it works.
$sqlstmt="SELECT * FROM dept WHERE shortname='$shortname'";

Thanks anyway


ben wrote:

> On Tue, 20 Mar 2001, Shahmat Dahlan wrote:
>
> > I'm trying to use mysql_result to extract data from one particular field
> > in a record. So I suppose the most viable solution is to use
> > mysql_result. But when I do this:
> >
> > $sqlstmt="SELECT * FROM dept WHERE shortname=$shortname";
> > $result=mysql_query($sqlstmt);
> > $temp=mysql_result($result,0,"shortname"); <-- This will give me an
> > error
> >
> > Did I do anything wrong here?
> >
> > Thanks in advance.
> > Shahmat
>
> The problem is that the query is wrong. Try
> $sqlstmt="SELECT * FROM dept WHERE shortname=\"$shortname\"";
>
> (you need to send quotes to mysql if the type of shortname in your
> database is text or the like, that's why there's the \"$var\")
>
> --
> Ben
> History is curious stuff You'd think by now we had enough Yet the fact
> remains I fear They make more of it every year.


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Re: [PHP-DB] mysql_result

2001-03-20 Thread Shahmat Dahlan

I've tried what you suggested and it works, thanks.


Russ Michell wrote:

> Try:
>
> $sqlstmt="SELECT * FROM dept WHERE shortname='$shortname'";
> //Include single quotes..
>
> Russ
>
> #---#
>
>  "Believe nothing - consider everything"
>
>   Russ Michell
>   Anglia Polytechnic University Webteam
>   http://gertrude.sipu.anglia.ac.uk/webteam
>   [EMAIL PROTECTED]
>   +44 (0)1223 363271 ext 2331
>
>   www.theruss.com
>
> #---#


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Re: [PHP-DB] Select where

2001-03-20 Thread boclair


- Original Message -
From: boclair <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, March 21, 2001 12:02 AM
Subject: [PHP-DB] Select where


> This is simple but  I cannot see where I am going wrong
>
> I have a table members with one of the fields
> status, varchar(10)
>
> The values may be active or retired or deceased or null
>
> If I run the select
>
> SELECT * FROM members WHERE status = 'deceased';
>
> I getMySQL said: You have an error in your SQL syntax near
> '\'deceased\';' at line 1
>
> Will somebody show me the correct syntax

___

> $query = "SELECT * FROM members where status='deceased'"
>
> Then just call the $query in your script
___

Thanks, I only gave the mySQL but the php scripting is


$deceased = mysql_query(SELECT * FROM members where
status=\'deceased\'");
 and later
while ($myrow = mysql_fetch_row($deceased))

MySQL now says in relation to the *while* line
Warning: Supplied argument is not a valid MySQL result resource

Tim Morris



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Re: [PHP-DB] How do I test on number of fields ?

2001-03-20 Thread Jason Stechschulte

On Mon, Mar 19, 2001 at 10:06:52PM -0800, Toke Herkild wrote:
> I've written this login rutine
> 
> But that doesn't work... I've tried to use mysql_num_rows too but I still
> get :
> Warning:  Supplied argument is not a valid MySQL result resource in
> g://web/herkild/html/loginout.php on line 113
> 
> if ($ACTION == 'Login') {
>   $query = "SELECT user.ID FROM user WHERE ";
>   $query .= "user.brugerID = $LoginBrugerID AND ";
>   $query .= "user.kodeord = $LoginKodeord;";
> 
>   $mysql_result = mysql_query($query, $mysql_link);
> 
> (line 113 >>)  if ($row = mysql_fetch_row($mysql_result)){
> setcookie("Herkild_Domain", $row[0]);
> Logged_in();
>   }
>   else {
> wronguser();
>   }
> }

It looks like you might be missing some quotes in your query.  After you
call mysql_query(), call mysql_error($mysql_link);  It will show you
your error in your query.

-- 
Jason Stechschulte
[EMAIL PROTECTED]
--
There are still some other things to do, so don't think if I didn't fix
your favorite bug that your bug report is in the bit bucket.  (It may be,
but don't think it.  :-)  Larry Wall in <[EMAIL PROTECTED]>

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Re: [PHP-DB] Select where

2001-03-20 Thread ben

On Wed, 21 Mar 2001, boclair wrote:

> This is simple but  I cannot see where I am going wrong
> 
> I have a table members with one of the fields
> status, varchar(10)
> 
> The values may be active or retired or deceased or null
> 
> If I run the select
> 
> SELECT * FROM members WHERE status = 'deceased';
> 
> I getMySQL said: You have an error in your SQL syntax near
> '\'deceased\';' at line 1
> 
> Will somebody show me the correct syntax
> 
> Tim Morris
> 
> 
> 

you need to pass the quotes to mysql, hence you have to backslash them in
the php code. Try :

$query="SELECT * FROM members WHERE status = \"deceased\"";

--
Ben
History is curious stuff You'd think by now we had enough Yet the fact
remains I fear They make more of it every year.


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Re: [PHP-DB] mysql_result

2001-03-20 Thread ben

On Tue, 20 Mar 2001, Shahmat Dahlan wrote:

> I'm trying to use mysql_result to extract data from one particular field
> in a record. So I suppose the most viable solution is to use
> mysql_result. But when I do this:
> 
> $sqlstmt="SELECT * FROM dept WHERE shortname=$shortname";
> $result=mysql_query($sqlstmt);
> $temp=mysql_result($result,0,"shortname"); <-- This will give me an
> error
> 
> Did I do anything wrong here?
> 
> Thanks in advance.
> Shahmat

The problem is that the query is wrong. Try
$sqlstmt="SELECT * FROM dept WHERE shortname=\"$shortname\"";

(you need to send quotes to mysql if the type of shortname in your
database is text or the like, that's why there's the \"$var\")

--
Ben
History is curious stuff You'd think by now we had enough Yet the fact
remains I fear They make more of it every year.


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[PHP-DB] Select where

2001-03-20 Thread boclair

This is simple but  I cannot see where I am going wrong

I have a table members with one of the fields
status, varchar(10)

The values may be active or retired or deceased or null

If I run the select

SELECT * FROM members WHERE status = 'deceased';

I getMySQL said: You have an error in your SQL syntax near
'\'deceased\';' at line 1

Will somebody show me the correct syntax

Tim Morris



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Re: [PHP-DB] More on strings

2001-03-20 Thread Darryl Friesen

>  $resultp = mysql_query("select Primaryid from primarycodes where Code =
> '".$row['Primaryexpertise']."'") or die (mysql_error());
>
>  $resultp = mysql_query("select Primaryid from primarycodes where Code
like
> '%$row[Primaryexpertise]%'") or die (mysql_error());
>
>  $pcodeid = mysql_fetch_array($resultp) or die("No pcodeid");

Two suggestions.:

First, stick the query into a variable instead of using it directly; that
way you can do some simple debugging (i.e. print the query to the page
instead of posting this email).  Second, take the array variable out of the
double quotes, like you did in the first line.

$sql = "select Primaryid from primarycodes where Code like'%" .
$row[Primaryexpertise] . "%'";
print $sql . "\n";
$resultp = mysql_query($sql) or die (mysql_error());


- Darryl

 --
  Darryl Friesen, B.Sc., Programmer/Analyst[EMAIL PROTECTED]
  Education & Research Technology Services, http://gollum.usask.ca/
  Department of Computing Services,
  University of Saskatchewan
 --
  "Go not to the Elves for counsel, for they will say both no and yes"




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[PHP-DB] mysql_result

2001-03-20 Thread Shahmat Dahlan

I'm trying to use mysql_result to extract data from one particular field
in a record. So I suppose the most viable solution is to use
mysql_result. But when I do this:

$sqlstmt="SELECT * FROM dept WHERE shortname=$shortname";
$result=mysql_query($sqlstmt);
$temp=mysql_result($result,0,"shortname"); <-- This will give me an
error

Did I do anything wrong here?

Thanks in advance.
Shahmat


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[PHP-DB] LO on postgreSQL HELP!

2001-03-20 Thread Patrick

Hallo!
I'm trying to read lo objects (images) from postgreSQL  using php's
pg_loreadall() function with the intention of put them dirctly on a html
document, but I can't. I also have problems about how to embed the result of
the pg_loread() on a html document.
Somebody can help me?
Thak you very much!



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RE: [PHP-DB] problems configuring php and postgres - "call to undefined function"

2001-03-20 Thread Steve Brett

get the postgrtesql source code and untar then run ./configure etc etc on
postgresql

then run ./configure in php as

./configure --with-pgsql=/path/to/untarred/source/files

e.g.

postgresql source untarred to /usr/src/postgresql

php configure options are:

./configure --with-pgsql=/usr/src/postgresql/ 

and not /usr/local/pgsql

php needs to see the source code not the install dir.

Steve

> -Original Message-
> From: Gary Huntress [mailto:[EMAIL PROTECTED]]
> Sent: 20 March 2001 12:39
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] problems configuring php and postgres - "call to
> undefined function"
> 
> 
> I just added phpinfo() to my test page and the "configure 
> command" section
> does NOT include --with-pgsqlclearly that is my problem
> 
> I just deleted my php-4.0.4 source directory and reinstalled 
> the tarball,
> ran ./configure --with-pgsql=/usr/local/pgsql --with-xml ; make ; make
> install, received no errors and still have the same error.
> 
> There is *something* that I'm not cleaning up properly..
> 
> Regards,
> 
> Gary
> 
> "Rasmus Lerdorf" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > What kind of binary install did you do?  An rpm?  If so, simply also
> > install the rpm-devel package and it will compile nicely.
> >
> > -Rasmus
> >
> > On Tue, 20 Mar 2001, Gary Huntress wrote:
> >
> > > I am trying to get postgresql (pg) support for php 
> working on my RH7
> box.  I
> > > had a working binary installation of postgres 7.03.  I 
> thought I had
> read
> > > that php4 had default support for pg, but my call to 
> pg_connect() gave
> me a
> > > "Call to undefined function" error.
> > >
> > > So, next I tried to re ./configure and
> add --with-pgsql=/usr/local/pgsql.
> > > That would not compile, bombing when it looked for 
> "postgres.h".   I did
> not
> > > have the pg source (since this was a binary installation) 
> so I installed
> > > that next (in /usr/src, if that matters)
> > >
> > > I deleted my existing binary installation of pg, and ran 
> ./configure;
> gmake;
> > > gmake install with no errors, and I added an entry in 
> /etc/ld.so.conf as
> > > stated in the pg INSTALL.   Pg is running fine with 
> several db and I can
> > > remotely connect, so I think thats ok.
> > >
> > > After finishing with the source installation of pg, I re- 
> ./configure
> php,
> > > make and get the same error looking for "postgres.h"
> > >
> > >  I then manually edited the ./configure file and added the
> include
> > > path to /usr/src/postgres/install  After this, I could
> successfully
> > > make php, then make install.
> > >
> > > I restarted apache, but I am still getting "Call to 
> undefined function".
> > >
> > > Any thoughts?   I'm pretty much out of ideas right now.
> > >
> > > Regards,
> > >
> > > Gary Huntress
> > >
> > >
> > >
> > > --
> > > PHP Database Mailing List (http://www.php.net/)
> > > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > > For additional commands, e-mail: [EMAIL PROTECTED]
> > > To contact the list administrators, e-mail: 
> [EMAIL PROTECTED]
> > >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> [EMAIL PROTECTED]
> >
> 
> 
> 
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Re: [PHP-DB] problems configuring php and postgres - "call to undefined function"

2001-03-20 Thread Gary Huntress

I just added phpinfo() to my test page and the "configure command" section
does NOT include --with-pgsqlclearly that is my problem

I just deleted my php-4.0.4 source directory and reinstalled the tarball,
ran ./configure --with-pgsql=/usr/local/pgsql --with-xml ; make ; make
install, received no errors and still have the same error.

There is *something* that I'm not cleaning up properly..

Regards,

Gary

"Rasmus Lerdorf" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> What kind of binary install did you do?  An rpm?  If so, simply also
> install the rpm-devel package and it will compile nicely.
>
> -Rasmus
>
> On Tue, 20 Mar 2001, Gary Huntress wrote:
>
> > I am trying to get postgresql (pg) support for php working on my RH7
box.  I
> > had a working binary installation of postgres 7.03.  I thought I had
read
> > that php4 had default support for pg, but my call to pg_connect() gave
me a
> > "Call to undefined function" error.
> >
> > So, next I tried to re ./configure and
add --with-pgsql=/usr/local/pgsql.
> > That would not compile, bombing when it looked for "postgres.h".   I did
not
> > have the pg source (since this was a binary installation) so I installed
> > that next (in /usr/src, if that matters)
> >
> > I deleted my existing binary installation of pg, and ran ./configure;
gmake;
> > gmake install with no errors, and I added an entry in /etc/ld.so.conf as
> > stated in the pg INSTALL.   Pg is running fine with several db and I can
> > remotely connect, so I think thats ok.
> >
> > After finishing with the source installation of pg, I re- ./configure
php,
> > make and get the same error looking for "postgres.h"
> >
> >  I then manually edited the ./configure file and added the
include
> > path to /usr/src/postgres/install  After this, I could
successfully
> > make php, then make install.
> >
> > I restarted apache, but I am still getting "Call to undefined function".
> >
> > Any thoughts?   I'm pretty much out of ideas right now.
> >
> > Regards,
> >
> > Gary Huntress
> >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > For additional commands, e-mail: [EMAIL PROTECTED]
> > To contact the list administrators, e-mail: [EMAIL PROTECTED]
> >
>
>
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RE: [PHP-DB] Expires Pages

2001-03-20 Thread Steve Brett

this should work ... in the top of the page, before any other headers are
sent



> -Original Message-
> From: Marcelo Pereira [mailto:[EMAIL PROTECTED]]
> Sent: 20 March 2001 12:32
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Expires Pages
> 
> 
> Hello,
> 
> I'd like the user of my site to log in/out, but, how can I 
> disallow the user to press the 'back button' of the browse 
> after logout, and back to the page ???
> 
> What I need:
> 
> - The user press 'logout';
>if (user press the 'back button' of the browser) then
>   show a page telling that the page expired just when the 
> user clicked at logout.
>fi
> 
> I was reading about cookies, I think I have to go in this 
> direction, but, how ???
> 
> Thanks in advance,
> 
> See ya.
> 
> Marcelo Pereira
> Campinas' University - Brazil
> 

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[PHP-DB] Expires Pages

2001-03-20 Thread Marcelo Pereira

Hello,

I'd like the user of my site to log in/out, but, how can I disallow the user to press 
the 'back button' of the browse after logout, and back to the page ???

What I need:

- The user press 'logout';
   if (user press the 'back button' of the browser) then
  show a page telling that the page expired just when the user clicked at logout.
   fi

I was reading about cookies, I think I have to go in this direction, but, how ???

Thanks in advance,

See ya.

Marcelo Pereira
Campinas' University - Brazil



Re: [PHP-DB] problems configuring php and postgres - "call to undefined function"

2001-03-20 Thread Gary Huntress

I had the default binary rpm that was installed with RH7.   I just installed
the postgres devel rpm, reran ./configure, make, make install and restarted
apache with the same result of "call to undefined function"

I noticed that the tutorial on
http://www.phpbuilder.com/columns/kevin20010314.php3 says that I need
php-pgsql-4.0.1pl2-9.i386.rpm however I am running php-4.0.4 and rpm
whined about dependencies and I didn't want to --force it.

I think that somehow I need to clean up and start from scratch.

Regards,

Gary H.

"Rasmus Lerdorf" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> What kind of binary install did you do?  An rpm?  If so, simply also
> install the rpm-devel package and it will compile nicely.
>
> -Rasmus
>
> On Tue, 20 Mar 2001, Gary Huntress wrote:
>
> > I am trying to get postgresql (pg) support for php working on my RH7
box.  I
> > had a working binary installation of postgres 7.03.  I thought I had
read
> > that php4 had default support for pg, but my call to pg_connect() gave
me a
> > "Call to undefined function" error.
> >
> > So, next I tried to re ./configure and
add --with-pgsql=/usr/local/pgsql.
> > That would not compile, bombing when it looked for "postgres.h".   I did
not
> > have the pg source (since this was a binary installation) so I installed
> > that next (in /usr/src, if that matters)
> >
> > I deleted my existing binary installation of pg, and ran ./configure;
gmake;
> > gmake install with no errors, and I added an entry in /etc/ld.so.conf as
> > stated in the pg INSTALL.   Pg is running fine with several db and I can
> > remotely connect, so I think thats ok.
> >
> > After finishing with the source installation of pg, I re- ./configure
php,
> > make and get the same error looking for "postgres.h"
> >
> >  I then manually edited the ./configure file and added the
include
> > path to /usr/src/postgres/install  After this, I could
successfully
> > make php, then make install.
> >
> > I restarted apache, but I am still getting "Call to undefined function".
> >
> > Any thoughts?   I'm pretty much out of ideas right now.
> >
> > Regards,
> >
> > Gary Huntress
> >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, e-mail: [EMAIL PROTECTED]
> > For additional commands, e-mail: [EMAIL PROTECTED]
> > To contact the list administrators, e-mail: [EMAIL PROTECTED]
> >
>
>
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[PHP-DB] No. of Users online right now?

2001-03-20 Thread Dan Eskildsen

Hi there,

This is a NewBie Question:
While out surfing I have often seen a message like:  "There are 17 users on
this site at the moment."

Can I do this with php?  Just to show how many people are on (browsing) the
site at that moment?  I run a community website.

Thanks

==
Regards from Denmark, Europe
Please cc your reply to [EMAIL PROTECTED]
==
I haven't lost my mind, I've got it backed up on TAPE somewhere...



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[PHP-DB] MySQL errors on win32

2001-03-20 Thread Dirk Bonenkamp - Bean IT

Hi all,

I just installed apache, php & mysql on a win98 laptop. I normally use
linux 

Things work fine, but when I execute more queries in one script, I get
error messages like:

Warning: MySQL: Unable to save result set in c:/program files/apache
group/apache/htdocs/network-club.com/agenda.php4 on line 14

I saw something in the FAQ about this, but that was when upgrading.

I use Win98 SE UK, Apache 1.3.14, php 4.0.4pl1, Mysql 3.23.22-beta.

TIA,

Dirk


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[PHP-DB] More on strings

2001-03-20 Thread Mick Lloyd

I thought I had it but I didn't! I can only get LIKE to perform queries that
allow me use the result for a subsequent query but for accuracy, I need to
use actual values. The first statement below works OK (at least I get a
Resource id #) - courtesy of help received here - but will not return
$pcodeid from the third statement - it dies giving me "No pcodeid". Using
the LIKE version (second one below) works fine and I can use the $pcodeid
array elements as I like. Any thoughts gratefully received.

 $resultp = mysql_query("select Primaryid from primarycodes where Code =
'".$row['Primaryexpertise']."'") or die (mysql_error());

 $resultp = mysql_query("select Primaryid from primarycodes where Code like
'%$row[Primaryexpertise]%'") or die (mysql_error());

 $pcodeid = mysql_fetch_array($resultp) or die("No pcodeid");


Regards

Mick Lloyd
[EMAIL PROTECTED]
Tel: +44 (0)1684 560224


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