[PHP-DB] mysql_insert_id - link_identifier

2002-01-28 Thread B. Verbeek

Hello,

How do I use the linkidentifier to get me the last_inserted_id
(mysql_insert_id) from a auto_increment-colomn?

Do I use:
>>
  $q_orders_id2 = "INSERT INTO orders(orders_datum, orders_sessid)";
  $q_orders_id2.= " VALUES ('$date', '$sid')";
  $r_orders_id2 = mysql_query($q_orders_id2, $sqllink_id);
  $orders_id = mysql_insert_id($sqllink_id);
<<

"$sqllink_id" must be the inserted value for the auto-increment-colomn named
"orders_id" and is placed in var "$orders_id"


Can anyone tell me if this is correct and what can be done to optimize?


regards
Bart



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[PHP-DB] mysql_insert_id - link_identifier

2002-01-28 Thread B. Verbeek



I already run the query without the value of the primary-key field:
>>
  $q_orders_id2 = "INSERT INTO orders(orders_datum, orders_sessid)";
  $q_orders_id2.= " VALUES ('$date', '$sid')";
  $r_orders_id2 = mysql_query($q_orders_id2, $sqllink_id);
  $orders_id = mysql_insert_id($sqllink_id);
<<

The primary-key field is: "orders_id" (not in above query).
Only "orders_datum" and "orders_sessid" are sent with the query.

When I call mysql_insert_id with tha vars: "$q_orders_id" or
"$r_orders_id" I get an warning (Warning: Supplied argument
is not a valid MySQL-Link resource in /home/httpd/blabla/file.php
on line 91) but not the wanted id.

In the manual it says that mysql_insert_id() is used with the
linkidentifier ($sqllink_id ? ) but not how to define the identifier
in mysql_query()...

>>How do I define a linkidentifier for a specific call to
  mysql_query() and then use mysql_insert_id() with that
  linkidentifier of the query?

regards
Bart


-Oorspronkelijk bericht-
Van: Alberto. Sartori [mailto:[EMAIL PROTECTED]]
Verzonden: maandag 28 januari 2002 13:11
Aan: [EMAIL PROTECTED]
Onderwerp: R: [PHP-DB] mysql_insert_id - link_identifier


Run the query without the value of the primary-key field.

-Messaggio originale-
Da: B. Verbeek [mailto:[EMAIL PROTECTED]]
Inviato: lunedì 28 gennaio 2002 13.08
A: Alberto. Sartori
Oggetto: RE: [PHP-DB] mysql_insert_id - link_identifier


? How do you mean ?
regards

-Oorspronkelijk bericht-
Van: Alberto. Sartori [mailto:[EMAIL PROTECTED]]
Verzonden: maandag 28 januari 2002 12:35
Aan: [EMAIL PROTECTED]
Onderwerp: R: [PHP-DB] mysql_insert_id - link_identifier


You can omit the primary key if the filed have auto-increment type

-Messaggio originale-
Da: B. Verbeek [mailto:[EMAIL PROTECTED]]
Inviato: lunedì 28 gennaio 2002 12.29
A: Php-Db-Help (E-mail)
Oggetto: [PHP-DB] mysql_insert_id - link_identifier


Hello,

How do I use the linkidentifier to get me the last_inserted_id
(mysql_insert_id) from a auto_increment-colomn?

Do I use:
>>
  $q_orders_id2 = "INSERT INTO orders(orders_datum, orders_sessid)";
  $q_orders_id2.= " VALUES ('$date', '$sid')";
  $r_orders_id2 = mysql_query($q_orders_id2, $sqllink_id);
  $orders_id = mysql_insert_id($sqllink_id);
<<

"$sqllink_id" must be the inserted value for the auto-increment-colomn named
"orders_id" and is placed in var "$orders_id"


Can anyone tell me if this is correct and what can be done to optimize?


regards
Bart



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[PHP-DB] session_registered variable problem

2002-01-28 Thread Markus Lervik


Hello!

I have a problem with session_registered variables. 
It seems that my sessions behave quite erratic. Well, 
actually just one variable, $query.

Somehow when I print the value of $query, it looks fine.
Then, when I change $query a bit and print it again, 
it looks fine.
Change it a third time, and it suddenly gets the value it
had in the beginning. 

Ideas as to how one would go about solving this little problem?

Cheers,
Markus

-- 
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Linux-administrator with a kungfoo grip
Vaasa City Library - Regional Library
[EMAIL PROTECTED]
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Re: [PHP-DB] mysql_insert_id - link_identifier

2002-01-28 Thread Jason Wong

On Monday 28 January 2002 20:18, B. Verbeek wrote:
> I already run the query without the value of the primary-key field:
>
>   $q_orders_id2 = "INSERT INTO orders(orders_datum, orders_sessid)";
>   $q_orders_id2.= " VALUES ('$date', '$sid')";
>   $r_orders_id2 = mysql_query($q_orders_id2, $sqllink_id);
>   $orders_id = mysql_insert_id($sqllink_id);
> <<
>
> The primary-key field is: "orders_id" (not in above query).
> Only "orders_datum" and "orders_sessid" are sent with the query.
>
> When I call mysql_insert_id with tha vars: "$q_orders_id" or
> "$r_orders_id" I get an warning (Warning: Supplied argument
> is not a valid MySQL-Link resource in /home/httpd/blabla/file.php
> on line 91) but not the wanted id.
>
> In the manual it says that mysql_insert_id() is used with the
> linkidentifier ($sqllink_id ? ) but not how to define the identifier
> in mysql_query()...
>
> >>How do I define a linkidentifier for a specific call to
>
>   mysql_query() and then use mysql_insert_id() with that
>   linkidentifier of the query?

When you initiate the connection to the db using mysql_connect() or 
mysql_pconnect() it returns a MySQL link identifier. 

In most cases you do NOT need to specify the link identifier when using 
mysql_query(). If you're only dealing with one database throughout your 
script then there's no reason for you to specify the link identifier because 
PHP automatically uses the most recently used one. That is it automatically 
chooses the correct link identifier to use.

So in your case what you want would probably be something like this:

  $q_orders_id2  = "INSERT INTO orders(orders_datum, orders_sessid)";
  $q_orders_id2 .= " VALUES ('$date', '$sid')";
  # for debugging, you may want to echo $q_order_id2
  $r_orders_id2 = mysql_query($q_orders_id2);
  $orders_id = mysql_insert_id();


hth
-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
I used to be Snow White, but I drifted.
-- Mae West
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Re: [PHP-DB] session_registered variable problem

2002-01-28 Thread Jason Wong

On Monday 28 January 2002 20:45, Markus Lervik wrote:
> Hello!
>
> I have a problem with session_registered variables.
> It seems that my sessions behave quite erratic. Well,
> actually just one variable, $query.
>
> Somehow when I print the value of $query, it looks fine.
> Then, when I change $query a bit and print it again,
> it looks fine.
> Change it a third time, and it suddenly gets the value it
> had in the beginning.
>
> Ideas as to how one would go about solving this little problem?

Could you post some code please?


-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
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RE: [PHP-DB] mysql_insert_id - link_identifier

2002-01-28 Thread B. Verbeek

Thanks Jason,

That was the answer I was looking for! It works like you suggested!!

Thanks
regards,

bart

-Oorspronkelijk bericht-
Van: Jason Wong [mailto:[EMAIL PROTECTED]]
Verzonden: maandag 28 januari 2002 13:52
Aan: [EMAIL PROTECTED]
Onderwerp: Re: [PHP-DB] mysql_insert_id - link_identifier


On Monday 28 January 2002 20:18, B. Verbeek wrote:
> I already run the query without the value of the primary-key field:
>
>   $q_orders_id2 = "INSERT INTO orders(orders_datum, orders_sessid)";
>   $q_orders_id2.= " VALUES ('$date', '$sid')";
>   $r_orders_id2 = mysql_query($q_orders_id2, $sqllink_id);
>   $orders_id = mysql_insert_id($sqllink_id);
> <<
>
> The primary-key field is: "orders_id" (not in above query).
> Only "orders_datum" and "orders_sessid" are sent with the query.
>
> When I call mysql_insert_id with tha vars: "$q_orders_id" or
> "$r_orders_id" I get an warning (Warning: Supplied argument
> is not a valid MySQL-Link resource in /home/httpd/blabla/file.php
> on line 91) but not the wanted id.
>
> In the manual it says that mysql_insert_id() is used with the
> linkidentifier ($sqllink_id ? ) but not how to define the identifier
> in mysql_query()...
>
> >>How do I define a linkidentifier for a specific call to
>
>   mysql_query() and then use mysql_insert_id() with that
>   linkidentifier of the query?

When you initiate the connection to the db using mysql_connect() or
mysql_pconnect() it returns a MySQL link identifier.

In most cases you do NOT need to specify the link identifier when using
mysql_query(). If you're only dealing with one database throughout your
script then there's no reason for you to specify the link identifier because
PHP automatically uses the most recently used one. That is it automatically
chooses the correct link identifier to use.

So in your case what you want would probably be something like this:

  $q_orders_id2  = "INSERT INTO orders(orders_datum, orders_sessid)";
  $q_orders_id2 .= " VALUES ('$date', '$sid')";
  # for debugging, you may want to echo $q_order_id2
  $r_orders_id2 = mysql_query($q_orders_id2);
  $orders_id = mysql_insert_id();


hth
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
I used to be Snow White, but I drifted.
-- Mae West
*/

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Re: [PHP-DB] session_registered variable problem

2002-01-28 Thread Markus Lervik

On Monday 28 January 2002 14:57, Jason Wong wrote:

> > Somehow when I print the value of $query, it looks fine.
> > Then, when I change $query a bit and print it again,
> > it looks fine.
> > Change it a third time, and it suddenly gets the value it
> > had in the beginning.

> Could you post some code please?

Well, for instance, this screws up the $query. It works if one clicks
once on the next-button, but clicking another time resets the query to

SELECT 

L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huomautuksia 
FROM uusi_lehtitaulu L, nimet N 
WHERE (L.Nimi_id=N.id AND  N.Nimi LIKE "asdf") 
LIMIT 0,20

if($go=="next") {
if($page < $nr_pages && $page >= 0) {
 $page++;
 $replace_with= (($page) * 20).",";
 echo "replace_with: " . $replace_with;
 $replace = (($page-1) * 20).",";
 echo "replace: " .$replace;
 $query=ereg_replace($replace,$replace_with,$query); 
}
}

It's the LIMIT 0,20 -> LIMIT 20,20 -> LIMIT 40,20 that doesn't work. Strange thing is, 
that
it worked about a week ago. : \ 
I can't really recall changing that part of the code at all at that point.

Cheers,
Markus

-- 
Markus Lervik
Linux-administrator with a kungfoo grip
Vaasa City Library - Regional Library
[EMAIL PROTECTED]
+358-6-325 3589 / +358-40-832 6709

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[PHP-DB] mySQL 4.0.2-alpha

2002-01-28 Thread Paul Stevens

Hi guys,

I'm running mySQL 3.23.43 at the mo' with PHP-GTK 0.5.0. Long and the short
of the matter is, anyone running mySQL 4.0.2-alpha in anywhere near a
production 
environment yet, and if so, how is it? My server is a Win2K Server SP2
box...so how steady are those windows binaries??? :-)

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[PHP-DB] Re: [PHP]please help

2002-01-28 Thread Lerp

Hi there, here's an upload form that might be what u are looking for:
Hope his helps you out :)
Joe


php page 1



Upload Photo:



php page 2 -- receiving/upload processing page

";
print $userfile_name . "";

// copy the file being posted
if(copy($userfile, "/ez/codesnipits/temprotatingimages/". $userfile_name)){
print "Your photo has been uploaded and renamed successfully." ;
}
else
{
print "No luck with the upload dude.";
}


# create a path for storing into database
$patharola = "temprotatingimages/". $userfile_name;
print $patharola;

// insert path into database here
// connect to db
$connectionToDBid = odbc_connect("cts", "jqwwode", "dfgh5");

# query statement
$sqlr = "INSERT INTO IMAGES (employee_id, imagepath) VALUES ($employeeid ,
'$patharola' )";
# execute the sql statement (query) on the connection made
$resultset = odbc_do($connectionToDBid, $sqlr);

?>








"Duky Yuen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I want to make something like when you are attaching something to an
> email in hotmail. I want to make something that searches your own
> computer for files on your harddisk that you want to upload to the
> server.
>
> But I don't know if this is possible. If this is possible.Can somebody
> tell me which function I should use? Or if someone have an example...
> please...
>
> Duky
>



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[PHP-DB] Re: Another dynamic sql.

2002-01-28 Thread Lerp

Your select options have no values associated with them! Not sure if you did
this on purpose or not, but no value will be sent to next page.

Joe :)


"Raymond Lilleodegard" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hi all!
>
> I have this form with some choices:
>
> 
> Enter how many cars you want:
> 
> ford
> bmw
> mercedes
> 
>
> And then I am trying to get the price out of a table in my database with
> this code:
>
>
> $sql = mysql_query("SELECT '$car' FROM varetabell where carid='$carid' ");
> $myrow= mysql_fetch_array($sql);
> $x = $myrow["$car"];
>
> $price = $x * $number;
>
>
> Shouldn't this work?  Or am I missing something here?
>
>
>
> Best regards Raymond
>
>



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RE: [PHP-DB] [PHP] PHP and MYSQL Security`

2002-01-28 Thread Special Design Services Inc.

What I do, because I am using a test  platform is to put the line with 
pconnect() in a small text file, place this file somewhere on the disk, 
outside of the web site root
and refer to that file with include() in my code.
This way, even if the PHP source code were compromised, the user name 
and password used to access the DB cannot be seen.




>
> 
>
> Subject:
>
> Re: [PHP] PHP and MYSQL Security`
> From:
>
> "Fred" <[EMAIL PROTECTED]>
> Date:
>
> Sun, 27 Jan 2002 16:59:31 -0800
> To:
>
> [EMAIL PROTECTED], [EMAIL PROTECTED]
>
>
>If this file has a .php extension remote users will not have access to the
>variables because the file is parsed by php and they never see the actual
>file contents when requesting the document via the web.  If you are
>concerned with users on localhost having access to the file, simply give it
>the correct permissions so that no one else has read access.
>
>If you are concerned about web users having access, if, for example, the php
>parser crashed and apache tried to pass the file through without parsing,
>you can put the default server, user and pass variables in the php.ini file
>which is not in the document root for apache.  Of course, this only works if
>all of your scripts use the same server, user and password.
>
>Fred
>
>Duky Yuen <[EMAIL PROTECTED]> wrote in message
>[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
>
>>How can I secure my username and password? In 1 of my files, it contains
>>the following:
>>
>>$conn = mysql_connect( "12.34.56.78", "username", "password");
>>mysql_select_db("database",$conn);
>>
>>What should I do, so people can't get this information?
>>
>>Duky
>>


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Re: [PHP-DB] session_registered variable problem

2002-01-28 Thread Jason Wong

On Monday 28 January 2002 21:05, Markus Lervik wrote:
> On Monday 28 January 2002 14:57, Jason Wong wrote:
> > > Somehow when I print the value of $query, it looks fine.
> > > Then, when I change $query a bit and print it again,
> > > it looks fine.
> > > Change it a third time, and it suddenly gets the value it
> > > had in the beginning.
> >
> > Could you post some code please?
>
> Well, for instance, this screws up the $query. It works if one clicks
> once on the next-button, but clicking another time resets the query to
>
> SELECT
>   L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huom
>autuksia FROM uusi_lehtitaulu L, nimet N
>   WHERE (L.Nimi_id=N.id AND  N.Nimi LIKE "asdf")
>   LIMIT 0,20
>
> if($go=="next") {
>   if($page < $nr_pages && $page >= 0) {
>$page++;
>$replace_with= (($page) * 20).",";
>echo "replace_with: " . $replace_with;
>$replace = (($page-1) * 20).",";
>echo "replace: " .$replace;
>$query=ereg_replace($replace,$replace_with,$query);
>   }
> }
>
> It's the LIMIT 0,20 -> LIMIT 20,20 -> LIMIT 40,20 that doesn't work.
> Strange thing is, that it worked about a week ago. : \
> I can't really recall changing that part of the code at all at that point.

Crikey, it looks terribly complicated for what looks like something to page 
through a set of results.

Wouldn't it be easier to do away with the ereg_replace()?

 if($go=="next") {
   if($page < $nr_pages && $page >= 0) {
 $page++;
 $start = $page * 20;   
   }
 }

 $qry = "SELECT
L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huom
autuksia FROM uusi_lehtitaulu L, nimet N
   WHERE (L.Nimi_id=N.id AND  N.Nimi LIKE "asdf")
LIMIT $start,20"


Even it this doesn't work, at least it's easier to understand :)

-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
The way to love anything is to realize that it might be lost.
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Re: [PHP-DB] session_registered variable problem

2002-01-28 Thread Markus Lervik

On Monday 28 January 2002 17:34, Jason Wong wrote:

>
> Crikey, it looks terribly complicated for what looks like something to page
> through a set of results.
>
> Wouldn't it be easier to do away with the ereg_replace()?
>
>  if($go=="next") {
>if($page < $nr_pages && $page >= 0) {
>  $page++;
>  $start = $page * 20;
>}
>  }
>
>  $qry = "SELECT
>
> L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huom
> autuksia FROM uusi_lehtitaulu L, nimet N
>WHERE (L.Nimi_id=N.id AND  N.Nimi LIKE "asdf")
>   LIMIT $start,20"
>
>
> Even it this doesn't work, at least it's easier to understand :)

Well, in this particular case it won't work at all, because the SQL-query is 
dynamically
created based on what the user wants to search for. It doesn't nessecary include all 
columns. The only static bit of the query is "FROM uusi_lehtitaulu L, nimet N" and 
"WHERE (L.Nimi_id=N.id" : ) It would be just as big a mess with a truckload of
concatenations, like 

$query = "SELECT " . $tables . " FROM uusi_lehtitaulu L, nimet N WHERE
(L.nimi_id=N.id" . $search . ") LIMIT $start,20";

Besides, I have a strange fetisch to do everything the hard way. : )


Cheers,
Markus

-- 
Markus Lervik
Linux-administrator with a kungfoo grip
Vaasa City Library - Regional Library
[EMAIL PROTECTED]
+358-6-325 3589 / +358-40-832 6709

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Re: [PHP-DB] session_registered variable problem

2002-01-28 Thread Jason Wong

On Monday 28 January 2002 23:45, Markus Lervik wrote:

> Well, in this particular case it won't work at all, because the SQL-query
> is dynamically created based on what the user wants to search for. It
> doesn't nessecary include all columns. The only static bit of the query is
> "FROM uusi_lehtitaulu L, nimet N" and "WHERE (L.Nimi_id=N.id" : ) It would
> be just as big a mess with a truckload of concatenations, like
>
> $query = "SELECT " . $tables . " FROM uusi_lehtitaulu L, nimet N WHERE
> (L.nimi_id=N.id" . $search . ") LIMIT $start,20";

I usually give the concatenations a miss if I can and just use:

  $query = "SELECT $tables FROM ... ";

> Besides, I have a strange fetisch to do everything the hard way. : )

Masochist :)

Anyway, as per the subject, how are you getting $page? I assume it is stored 
in the session. Is $page behaving correctly when you go from one page to the 
next? Could you post the relevant code? 


-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
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and without any visible reason.
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Re: [PHP-DB] session_registered variable problem

2002-01-28 Thread Markus Lervik

On Monday 28 January 2002 18:06, Jason Wong wrote:

> > Besides, I have a strange fetisch to do everything the hard way. : )
> Masochist :)

:D

> Anyway, as per the subject, how are you getting $page? I assume it is
> stored in the session. Is $page behaving correctly when you go from one
> page to the next? Could you post the relevant code?

$page behaves just fine :

first time around (haven't got to the nav() -function yet..) :

query from nav(): 
SELECT 

L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huomautuksia 
FROM uusi_lehtitaulu L, nimet N 
WHERE L.Nimi_id=N.id 
LIMIT 0,20

//this is from the print_table -function

query: 
SELECT 

L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huomautuksia 
FROM uusi_lehtitaulu L, nimet N 
WHERE L.Nimi_id=N.id 
LIMIT 0,20

page: 0   <- note! bells! whistles! 
nr_pages: 230


We hit the next-button, and we get...

replace_with: 20,
replace: 0,

query from nav(): 
SELECT 

L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huomautuksia 
FROM uusi_lehtitaulu L, nimet N 
WHERE L.Nimi_id=N.id 
LIMIT 20,20

query: 
SELECT 

L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huomautuksia 
FROM uusi_lehtitaulu L, nimet N 
WHERE L.Nimi_id=N.id 
LIMIT 20,20

page: 1 <- bells 'n' whistles again!
nr_pages: 230


Looks quite good. We hit the next-button again :

with: 40,
what: 20,

query from nav(): 
SELECT 

L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huomautuksia 
FROM uusi_lehtitaulu L, nimet N 
WHERE L.Nimi_id=N.id 
LIMIT 0,20

query: 
SELECT 

L.Tunnus,N.Nimi,L.ISSN,L.Vuosi,L.Lisätietoja,L.Vuosikerta,L.Numerot,L.Huomautuksia 
FROM uusi_lehtitaulu L, nimet N 
WHERE L.Nimi_id=N.id 
LIMIT 0,20

page: 2 <- yup, it's incrementing, all right...
nr_pages: 230

Whack! Doesn't work any more...  It's starting to drive me mad. 


Cheers,
Markus

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Vaasa City Library - Regional Library
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Re: [PHP-DB] session_registered variable problem

2002-01-28 Thread Jason Wong

On Tuesday 29 January 2002 00:24, Markus Lervik wrote:
> On Monday 28 January 2002 18:06, Jason Wong wrote:

> $page behaves just fine :

[snip]

> page: 2 <- yup, it's incrementing, all right...
> nr_pages: 230
>
> Whack! Doesn't work any more...  It's starting to drive me mad.

Me too. Post the full code so we can all revel in this madness!


-- 
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk

/*
"Why must you tell me all your secrets when it's hard enough to love
you knowing nothing?"
-- Lloyd Cole and the Commotions
*/

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[PHP-DB] UPDATE query to add to a field

2002-01-28 Thread Adv. Systems Design


hello all:

I need to be able to update a field in MySQL, the catch is that I have to add on to 
text that is already there and I have to be able to do it within MySQL (phpMyAdmin). 
My first idea was to do: 

SET prod_desc = prod_desc + "more info to tack on to end"

but this is a mathematical function which returns 0!

Any ideas?

Thanks



Make it idiot-proof and someone will make a better idiot.


-
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Yahoo! Auctions Great stuff seeking new owners! Bid now!


Re: [PHP-DB] UPDATE query to add to a field

2002-01-28 Thread Paul DuBois

At 9:06 -0800 1/28/02, Adv. Systems Design wrote:
>hello all:
>
>I need to be able to update a field in MySQL, the catch is that I 
>have to add on to text that is already there and I have to be able 
>to do it within MySQL (phpMyAdmin). My first idea was to do:
>
>SET prod_desc = prod_desc + "more info to tack on to end"
>
>but this is a mathematical function which returns 0!

Use CONCAT().

>Any ideas?
>
>Thanks


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Re: [PHP-DB] UPDATE query to add to a field

2002-01-28 Thread Joffrey van Wageningen

> SET prod_desc = prod_desc + "more info to tack on to end"
> 
> but this is a mathematical function which returns 0!

update table set field = concat(field, "more info") where id = 1;

find the string functions in the mysql manual...

with kind regards,
Joffrey van Wageningen


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Re: [PHP-DB] UPDATE query to add to a field

2002-01-28 Thread DL Neil

Hello Adv. Systems Design,

> I need to be able to update a field in MySQL, the catch is that I have to add on to 
>text that is already there
and I have to be able to do it within MySQL (phpMyAdmin). My first idea was to do:
>
> SET prod_desc = prod_desc + "more info to tack on to end"
>
> but this is a mathematical function which returns 0!
>
> Any ideas?


SET prod_desc = CONCAT( prod_desc, "more info to tack on to end" )

Manual: 6.3.2  String Functions

Regards,
=dn



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RE: [PHP-DB] [PHP] PHP and MYSQL Security`

2002-01-28 Thread Peter Adams \(IKN\)

I've figured out a fairly secure program structure.  Here's one option
(I'm sure there's as many ways to accomplish similar security as there
are people on this list):

First, a little info about the environment.  It's a Linux OS running
Apache Web Server.  Multi-user environment providing hosting to multiple
domains.  Development is done on Windows boxes.

Now, to accomplish security and keep it relatively well hidden took some
doing.  First, I use what I call 'control files'.  These are the only
files in the Web accessible directory tree (i.e.
www.interkan.net/News/index.phtml).  These files contain only code to
process submitted commands (or default ones should no command be
submitted) and include the proper files (config module which is where
the mySQL access info is stored, global code libraries, and the actual
code modules to handle submitted data).

The included modules are all kept in a PHP include directory in the
appropriate user directory (i.e. /home/user/php-inc/).  Due to
restrictions, we have to have the files themselves with 644 permissions
(so the Web server can read them), but the directory permissions for
php-inc and php-inc/ are set to 711.  The permissions work out
that no one can read the files unless they (1) know the exact path and
filename and (2) have shell access to the server (the only people that
have that are employees).

This helps in a couple ways.  If the PHP process ever dies, all someone
will see when going a PHP file is the file comment block, the file
include information (not necessarily good, but they'd have to get into
the server with a shell account first), and some if and switch
statements.  It also narrows down any security breaches to someone who
had access to the system, instead of the entire Internet community.
__
Peter Adams[EMAIL PROTECTED]
Web Developer  http://www.interkan.net
InterKan.Net, Inc. (785) 565-0991



> -Original Message-
> From: Duky Yuen [mailto:[EMAIL PROTECTED]] 
> Sent: Sunday, January 27, 2002 6:38 PM
> To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
> Subject: [PHP-DB] [PHP] PHP and MYSQL Security`
> 
> 
> How can I secure my username and password? In 1 of my files, 
> it contains the following:
> 
> $conn = mysql_connect( "12.34.56.78", "username", "password");
> mysql_select_db("database",$conn);
> 
> What should I do, so people can't get this information?
> 
> Duky
> 
> 
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RE: [PHP-DB] Re: [PHP] PHP and MYSQL Security`

2002-01-28 Thread Peter Adams \(IKN\)

> >If this file has a .php extension remote users will not have 
> access to 
> >the variables because the file is parsed by php and they 
> never see the 
> >actual file contents when requesting the document via the 
> web.  If you 
> >are concerned with users on localhost having access to the 
> file, simply 
> >give it the correct permissions so that no one else has read access.
> 
> No so easy.  The server itself must have read access.  If 
> other users on the local host can install scripts that the 
> server executes, any of those scripts can read the text of 
> your scripts.
> 
> What then? You're hosed.

That's the beauty of running safe mode.  With that engaged, scripts
can't access files owned by a different username.  There are ways around
it (we have a special exception set up for an app I've been working on
that allows it to include a file based on the domain it was called from
then include the proper templates, which are generally owned by a
different user), but it's difficult without shell access.
__
Peter Adams[EMAIL PROTECTED]
Web Developer  http://www.interkan.net
InterKan.Net, Inc. (785) 565-0991


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[PHP-DB] Re: [PHP] File uploading like hotmail

2002-01-28 Thread qartis

I think you might be looking for  HTML form element.


"Duky Yuen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I want to make something like when you are attaching something to an
> email in hotmail. I want to make something that searches your own
> computer for files on your harddisk that you want to upload to the
> server.
>
> But I don't know if this is possible. If this is possible.Can somebody
> tell me which function I should use? Or if someone have an example...
> please...
>
> Duky
>



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[PHP-DB] Re: uploading

2002-01-28 Thread Thomas \"omega\" Henning

thanks alot it helped
"Thomas "Omega" Henning" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hello how can i upload something from my hdd to the webserver using php?
>
> note: don't have ftp on the machine!!
>
> Thanks
>
> Thomas "omega" Henning
>
>



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[PHP-DB] storing and retrieving arrays in mysql

2002-01-28 Thread Corey Eiseman

Hi folks, I've got a question hopefully someone can shed some light on for
me.

I'm building an online store for a client, and one of the things he wants is
to organize his products into categories and subcategories.. not so unusual,
but the kicker is he wants to be able to associate a subcategory with more
than one category.

I was thinking that I should be able to easily serialize an array of cat_IDs
and store it, but my concern is that this will sacrifice a great deal of
flexibility when retrieving the data. For instance when I want to get the
subcategories in a single category, I would pretty much have to select ALL
the rows in the subcategory table, unserialize the category array for each
row, and then check each to see if the cat_ID is in the array..?

That just feels inefficient to me, and I'm almost certain I must be
overlooking something simpler..

Also, I don't think I can use a SET data type because I want to be able to
add values to the set (categories) dynamically in the future. But maybe I'm
wrong and there is a way to do that...?

Anyway I thought I'd throw it to those more experienced than me before I
started coding. Anyone have any ideas?

Thanks in advance,

Corey Eiseman
Infinite Orange Incorporated
http://infiniteorange.com/




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Re: [PHP-DB] storing and retrieving arrays in mysql

2002-01-28 Thread olinux

I've done something similar by building a
multidimensional array of categories and subcats in
one query and then Loop thru this with a for each to
build the category structure.

Categories table looks like this
CatID | ParentID | CategoryName

Array is 
$category_menu[$ParentID][$CatID]

I'm sure there's a better way, but this seems to work
OK.

I will email code sample later [off-list]. I would be
happy to share code with others as well.

olinux

--- Corey Eiseman <[EMAIL PROTECTED]> wrote:
> Hi folks, I've got a question hopefully someone can
> shed some light on for
> me.
> 
> I'm building an online store for a client, and one
> of the things he wants is
> to organize his products into categories and
> subcategories.. not so unusual,
> but the kicker is he wants to be able to associate a
> subcategory with more
> than one category.
> 
> I was thinking that I should be able to easily
> serialize an array of cat_IDs
> and store it, but my concern is that this will
> sacrifice a great deal of
> flexibility when retrieving the data. For instance
> when I want to get the
> subcategories in a single category, I would pretty
> much have to select ALL
> the rows in the subcategory table, unserialize the
> category array for each
> row, and then check each to see if the cat_ID is in
> the array..?
> 
> That just feels inefficient to me, and I'm almost
> certain I must be
> overlooking something simpler..
> 
> Also, I don't think I can use a SET data type
> because I want to be able to
> add values to the set (categories) dynamically in
> the future. But maybe I'm
> wrong and there is a way to do that...?
> 
> Anyway I thought I'd throw it to those more
> experienced than me before I
> started coding. Anyone have any ideas?
> 
> Thanks in advance,
> 
> Corey Eiseman
> Infinite Orange Incorporated
> http://infiniteorange.com/
> 
> 
> 
> 
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> [EMAIL PROTECTED]
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> [EMAIL PROTECTED]
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> [EMAIL PROTECTED]
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RE: [PHP-DB] storing and retrieving arrays in mysql

2002-01-28 Thread Peter Adams \(IKN\)

I've been working on something like this since about September/October.
It's not too hard, but it did take more coding than I thought it would.
Now, there's a couple of ways you could do this.  I chose the cheap (in
terms of the number of DB tables) way out.

The way I have it set up is this.  You have certain categories ("Main"
categories) that have a parent category of ',0,'  (I found I had to
enclose the categoryIDs in commas to prevent accidentally selecting a
category that matched any of the digits of any other categoryID in the
same order) then I have the subcategories which have multiple parents.

The form to add a new product gathers all the information I need --
category name/title, description, whether it should be displayed and the
parent category(s), with the parent categorys being displayed in the
form of checkboxes.  This lets the admin select multiple categories for
the parent (we have a similar set up for the products since a product
can belong in multiple categories).

The checkboxes themselves are named p_id (I think) and are an array,
thus I print the name value as NAME='p_id[" . $cid . "]', where $cid is
the variable containing the category ID (I actually have it in a
recordset array, of course).  Then, it just prints a p_id checkbox for
each possible parent (obviously when editing the category ID of the one
being edited isn't printed).  When editing, those categories to which
the category has as parents are checked.

When the form is submitted, it puts the posted variable values (the
array of checkboxes) into a comma-delimited string.  Whenever I query
the database for subcategories, unfortunately, I have to make sure I
remember to check for parent LIKE '%,ID,%' so I don't pick up one I
don't want (i.e. if I were to say LIKE '%ID%' and I was looking for ID =
20 and had that and ID =200, subcategories with parents of category 20
and 200 would end up returned).  That's how I handled it.  I'm probably
going to upgrade to this next option later on, though.

Create a table for the categories and then a table that determines a
category's children.  Basically, it's just two fields: categoryID and
parent -- with both fields making up the primary key.  Then, for each
parent category they checked on the form, insert the new category's ID
in the categoryID field and the parent's ID into the parent field.

That's probably the best way since it would probably cut down on space
(I use a TEXT field currently, whereas the second option could be done
with two MEDIUMINT fields).

If you want some code samples, e-mail me privately.  If you want to see
the script (version 1.5.23 -- we've got 2.0 almost finished up now) in
action, we've got one of our clients playing guinea pig at
www.4dacres.com.

__
Peter Adams[EMAIL PROTECTED]
Web Developer  http://www.interkan.net
InterKan.Net, Inc. (785) 565-0991

> -Original Message-
> From: Corey Eiseman [mailto:[EMAIL PROTECTED]] 
> Sent: Monday, January 28, 2002 2:56 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] storing and retrieving arrays in mysql
> 
> 
> Hi folks, I've got a question hopefully someone can shed some 
> light on for me.
> 
> I'm building an online store for a client, and one of the 
> things he wants is to organize his products into categories 
> and subcategories.. not so unusual, but the kicker is he 
> wants to be able to associate a subcategory with more than 
> one category.
> 
> I was thinking that I should be able to easily serialize an 
> array of cat_IDs and store it, but my concern is that this 
> will sacrifice a great deal of flexibility when retrieving 
> the data. For instance when I want to get the subcategories 
> in a single category, I would pretty much have to select ALL 
> the rows in the subcategory table, unserialize the category 
> array for each row, and then check each to see if the cat_ID 
> is in the array..?
> 
> That just feels inefficient to me, and I'm almost certain I 
> must be overlooking something simpler..
> 
> Also, I don't think I can use a SET data type because I want 
> to be able to add values to the set (categories) dynamically 
> in the future. But maybe I'm wrong and there is a way to do that...?
> 
> Anyway I thought I'd throw it to those more experienced than 
> me before I started coding. Anyone have any ideas?
> 
> Thanks in advance,
> 
> Corey Eiseman
> Infinite Orange Incorporated
> http://infiniteorange.com/
> 
> 
> 
> 
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[PHP-DB] How to upload image?

2002-01-28 Thread Chris Payne

Hi there everyone,

I'm writing an update utility for a database system, but I need to give the user the 
option to select a file from their machine to upload to the server (Not to the DB 
itself, but an images dir), what is the best way of doing this in PHP so that it 
brings up a file requestor too?

Thanks for your help everyone :-)

Chris



Re: [PHP-DB] How to upload image?

2002-01-28 Thread David Sullivan

For the form, use 
and as for handling where the file get saved on the server-side, see:
http://www.php.net/manual/en/features.file-upload.php


On January 28, 2002 05:04 pm, you wrote:
> Hi there everyone,
>
> I'm writing an update utility for a database system, but I need to give the
> user the option to select a file from their machine to upload to the server
> (Not to the DB itself, but an images dir), what is the best way of doing
> this in PHP so that it brings up a file requestor too?
>
> Thanks for your help everyone :-)
>
> Chris

- Dave

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[PHP-DB] Date function

2002-01-28 Thread Sridhar Moparthy

Hi All,

Is there any function in PHP4.X or in Oracle8i to calculate number of
Business days ( excluding Saturday and Sunday)  between two given dates?

Thanks in advance.
Sridhar.

-Original Message-
From: qartis [mailto:[EMAIL PROTECTED]]
Sent: Monday, January 28, 2002 1:38 PM
To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Subject: [PHP-DB] Re: [PHP] File uploading like hotmail


I think you might be looking for  HTML form element.


"Duky Yuen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I want to make something like when you are attaching something to an
> email in hotmail. I want to make something that searches your own
> computer for files on your harddisk that you want to upload to the
> server.
>
> But I don't know if this is possible. If this is possible.Can somebody
> tell me which function I should use? Or if someone have an example...
> please...
>
> Duky
>



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[PHP-DB] ensuring unique field value in MySQL using PHP

2002-01-28 Thread Janet Valade

I have a form in which users create their own login name. The code for
storing the login name they created does something like the following:

$userlogin is the name the user typed in the form
$sql = "select loginname from login where loginname='$userlogin'";
$result=mysql_query($sql);
$num=mysql_num_rows($result);
if ($num > 0)
{
   echo "Sorry, that login name is already taken, Try another";
   shows form again;
}
else
   insert new record into database;
   echo "okay, your new account is added";

I am wondering if it is possible that two people could hit the submit button
at the exact same time with the same login name so that the same login name
could get entered twice? Seems unlikely that this would happen by accident,
but is this something a hacker could do on purpose?

Thanks,

Janet



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[PHP-DB] RE: ensuring unique field value in MySQL using PHP

2002-01-28 Thread Adam Royle

Checking a small database for username/password combination would happen so
quick, it would be nearly impossible for two usernames to be entered in.
Your script should work properly, but to make sure no duplicates are
entered, you can change the column definition using the "ALTER columnName"
command to make sure there are no duplicates. Look in the mySQL
documentation (www.mysql.com) to find the correct command.

Adam

-Original Message-
From: Janet Valade [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, January 29, 2002 1:24 PM
To: [EMAIL PROTECTED]
Subject: ensuring unique field value in MySQL using PHP


I have a form in which users create their own login name. The code for
storing the login name they created does something like the following:

$userlogin is the name the user typed in the form
$sql = "select loginname from login where loginname='$userlogin'";
$result=mysql_query($sql);
$num=mysql_num_rows($result);
if ($num > 0)
{
   echo "Sorry, that login name is already taken, Try another";
   shows form again;
}
else
   insert new record into database;
   echo "okay, your new account is added";

I am wondering if it is possible that two people could hit the submit button
at the exact same time with the same login name so that the same login name
could get entered twice? Seems unlikely that this would happen by accident,
but is this something a hacker could do on purpose?

Thanks,

Janet




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Re: [PHP-DB] Re: Moving from MySQL to MSSQL Server 2000

2002-01-28 Thread Manuel Lemos

Hello,

Boaz Yahav wrote:
> 
> I don't think you realize the nature of the site. Its an auction site
> and not a retail site.
> Prices need to be updated on thousands of clients simultaneously. Our
> goal is to
> refresh the data on the client every 3-10 seconds while having 1500
> online open
> auctions.
> 
> This is quite a mess :)

humm I am afraid that goal may be a bit unrealistic for your budget.
I suggest that you cache as much content as you can to avoid hitting the
database.

One technique that I use in the PHP Classes site is to cache content
indefinitly. Then when the data that defines that content is changed I
call a method of my cache file class to void the cache, so it will force
the cached data if and when it is needed on the next access for a page
that uses that. This way I don't retard the database update process and
the cache is only generated if it will ever be need. So I avoid having
situations where more than one update is made to the database before the
cached data needs to be redisplayed.

For making your site static as much as possible, I usually recommend the
"page fault" technique. That consists of static pages that are deleted
when their content is updated. When there is a request for them, since
when they are missing the server redirects the request to your error
document. That is where you regenarate the page file with upto date
data. While the page is upto date it remains in the disk and is served
as static page at the server top speed (no scripting language nor
database to clutter the access).

Regards,
Manuel Lemos

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Re: [PHP-DB] Re: Moving from MySQL to MSSQL Server 2000

2002-01-28 Thread Manuel Lemos

Hello,

Boaz Yahav wrote:
> 
> Fail Over is not for when the cluster is too busy, it's for when the
> server dies, or when someone killed it's power, or when you loose your
> network, or when you get a blue screen (assuming you are on windows :).

Ok, it is the same. When the server is dead, the response is the same:
none.


> Then, in about 30 seconds the cluster will do a fail over to the 2nd
> node and the site will continue to work as if nothing happens... it's
> pretty amazing to see this in action :)

I had requests to implement that in Metabase. The problem is that PHP
database connection functions do not let you define the connection
timeout parameters or check if the current persistent connection is
still alive. Anyway, nothing that a good middleware could not do better
for you.

Regards,
Manuel Lemos

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[PHP-DB] Date operations.

2002-01-28 Thread Garry Optland

Hi All,

I have a problem working out a suitable algorithm either in PHP or MySQL. 

Basically I have a DB that keeps track of breeding records. Each record 
has a paired data, and a split-up date. 

I need to generate some statistics to work out average numbers of pairs 
per month, averaged on a daily basis, for a given start and stop date, 
typically a year or year-to-date.

All the algorithms I can think of are messy, where I have to loop through 
all the breeding records for every day of the year, and count how many 
pairs are breeding by seeing if the date is between the start and stop 
dates, and then average that on a monthly basis. I can't see 
that scaling very well, as there might be several hundred breeding records 
for a given year, multiplied by 365 days.

Has anyone any hints/pointers for an efficient way to do this?

Regards,
Garry.


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