Re: [PHP-DB] Shift_JIS Character Set

2003-12-12 Thread Jason Wong
On Friday 12 December 2003 14:47, Ng Hwee Hwee wrote:

 i'm doing a form (form.php)
 1) user comes to form.php
 2) user enters details and press submit
 3) details get passed to form action verify.php
 4) verify.php puts all the $_POST into an array $formVars
 5) verify.php session_register(formVars)
 6) if there is an error, verify.php sends user back to form.php
 7) form.php echos all the $formVars (i.e. details that user has previously
 entered) in their respective textfields, together with the error messages

 i need to stripslashes at Step7 because as you have mentioned in a previous
 post on this list, magic_quotes_gpc added slashes to all my $_POST. Thus,
 since I need to echo them, I need to stripslashes.. I tried echoing the
 $formVars without the stripslashes and I see all the slashes before the
 quotes and backslashes etc.. so, by right, with English characters, i still
 need to stripslashes.

OK, if you *know* that the data has been through addslashes() (or equivalent 
eg magic_quotes_gpc) then doing a strip_slashes() should revert the data to 
its original form.

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Re: [PHP-DB] Shift_JIS Character Set

2003-12-12 Thread Ng Hwee Hwee
yes, and that's why i'm appealing for help on how to stripslashes when it
comes to the Shift_JIS character set...

will really appreciate advises! thanx lots!


- Original Message -
 OK, if you *know* that the data has been through addslashes() (or
equivalent
 eg magic_quotes_gpc) then doing a strip_slashes() should revert the data
to
 its original form.

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 Jason Wong - Gremlins Associates - www.gremlins.biz
 Open Source Software Systems Integrators
 * Web Design  Hosting * Internet  Intranet Applications Development *

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RE: [PHP-DB] NEXTVAL Question

2003-12-12 Thread N . A . Morgan
put the schema name before the sequence object you need to increment, i.e.

$sql = select SCHEMA.TEAM_SEQ.NEXTVAL as \nextval\ from sys.dual;

Regards,
Neil Morgan

-Original Message-
From: Paul Miller [mailto:[EMAIL PROTECTED] 
Sent: 12 December 2003 00:05
To: [EMAIL PROTECTED]
Subject: [PHP-DB] NEXTVAL Question


Does anyone know how to make this work???

$sql = select TEAM_SEQ.NEXTVAL as \nextval\ from sys.dual;

I get the following error in PHP:

ociexecute(): OCIStmtExecute: ORA-00903: invalid table name

But when I use my SQL tool to hit the DB will all the same parameters, it
works just fine.

I have also tried:
$sql = select TEAM_SEQ.NEXTVAL from sys.dual;
$sql = select TEAM_SEQ.NEXTVAL from dual;

all with the same results.

I am running Red Hat AS, PHP 4.3, OCI8 Functions and Oracle 9i.

- Paul

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[PHP-DB] Need help!!!

2003-12-12 Thread irinchiang


Hi..

Anyone can help me with this script???I kept getting error saying i did not 
define the variable $tutor_id and whenever I define it as $tutor_id = $_POST
['tutor_id'], I got error saying there was an Undefined index!
What was exactly the problem??
Need help desperately , greatly appreciate any suggestions/help/solutions 
given...

---

===edit.php 
 
$dsn = mysql://root:[EMAIL PROTECTED]/tb1; 
$db = DB::connect ($dsn); 

   if (DB::isError ($db)) 
   die ($db-getMessage()); 

$action = $_GET[action]; 

if($action == delete) 
{ 
$sql = mysql_query (DELETE FROM tutor WHERE tutor_id='$tutor_id'); 
} 

if($action == edit) 
{ 
$sql = mysql_query (SELECT * FROM tutor WHERE tutor_id 
='$tutor_id');**ERROR HERE!!* 

$row = mysql_fetch_array($sql); 

echo form name=\edit_tutor\ method=\post\ action=\edit2.php\; 
echo Tutor ID : .$row['tutor_id'].br; 
echo input type=\hidden\ name=\tutor_id\ value=\.$row 
['tutor_id'].\; 
echo Name : input name=\tutor_name\ type=\text\ value=\.$row 
['tutor_name'].\br; 
echo Contact No : input type=\text\ name=\tutor_contact\ value=\.$row 
['tutor_contact'].\br; 
echo E-mail : input type=\text\ name=\tutor_email\ value=\.$row 
['tutor_email'].\br; 
echo Profile : input type=\text\ name=\tutor_profile\ value=\.$row 
['tutor_profile'].\br; 
echo input type=\submit\ name=\Submit\ value=\Update\; 
echo /form; 

} 
? 

$result = $db-query($sql); 

if( DB::isError($result) ) { 
die ($result-getMessage()); 
} 
---


Irin.

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RE: [PHP-DB] Need help!!!

2003-12-12 Thread Griffiths, Daniel
how are you getting the $tutor_id var?, in your script you are using $action = 
$_GET[action]; so you will need to define $tutor_id like $tutor_id = 
$_GET['tutor_id'], or just use $_GET['tutor_id'].

a form can only put variables in the POST array OR the GET array, and if your using 
hyperlinks to access this script then they will of course be GET vars. 

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: 12 December 2003 09:38
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Need help!!!




Hi..

Anyone can help me with this script???I kept getting error saying i did not 
define the variable $tutor_id and whenever I define it as $tutor_id = $_POST
['tutor_id'], I got error saying there was an Undefined index!
What was exactly the problem??
Need help desperately , greatly appreciate any suggestions/help/solutions 
given...

---

===edit.php 
 
$dsn = mysql://root:[EMAIL PROTECTED]/tb1; 
$db = DB::connect ($dsn); 

   if (DB::isError ($db)) 
   die ($db-getMessage()); 

$action = $_GET[action]; 

if($action == delete) 
{ 
$sql = mysql_query (DELETE FROM tutor WHERE tutor_id='$tutor_id'); 
} 

if($action == edit) 
{ 
$sql = mysql_query (SELECT * FROM tutor WHERE tutor_id 
='$tutor_id');**ERROR HERE!!* 

$row = mysql_fetch_array($sql); 

echo form name=\edit_tutor\ method=\post\ action=\edit2.php\; 
echo Tutor ID : .$row['tutor_id'].br; 
echo input type=\hidden\ name=\tutor_id\ value=\.$row 
['tutor_id'].\; 
echo Name : input name=\tutor_name\ type=\text\ value=\.$row 
['tutor_name'].\br; 
echo Contact No : input type=\text\ name=\tutor_contact\ value=\.$row 
['tutor_contact'].\br; 
echo E-mail : input type=\text\ name=\tutor_email\ value=\.$row 
['tutor_email'].\br; 
echo Profile : input type=\text\ name=\tutor_profile\ value=\.$row 
['tutor_profile'].\br; 
echo input type=\submit\ name=\Submit\ value=\Update\; 
echo /form; 

} 
? 

$result = $db-query($sql); 

if( DB::isError($result) ) { 
die ($result-getMessage()); 
} 
---


Irin.

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RE: [PHP-DB] Need help!!!

2003-12-12 Thread irinchiang


Yes I'm using a hyperlink to access this page (see below)
echo tda href=\edit.php?id=.$row
[tutor_id].action=edit\Modify/a/td;

and so I defined my variable $tutor_id as $tutor_id = $_GET['tutor_id'];
but i still got the error: Undefined index: tutor_id 
in /usr/local/apache/htdocs/tutor/edit.php 

WHat could be the problem?;(


---

how are you getting the $tutor_id var?, in your script you are using $action 
= $_GET[action]; so you will need to define $tutor_id like $tutor_id = $_GET
['tutor_id'], or just use $_GET['tutor_id']. 

a form can only put variables in the POST array OR the GET array, and if your 
using hyperlinks to access this script then they will of course be GET vars. 

-Original Message- 
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 
Sent: 12 December 2003 09:38 
To: [EMAIL PROTECTED] 
Subject: [PHP-DB] Need help!!! 




Hi.. 

Anyone can help me with this script???I kept getting error saying i did not 
define the variable $tutor_id and whenever I define it as $tutor_id = $_POST 
['tutor_id'], I got error saying there was an Undefined index! 
What was exactly the problem?? 
Need help desperately , greatly appreciate any suggestions/help/solutions 
given... 

---
 

===edit.php 
 
$dsn = mysql://root:[EMAIL PROTECTED]/tb1; 
$db = DB::connect ($dsn); 

   if (DB::isError ($db)) 
   die ($db-getMessage()); 

$action = $_GET[action]; 

if($action == delete) 
{ 
$sql = mysql_query (DELETE FROM tutor WHERE tutor_id='$tutor_id'); 
} 

if($action == edit) 
{ 
$sql = mysql_query (SELECT * FROM tutor WHERE tutor_id 
='$tutor_id');**ERROR HERE!!* 

$row = mysql_fetch_array($sql); 

echo form name=\edit_tutor\ method=\post\ action=\edit2.php\; 
echo Tutor ID : .$row['tutor_id'].br; 
echo input type=\hidden\ name=\tutor_id\ value=\.$row 
['tutor_id'].\; 
echo Name : input name=\tutor_name\ type=\text\ value=\.$row 
['tutor_name'].\br; 
echo Contact No : input type=\text\ name=\tutor_contact\ value=\.$row 
['tutor_contact'].\br; 
echo E-mail : input type=\text\ name=\tutor_email\ value=\.$row 
['tutor_email'].\br; 
echo Profile : input type=\text\ name=\tutor_profile\ value=\.$row 
['tutor_profile'].\br; 
echo input type=\submit\ name=\Submit\ value=\Update\; 
echo /form; 

} 
? 

$result = $db-query($sql); 

if( DB::isError($result) ) { 
die ($result-getMessage()); 
} 
---
 


Irin. 

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[PHP-DB] Thanks alot Griffiths!=)

2003-12-12 Thread irinchiang


Thanks alot for your help! That was very careless of me..;p I was stil quite a 
newbie to php-db coding...

By the way one last favour to ask of u hope u dun mind...

I have a script which was used to INSERT data into the database:
(See script below):

But everytime when I execute it I got the error DB constraint

I think the problem lies here:

$sql = INSERT INTO tutor (tutor_name, tutor_contact, tutor_email, 
tutor_profile)
VALUES ($tutor_name,$tutor_contact,$tutor_email,$tutor_profile);

Whenever I single quote the string value:
 VALUES ($tutor_name','$tutor_contact','$tutor_email','$tutor_profile');
it actually insert a NULL value into the DB though I did enter values into the 
text field.But if I dun quote it, I got the error:
DB constraint. By right I ought to quote it right??
I have gone thru my code umpteen times but to no avail. 
Where have i gone wrong??? ;(

---

$dsn = mysql://root:[EMAIL PROTECTED]/table1;

$db = DB::connect ($dsn);
   if (DB::isError ($db))
   die ($db-getMessage());

//create variables ready to go into DB

$tutor_name = $db-quote($_POST[tutor_name]);
$tutor_contact = $db-quote($_POST[tutor_contact]);
$tutor_email = $db-quote($_POST[tutor_email]);
$tutor_profile = $db-quote($_POST[tutor_profile]);

$sql = INSERT INTO tutor (tutor_name, tutor_contact, tutor_email, 
tutor_profile)
VALUES ($tutor_name,$tutor_contact,$tutor_email,$tutor_profile);

$result = $db-query($sql);
  if( DB::isError($result) ) {
die ($result-getMessage());
}
?HTML
head
titleAdd Tutor/title

!-- **CSS and HTML** --

/head
body
div align=center
div id=pagecontent5
?php if( $_POST[submit] != Save )
{
?

div align=center
form name=classinfo action=?php echo $_SERVER[PHP_SELF]; ?  
method=post
input type=hidden name=form_submitted value=1

table name=add_tutor  border=0px cellpadding=0 cellspacing=2 
width=900tr
 td class=lighter width=200Name:/td
 td class=lighter width=500Input type=text size=50 
name=tutor_name/td
   /tr
tr
td class=darker width=150Contact No:/td
td class=darker width=150Input type=text size=30 
name=tutor_contact/td
   /tr
tr
td class=lighter width=150Email:/td
 td class=lighter width=150Input type=text size=50 
name=tutor_email/td
   /tr

tr
td class=darker width=150Profile:/td
 td class=darker width=150TEXTAREA class=textarea 
NAME=tutor_profile ROWS=6 COLS=70
/TEXTAREA
/td
/tr
tr
td class=lighter width=150/td
 td class=lighter width=150a href=teacher_summary.php
div align=centerinput type=submit name=submit  
value=Save/div/a/td
   /tr
/table
/form
br


?php } else {?
table border=0
tr
td class=darker width=300Teacher:/td
td class=darker width=600
?
echo $_POST[tutor_name];
?
/td
/tr
trtd class=lighter width=300Contact No:/td
td class=lighter width=600
?
echo $_POST[tutor_contact];
?
/td
/tr
tr
td class=darker width=300E-mail Address:/td
td class=darker width=600
?
echo $_POST[tutor_email];
?
/td
/tr
tr
td class=lighter width=300Profile:/td
td class=lighter width=600
?
echo $_POST[tutor_profile];
?
/td
/tr
/body
/html
/div
/div
?}?

---
(I'm terribly sorry if the code is kinda messy)

All help are greatly appreciated!

Irin.

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Re: [PHP-DB] pg_result_error()

2003-12-12 Thread Martin Marques
El Vie 12 Dic 2003 00:09, Gerard Samuel escribió:
 What good is this function?
 A quick example of the wall Im running into -
 $sql = 'INSERT INTO .';
 $result = pg_query($conn_id, $sql);
 if ($result === false)
 {
 var_dump( pg_result_error( $result ) );

I would use here this:
die(pg_result_error( $result ));

 }

 According to the manual, pg_result_error takes the result resource.
 If that resource is boolean false for one reason or another, then
 pg_result_error isn't useful.
 Anyone has any other ideas, besides using
 pg_last_error()?

Did you try it?

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Re: [PHP-DB] More of a question rather than seeking code. :)

2003-12-12 Thread Jason Wong
On Friday 12 December 2003 20:15, JeRRy wrote:

[snip]

 I just recently added a flag in the database so if the
 email bounces I manully flag the account as a bad
 email so in future emails won't be sent to that users
 email unless they correct their email.  This is time
 consuming.

[snip]

Do yourself a big favour and setup something using a real mailing-list 
software. Any decent mailing-list manager will handle bounces automatically. 
Sustained bounces from an email address will eventually lead it to being 
suspended or unsubscribed automatically.

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Open Source Software Systems Integrators
* Web Design  Hosting * Internet  Intranet Applications Development *
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[PHP-DB] anyone trickier than 'LIKE' ?

2003-12-12 Thread Hadi
Hi all,
I have this data (eg.) ..big fat buddy... on the database . If
a person enter big buddy in the search form , the ...big fat
buddy. data will not be found . Any suggestion ?
Thanks in advance.


select * from table where data like %keywords%;

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Re: [PHP-DB] anyone trickier than 'LIKE' ?

2003-12-12 Thread Larry E . Ullman
I have this data (eg.) ..big fat buddy... on the 
database . If
a person enter big buddy in the search form , the ...big fat
buddy. data will not be found . Any suggestion ?
select * from table where data like %keywords%;
You can break up the search terms into words and do
...data LIKE '%$keyword[0]%' OR data LIKE '%$keyword[1]%'...
This will be slow though. You could also do full text searches if you 
are using MySQL. Have good indexes for whichever route you go.

Larry

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RE: [PHP-DB] anyone trickier than 'LIKE' ?

2003-12-12 Thread N . A . Morgan
$keywords = str_replace(' ','%',$keywords);
select * from table where data like %keywords%;

-Original Message-
From: Hadi [mailto:[EMAIL PROTECTED] 
Sent: 12 December 2003 13:21
To: [EMAIL PROTECTED]
Subject: [PHP-DB] anyone trickier than 'LIKE' ?


Hi all,
I have this data (eg.) ..big fat buddy... on the database . If
a person enter big buddy in the search form , the ...big fat
buddy. data will not be found . Any suggestion ? Thanks in advance.


select * from table where data like %keywords%;

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Re: [PHP-DB] pg_result_error()

2003-12-12 Thread Gerard Samuel
Im going to CC this to the PostgreSQL list also.

On Friday 12 December 2003 06:44 am, Martin Marques wrote:
 El Vie 12 Dic 2003 00:09, Gerard Samuel escribió:
  What good is this function?
  A quick example of the wall Im running into -
  $sql = 'INSERT INTO .';
  $result = pg_query($conn_id, $sql);
  if ($result === false)
  {
  var_dump( pg_result_error( $result ) );

 I would use here this:
   die(pg_result_error( $result ));

That is fine and all, but my original example was just an example of the non 
functionality of pg_result_error(), not how to handle errors when a query 
fails.
But for arguement sake, lets use your example in some dummy code[0].
$result is still boolean false, and pg_result_error() will still return an 
empty string, and using die, would just die, with no report of what happened.
Then whats the use of pg_result_error().

  According to the manual, pg_result_error takes the result resource.
  If that resource is boolean false for one reason or another, then
  pg_result_error isn't useful.
  Anyone has any other ideas, besides using
  pg_last_error()?

 Did you try it?

Yes I've tried it.  In my DB class, Im currently using both pg_result_error() 
and pg_last_error(), with pg_last_error() being secondary (a fall back) to 
pg_result_error().
Because according to the manual -
http://us2.php.net/manual/en/function.pg-last-error.php
--quote--
Error messages may be overwritten by internal PostgreSQL(libpq) function 
calls. It may not return appropriate error message, if multiple errors are 
occured inside a PostgreSQL module function.

Use pg_result_error(), pg_result_status() and pg_connection_status() for 
better error handling.
--quote--

So again, I beg the question.
What good is pg_result_error(), when you *must* feed it boolean false and it 
returns an empty string??

[0]
?php

$conn = pg_connect(dbname=foo user=bar password=pass);

$sql = 'SELECT * FROM flagss';  // flags was misspelled
$result = @pg_query($conn, $sql);  // returns false
if ($result === false)
{
die( pg_result_error($result) );  // results in an empty page
}

?

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Re: [PHP-DB] pg_result_error()

2003-12-12 Thread Martin Marques
El Vie 12 Dic 2003 11:19, Gerard Samuel escribió:
 Im going to CC this to the PostgreSQL list also.

 On Friday 12 December 2003 06:44 am, Martin Marques wrote:
  El Vie 12 Dic 2003 00:09, Gerard Samuel escribió:
   What good is this function?
   A quick example of the wall Im running into -
   $sql = 'INSERT INTO .';
   $result = pg_query($conn_id, $sql);
   if ($result === false)
   {
   var_dump( pg_result_error( $result ) );
 
  I would use here this:
  die(pg_result_error( $result ));

 That is fine and all, but my original example was just an example of the
 non functionality of pg_result_error(), not how to handle errors when a
 query fails.
 But for arguement sake, lets use your example in some dummy code[0].
 $result is still boolean false, and pg_result_error() will still return an
 empty string, and using die, would just die, with no report of what
 happened. Then whats the use of pg_result_error().

Looks like you are totally right. Tried it and it works horrible. Any idea on 
why this is like this?

P.D.: I had to pass my php4 in Debian to unstable to get a workable (with the 
newer capabilities) version. Very annoing, especially becuase I had to pass 
apache to unstable as well. :-(

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[PHP-DB] MSSQL Error message

2003-12-12 Thread Liana Cristine Leopold

 
Hi

I´m using a connection using php 4.3.3 with sql server  ... and i´m having a error 
message :
mssql error: [: Connection error to server '192.0.0.6\binotto' with user 'LUA']  ..

The user 'LUA' is the dbowner ...

anyone can help me???

Sorry my bad english ...

Liana Cristine Leopold

RE: [PHP-DB] Cronjob

2003-12-12 Thread Daevid Vincent
One thing I might suggest is to queue up your inserts, then use the extended
insert syntax to populate several records at once in a single INSERT instead
of 12 separate ones. I believe you can alter the /etc/my.cnf file to play
more 'nice' as far as priorities and memory, but I've never touched it
  

 -Original Message-
 From: Ryan Marks [mailto:[EMAIL PROTECTED] 
 Sent: Thursday, December 11, 2003 6:45 PM
 To: PHP-DB
 Subject: [PHP-DB] Cronjob
 
 Hello all,
 
 I have a cronjob that runs a PHP script against MySQL.  The 
 script takes 5
 minutes to process 4000 records (approximately 12 records per 
 second).  Here
 are the top two entries from top:
   PID USER PRI  NI  SIZE  RSS SHARE STAT %CPU %MEM   TIME 
 CPU COMMAND
  8671 mysql 25   0 20172  19M  2000 R97.8  3.9   2:02 
   0 mysqld
  8670 appmgr15   0  2088 2088  1424 S 2.2  0.4   0:03   0 php
 
 Is there a way that I can slow down the script so mysqld does 
 not bog the
 machine down.
 
 Here is a snippet of code.  The function is called by the cronjob.
 snip
 function rank_mailbox_current($company_ID){
   $query = select app_ID from mailbox where company_ID = 
 \$company_ID\
   and deleted_date = 0 order by mailbox.ID desc;
   $result=query_database($query);
 
   while ($row = mysql_fetch_assoc($result)){
   $app_ID = $row[app_ID];
   $rank_of_app=rank_app($app_ID,$company_ID);
   }
 }
 /snip
 
 Server: P-IV 2.4GHz, 512 DDRAM, RedHat 9
 PHP Version 4.3.1
 MySQL 4.0.3
 
 Thanks,
 Ryan
 
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[PHP-DB] Re: [PHP] [PHP-DB] pg_result_error()

2003-12-12 Thread Gerard Samuel
On Friday 12 December 2003 10:24 am, Martin Marques wrote:
  That is fine and all, but my original example was just an example of the
  non functionality of pg_result_error(), not how to handle errors when a
  query fails.
  But for arguement sake, lets use your example in some dummy code[0].
  $result is still boolean false, and pg_result_error() will still return
  an empty string, and using die, would just die, with no report of what
  happened. Then whats the use of pg_result_error().

 Looks like you are totally right. Tried it and it works horrible. Any idea
 on why this is like this?


Seems like Im not the only one who thought this function is useless...
http://bugs.php.net/bug.php?id=18747
So Im going to modify my code to not use this function, as it is really a 
total waste of time at the moment.
Thanks for the chat...

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