RE: [PHP] PHP- something tha i don't understand

2001-07-17 Thread Boget, Chris

 8:$result = mysql_db_query(users, $query); 
 9:$r=mysql_fetch_array($result); 
 10: $count=$r[count];
 And the error message is 
 Warning: Supplied argument is not a valid MySQL result 
 resource in 
 c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 9

You need to validate that the query returned a value.  Your query
failed somewhere...

Chris



Re: [PHP] php- something that i don't understand

2001-07-17 Thread Christopher Ostmo

Yassel Omar Izquierdo Souchay pressed the little lettered thingies in this order...

 Hey guys 
 here is my code
 
 4: mysql_connect() or die (Problemas conectandose a la base de datos); 5:
 $query=select * from info where FirstName='$FirstName' and 6:
 LastName='$LastName' and email='$email'; 7: $result =
 mysql_db_query(users, $query); *8: $r=mysql_fetch_array($result);
  9: $count=$r[count];
 
 and i follow redcieving this message
 
 Warning: Supplied argument is not a valid MySQL result resource in
 c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8
 
 

Change line 7 to this:
$result = mysql_db_query(users, $query) or die(mysql_error());

What is the error after you do that?

Christopher Ostmo
a.k.a. [EMAIL PROTECTED]
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RE: [PHP] php- something that i don't understand

2001-07-17 Thread Sam Masiello


Pardon my ignorance if you can really do it this way (because I have never
seen it done like this):

In line 4, I wasn't aware that you could call mysql_connect() without any
parameters.  Can you?

HTH

Sam Masiello
Software Quality Assurance Engineer
Synacor
(716) 853-1362 x289
[EMAIL PROTECTED]

 -Original Message-
From:   Yassel Omar Izquierdo Souchay [mailto:[EMAIL PROTECTED]]
Sent:   Tuesday, July 17, 2001 4:38 PM
To: [EMAIL PROTECTED]
Subject:[PHP] php- something that i don't understand

Hey guys
here is my code

4: mysql_connect() or die (Problemas conectandose a la base de datos);
5: $query=select * from info where FirstName='$FirstName' and
6: LastName='$LastName' and email='$email';
7: $result = mysql_db_query(users, $query);
*8: $r=mysql_fetch_array($result); 
9: $count=$r[count];

and i follow redcieving this message

Warning: Supplied argument is not a valid MySQL result resource in
c:\inetpub\wwwroot\PhpAndMysqlTest\test2\reg1.php3 on line 8


thanks for your answers
Yassel



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