Hi!
Why dont u try this,
if(mysql_num_rows($var1)>0)
{
mysql_free_result($var1);
delete;
}
else
{
echo("record doesnot exists");
}
cheers,
Jayasimhan A
----- Original Message -----
From: Palafox, Alfredo <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, January 23, 2002 11:35 PM
Subject: fecth_array error
> Hi,
>
> I'm new to mysql database design, I have a PHP script to DELETE records
but
> before that I need to be sure that the record exists.
>
> My script goes like this:
>
> line 6 $var1 = " select * from table.field where id_record = '$record' ";
> line 7 if( $var2 = mysql_fetch_array($var1) )
> line 8 {
> line 9 deleting ...
> line 10 }
> line 11 else
> line 12 {
> line 13 record does not exit ...
> line 14 "It is not a valid record"
> line 15 }
>
> So when I test it with a record I know it doesn't exist, the script gives
me
> the message ("It is not a valid record") and this : Warning: Supplied
> argument is not a valid MySQL result resource in
> c:\apache\htdocs\nuevo\icum\eliminar.php on line 7
>
> Any help will be appreciated
>
> Alfredo Palafox
>
>
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