RE: [PHP-DB] mysql_query troubles

[Karl Phillipson]

Brad,

Have you defined the database variable you are connecting to?

ie: $DB="your_database_name"; and then connecting:

if(!($testResult = mysql_query($DB, "SELECT * FROM login_table where Pass =
password('$password')")))

try the code snippet below and see if it spits an error at you -  you will
need to define the $DB variable.


<?
//error reporting function
function DisplayErrMsg ( $message )
{
sprintf("<BLOCKQUOTE><BLOCKQUOTE><BLOCKQUOTE><H3>%s</H3></BLOCKQUOTE></BLOCK
QUOTE></BLOCKQUOTE>\n", $message);
}

//begin query 

if(!($testResult = mysql_query($DB, "SELECT * FROM login_table where Pass =
password('$password')")))
                        //if we get an error print it
                        {
                        DisplayErrMsg(sprintf("internal error %d:%s\n",
                        mysql_errno(), mysql_error()));
                        exit();
                        }

$num_rows = mysql_num_rows($testResult);
print "$num_rows";
exit();
mysql_free_result($testResult);

?>







-----Original Message-----
From: Brad Wright [mailto:[EMAIL PROTECTED]]
Sent: 17 July 2001 06:31
To: [EMAIL PROTECTED]
Subject: [PHP-DB] mysql_query troubles


Hi all',
I have a mysql_query that reads:

 $testResult = mysql_query("SELECT * FROM login_table where Pass =
password('$password')") or die ("ouch");

$num_rows = mysql_num_rows($testResult);


the problem is that I keep getting a value of 0 for $num_rows when I know
the table has 1 entry for PAss which equals password('$password'). The
password I have used is just 'a' (before pasing throught the password
function in the insert).


I DO have a table named login_table
I KNOW that there is an entry that matches 'password('$password)'.. ( i
checked using phpMyAdmin)


can anyone help??????


Thanks in advance,
brad


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