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php-db | 2001072

RE: [PHP-DB] mysql_query troubles From: Karl Phillipson (Karl <email protected>)
Date: 07/17/01

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:bradwright <email protected>]
Sent: 17 July 2001 06:31
To: php-db <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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