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download8000 -
Joined: 28 Dec 2009 Posts: 1
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Posted: Mon Dec 28, 2009 2:53 am Post subject: PHP&MySql Function Problems? |
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I'm getting an erro when trying to count the number of records obtained from a mysql db. the error states:
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\Program Files\Abyss Web Server\htdocs\login.php on line 6
the code is:
$db = mysql_connect("localhost", "root");
mysql_select_db("itjobhunt", $db);
$sql = "Select * FROM agency WHERE agy_ID = '" + $agy_Account + "'";
$result = mysql_query($sql);
$results = mysql_num_rows($result);
echo "<br>$results<br>";
I have an idea that this may be to do with the setup of PHP but may be completely wrong. Any idea's?? |
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DonQuichote -
Joined: 24 Dec 2006 Posts: 68 Location: The Netherlands
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Posted: Mon Dec 28, 2009 5:44 pm Post subject: Check errors |
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You do not check for any errors in your code. So don't ask us what is wrong, ask your system. The mysql_error function has the message. You may not even get a connection (check that you get a resource). If you do get a connection, every call to a mysql_* function should be followed by a check on mysql_errno(), and if it returns anything else than zero, check the error message in mysql_error(). Furthermore, the variable $agy_Account could be empty. If this is a case of ancient code that relies on automatically created variables from post parameters or even magic quotes, update the code.
But if you have this code on a public server,
- implement error handling. Really. And that means communicating the technical exact error messages to you, not to the visitor of the web site.
- never, never use unfiltered user input in queries. Never.
- switch display of warnings and errors OFF on the live server, and ON on your development machine. |
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badai -
Joined: 24 Apr 2003 Posts: 82
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Posted: Sat Jan 02, 2010 4:22 pm Post subject: |
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long way to go...
seem like you confusing javascript with php. + for joining string in script, . (dot) for php.
the error tells u mysql_num_rows have problem executing with given parameter (the $result).
also, try not to reuse variable name. you should use something like this for number of result:
$num = mysql_num_rows($result)
that way, you can access $result later.
but i don't put too much hope u understand anything i said anyway.
good luck man. |
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Toasty -
Joined: 21 Feb 2008 Posts: 298 Location: Chicago, IL
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Posted: Mon Jan 04, 2010 9:39 pm Post subject: |
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Code is messed up. Fix the code via the errors that PHP is spitting out like crazy, and I'll help you figure out the mysql portion. I assume you're not getting a connection to the database "using password: no". _________________ Audit the secure configuration of your server headers! |
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Toasty -
Joined: 21 Feb 2008 Posts: 298 Location: Chicago, IL
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