Member Avatar for twelvetwelve

Hi I’m new to PHP and MySQL and I’m trying to return a MySQL query using a PHP variable that has been passed to it. I currently get this error:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\moodle\logicquiz\quiz.php on line 57

Error performing query: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''venn'' at line 1

Where "venn is the table name"

The code works when I put a table name in instead of $quiz_name, so I'm guessing there is a problem with my syntax as the error message suggests, any suggestions as to what I'm doing wrong would be gratefully appreciated.

Many Thanks
NH

//gets quiz name from URL passed in code in previous page $quiz_name = $_GET['quiz']; //counts number of questions in a quiz table $result = mysql_query("SELECT * FROM '$quiz_name'"); $num_rows = mysql_num_rows($result12); if (!$result) { exit ('<p> Error performing query: ' . mysql_error() . '<p/>'); }
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  • Latest Post Latest Post by nav33n
Member Avatar for nav33n

$result = mysql_query("SELECT * FROM '$quiz_name'");
$num_rows = mysql_num_rows($result12);

$result12 should be $result.

Member Avatar for twelvetwelve

sorry I posted my code wrong

Even when I change it it still gets the same error, my actual code is

//gets quiz name from URL passed in code in previous page $quiz_name = $_GET['quiz']; //counts number of questions in a quiz table $result2 = mysql_query("SELECT * FROM '$quiz_name'"); $num_rows = mysql_num_rows($result2); if (!$result2) { exit ('<p> Error performaing query: ' . mysql_error() . '<p/>'); }

NH

Member Avatar for nav33n

Do you have the database connection in your script ? And Check if $quiz_name isn't empty.

Member Avatar for twelvetwelve

At the top of my code above all this is a database connection. When I run the code it will echo the variable $quiz_name with the correct value. so I would guess that means it's not empty. Also whenn I replace $quiz_name with venn (the table name) it runs fine, so I assume it's connecting to the database fine.

Thanks for your help on this.

//gets quiz name from URL passed in code in previous page $quiz_name = $_GET['quiz']; //counts number of questions in a quiz table $result2 = mysql_query("SELECT * FROM '$quiz_name'"); $num_rows = mysql_num_rows($result2); if (!$result2) { exit ('<p> Error performing query: ' . mysql_error() . '<p/>'); echo '<table border="1" align="center">'; echo '<tr>'; echo '<td>'; echo '<h1>Quiz Name:</h1>'; echo '</td>'; echo '<td>'; echo '<h2>' .$quiz_name. '</h2>'; echo '</td>'; echo '</tr>'; echo '<tr>'; echo '<td>'; echo '<h1>Number of Questions:</h1>'; echo '</td>'; echo '<td>'; echo '<h2>' .$num_rows. '</h2>'; echo '</td>'; echo '</tr>'; echo '<tr>'; echo '<td>'; echo '</td>'; echo '<form action="quiz.php'.$quiz_name.'" method="get">'; echo '<td>'; echo '<input type="submit" value="Start Quiz"'; echo '</td>'; echo '</form>'; echo '</tr>'; echo '</table>'; ?>
Member Avatar for nav33n

Try this.

mysql_query("SELECT * FROM $quiz_name");

If this doesn't work, print out the query and run it in phpmyadmin/mysql console. See what the problem is.

Member Avatar for twelvetwelve

Thank you very much for your help. That solved it!

This is the code that works

$result2 = mysql_query("SELECT * FROM $quiz_name"); $num_rows = @mysql_num_rows($result2); if (!$result2) { exit ('<p> Error performing query: ' . mysql_error() . '<p/>');}

Many thanks once again
:)

Member Avatar for nav33n

You are welcome! Btw, php parser doesn't parse a variable if you put it within ' '. More about it here .

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