Mysql_fetch_array supplied argument is not a valid MYSQL result

Question

How do I fix the "Mysql_fetch_array supplied argument is not a valid MYSQL result" error?

This is the code that had the error:

<?php
    require "connect.php";
    $query = mysql_query("SELECT author, date 
                            FROM articles 
                           WHERE id = ' . $id . '") or die(mysql_error());
    $runrows = mysql_fetch_array('$query');

     $author = $runrows['author'];
     $date = $runrows['date'];        
?>

Looks like an SQL injection point :) (Just a guess and totally off topic, just want to warn you :) ) — PeeHaa
– PeeHaa, Commented Jun 28, 2011 at 16:16
You may want to check about SQL Injection and Security measures to be followed in PHP — Balanivash
– Balanivash, Commented Jun 28, 2011 at 16:19
Short answer: WHERE id = " .((int)$id) and also see my presentation slideshare.net/billkarwin/sql-injection-myths-and-fallacies — Bill Karwin
– Bill Karwin, Commented Jun 28, 2011 at 16:21
possible duplicate of mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in php — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Sep 16, 2011 at 17:52

Balanivash · Accepted Answer · 2011-06-28 16:17:50Z

3

  $query = mysql_query("SELECT author, date FROM articles WHERE id = ' . $id . '") or die(mysql_error());
  $runrows = mysql_fetch_array($query);

There should be no quotes for $query. Its a variable, not a string.

answered Jun 28, 2011 at 16:17

Balanivash

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1 Comment

Bill Karwin Over a year ago

+1 Right, in fact $query is a resource, therefore it makes no sense to make it part of an expression or interpolation.

PeeHaa · Accepted Answer · 2011-06-28 16:22:33Z

1

$runrows = mysql_fetch_array('$query');

Should be

$runrows = mysql_fetch_array($query);

or

$runrows = mysql_fetch_array("$query");

Using '$variable' just prints out $variable as text. You want to use the value of the variable.

SQL Injection is when you use that $id variable from user interaction.

E.g. you have the following URL:

http://example.com/articles.php?id=1

If you just add the id variable to your query people can inject code in the variable.

http://en.wikipedia.org/wiki/SQL_injection

To prevent this you can simply do: $id = mysql_real_escape_string($id);

Or use prepared statements if it is supported.

http://en.wikipedia.org/wiki/Prepared_statements#Parameterized_statements

edited Jun 28, 2011 at 16:22

answered Jun 28, 2011 at 16:17

PeeHaa

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Comments

afaolek · Accepted Answer · 2011-06-28 16:18:52Z

0

Remove the dots and use something like this in the third line

$query = mysql_query("SELECT author, date FROM articles WHERE id = '$id'") or die(mysql_error());

answered Jun 28, 2011 at 16:18

afaolek

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Comments

Rahul · Accepted Answer · 2012-05-10 19:59:50Z

0

This has happened to me when I had not closed the connection using mysql_close($conn). I guess the socket between from the db was still open, and next time when i again connected to the same db, this particular error came. There were no other issues. Once I explicitly closed the connection and then again restarted the connection, this issue was resolved. Hope it helps.

answered May 10, 2012 at 19:59

Rahul

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