Mysql_fetch_array supplied argument is not a valid MYSQL result

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:

<开发者_StackOverflow社区;?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'];        
?>

  $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.

$runrows = mysql_fetch_array('$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

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());

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.