Warning: mysql_fetch_array()

Question

Encountered with the following warning when trying to access the page:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/designand/www/www/random.php on line 13

Everything was working fine when I was testing it on XAMPP.

<?php
$db_hostname = "localhost";
$db_username = "root";
$db_name = "links";
$db_pass = "xxx";


$dbh = mysql_connect ($db_hostname, $db_username, $db_pass) or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db($db_name) or die(mysql_error());

$num_displayed = 1 ;
$result = mysql_query ("SELECT * FROM 'links' ORDER BY RAND() LIMIT $num_displayed"); 
while($row = mysql_fetch_array( $result )) 

{
    echo "<a href=\"" . $row["link"] . "\"><img src=\"" . $row["image"] . "\" border=0 alt=\"\"></a>" ;
}
mysql_close($dbh);
?>

what differences are there between your setup on XAMPP and your setup on the server? — Scott M.
– Scott M., Commented Feb 9, 2011 at 1:02
Try to echo exact query which you're trying to perform. — zerkms
– zerkms, Commented Feb 9, 2011 at 1:02
possible duplicate of Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result — dgw
– dgw, Commented Aug 8, 2012 at 15:23

Michael Madsen · Accepted Answer · 2011-02-09 01:03:13Z

3

An error like that almost always means there's a problem with your query.

To find out what error message MySQL returns, you can put or die(mysql_error()) just before the semicolon after your mysql_query call. You may also want to print the exact query text you send to MySQL, as this may help you to see the actual problem.

Given that I don't know how much you may have anonymized this example, I can't be sure if this is the actual error, but it looks like you've surrounded your table name with apostrophes ('). This is incorrect; the correct character for escaping table and column names is `.

answered Feb 9, 2011 at 1:03

Michael Madsen

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2 Comments

zerkms Over a year ago

Also it is possible that $num_displayed is not defined.

cbeezy Over a year ago

You are correct. Apostrophes (') were killing it. (`) fixed it immediately.

Mārtiņš Briedis · Accepted Answer · 2011-02-09 01:02:05Z

0

"supplied argument is not a valid MySQL result resource" means that the query didn't return a valid resource. It failed. To view the error just print out mysql_error() after mysql_query().

Could be that the table doesn't exist. Did you check it?

The second thinkg - ORDER BY RAND() is bad! You should think of different ways on how to shuffle the result. Just google it, there are a lot of other ways to do it - ORDER BY RAND() @ Google

answered Feb 9, 2011 at 1:02

Mārtiņš Briedis

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1 Comment

zerkms Over a year ago

"The second thinkg - ORDER BY RAND() is bad!" --- he is newbie. He cannot even perform simplest query, and you're trying to teach him about performant randomizing :-S "that the table" --- that "table" doesn't exist, since table cannot be specified with single quotes around.

Scoobler · Accepted Answer · 2011-02-09 01:03:09Z

0

Try changing this line:

$result = mysql_query ("SELECT * FROM 'links' ORDER BY RAND() LIMIT $num_displayed");

To:

$result = mysql_query ("SELECT * FROM 'links' ORDER BY RAND() LIMIT $num_displayed") or die (echo mysql_error());

It seems like the SQL is failing to return a valid response. Adding the above should show you the last MySQL error and help point you in the correct direction.

answered Feb 9, 2011 at 1:03

Scoobler

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Comments

Cesar · Accepted Answer · 2011-02-09 01:03:40Z

0

Please add this code to get if the mysql is throwing any errors:

  $result = mysql_query('SELECT * WHERE 1=1');
    if (!$result) {
        die('Invalid query: ' . mysql_error());
    }

while(....

answered Feb 9, 2011 at 1:03

Cesar

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