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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result reso

achilles1971

By achilles1971
in PHP Coding Help

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achilles1971 Newbie

achilles1971

Posted
Posted

In the following code, at the mysql query immediately following the <!--end accordianButton div-->,

 

Some of the rows will echo out the content, while some of the rows will throw the Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in... error.  Any idea why?

 

Go here to see what I mean:  http://www.chalmerscommunitychurch.com/P2P_archives.php

 

 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

<title>Pastor to People Blog Archives</title>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"> </script>

<script type="text/javascript" src="js/javascript.js"> </script>

<style>

.accordionButton {

width: 100%;

height:30px;

float: left;

background: url(../images/button.png);

border-bottom: 1px solid #FFFFFF;

cursor: pointer;

}

 

.accordionContent {

width: 100%;

float: left;

display: none;

 

}

</style>

</head>

 

<body>

<?php

require("include.php");

$con = mysql_connect("$db_host","$db_username","$db_pass");

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }

 

mysql_select_db("chalmers_db", $con);

 

 

         

            $result = mysql_query("SELECT title, date FROM blog");

            while($row = mysql_fetch_array($result))

          {

  $title=$row['title'];

  $sqldate=$row['date'];

  $date=date('m-d-Y',strtotime($sqldate));

            ?>

            <!--start accordionButton div-->

            <div class="accordionButton"><?php echo $title;?>, <?php echo $date;?>

            </div>

            <!--end accordianButton div-->

           

               

                    <?php

 

$query = mysql_query("SELECT content FROM blog WHERE title = '".$title."' ORDER BY date DESC");

while ($row = mysql_fetch_array($query)){

$content = $row['content'];

 

?>

<!--start accordionContent div-->

<div class="accordionContent" align="justify">

<?php echo $content;?>

 

</div>

<!--end accordionContent div-->

<?php

}

?>                              

           

               

              <?php } ?>

      <p><p><p><p><a href="http://www.chalmerscommunitychurch.com"><h3>Back to Chalmers Community Church</h3></a>

      </body>

</html>

joel24 Member

joel24

Posted
Posted

if it's working only sometimes, then there must be no blog entries for some... check to see if results returned with mysql_num_rows()

 

                        $query = mysql_query("SELECT content FROM blog WHERE title = '".$title."' ORDER BY date DESC");
if (mysql_num_rows($query) {
#now do the fetch
                        while ($row = mysql_fetch_array($query)){
#etc etc
}

trq Prolific Member

trq

Posted
Posted

Before checking anything with [mysql_num_rows[/m] you should check your query actually succeeded. The general syntax is:

 

if ($result = mysql_query($sql)) {
  if (mysql_num_rows($result)) {
    // $result contains data, use it
  } else {
    // no records found
  }
} else {
  // query failed
}

Psycho Prolific Member

Psycho

Posted
Posted

if it's working only sometimes, then there must be no blog entries for some... check to see if results returned with mysql_num_rows()

 

Although thorpe didn't state it explicitly, he was trying to allude to the fact that the reason for the error is that the query is failing - not that there were no results. Simply doing a mysql_num_rows() on a failed query will generate the same type of error. That error would not occur if the query succeeded with 0 results.

PFMaBiSmAd Advanced Member

PFMaBiSmAd

Posted
Posted

I'm going to guess that some of the titles in $title probably contain some sql special characters that need to be escaped before being put into the inner query.

 

Faux Edit: There's actually no need for two queries. You can use one to accomplish the same thing. Just detect when the title changes to close out a previous section and output a new heading.

 

Real Edit2: Visiting the link in the 1st post in the thread confirms that there are ' in the titles that fail.

The Little Guy Prolific Member

The Little Guy

Posted
Posted

I see your inserting into the database without escaping your strings. I have a feeling the rows that are failing have a quote in them and the ones that work don't have a quote in them.

 

This could be the problem...

PFMaBiSmAd Advanced Member

PFMaBiSmAd

Posted
Posted

Your php code using one query -

 

<?php
require("include.php");
$con = mysql_connect("$db_host","$db_username","$db_pass");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("chalmers_db", $con);

$heading = null; // remember the last heading
$result = mysql_query("SELECT title, date, content FROM blog ORDER BY date DESC");
while($row = mysql_fetch_array($result)){
if($heading != $row['title']){
	// a new heading detected, start a new section here...
	$heading = $row['title']; // save the new title
	$date=date('m-d-Y',strtotime($row['date']));
	?>
	<!--start accordionButton div-->
	<div class="accordionButton"><?php echo "{$row['title']}, $date";?>
	</div>
	<!--end accordianButton div-->
	<?php
}
// handle each piece of data here...
?>
<!--start accordionContent div-->
<div class="accordionContent" align="justify"><?php echo $row['content'];?>
</div>
<!--end accordionContent div-->	
<?php
}
?>

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