MySQL Into HTML Table With PHP

Question

Using the example I found here I'm using the code below to create a table with PHP. However when I run this, I receive the following error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /homepages/2/d333603417/htdocs/locationsconsole.php on line 94

Line 94 of my code is while (($row = mysql_fetch_assoc($query)) !== false) {

Could someone perhaps tell me please, have I interpreted the code wrongly, or is there an error with code. I just wondered whether someone could perhaps take a look at this please and provide a little guidance on how I can rectify this issue.

<form name="locationsconsole" id="locationsconsole" method="post" action="locationsaction.php">
<?php
$query = "SELECT  l.*, COUNT(f.locationid) totalfinds FROM detectinglocations l LEFT JOIN finds f ON f.locationid = l.locationid WHERE l.userid = '27'  GROUP BY l.locationname";
?>

<table>
<thead>
<tr>
    <th width="62"><div align="center">Location Name</div></th>
    <th width="120"><div align="left">Location Address</div></th>
    <th width="81"><div align="center">No. Of Finds Made </div></th>                            
</tr>
</thead>
<tbody>
<?php
    while (($row = mysql_fetch_assoc($query)) !== false) {
?>
<tr>
<td><?php echo $row['locationname']; ?></td>
  <td><?php echo $row['returnedaddress']; ?></td>
    <td><?php echo $row['totalfinds']; ?></td>
    </tr>
<?php
}
?>
</tbody>
</table>
</form>

What is table 1? Shouldn't you have an "detectinglocations AS 1"? — Sablefoste
– Sablefoste, Commented Jul 5, 2012 at 17:21
@SableFoste: It's actually the letter "L". As for the... AS... it's implicit :) — Ry-
– Ry- ♦, Commented Jul 5, 2012 at 17:23

Ry- · Accepted Answer · 2012-07-05 17:20:07Z

1

You need to actually perform the query before fetching the results, using mysql_query:

$sql = "SELECT  l.*, COUNT(f.locationid) totalfinds FROM detectinglocations l LEFT JOIN finds f ON f.locationid = l.locationid WHERE l.userid = '27'  GROUP BY l.locationname";
$query = mysql_query($sql);

However, please note that the mysql_ extensions are discouraged and will be removed in future versions of PHP, so I suggest you switch to PDO or MySQLi if at all possible.

answered Jul 5, 2012 at 17:20

Ry-♦

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1 Comment

IRHM Over a year ago

Hi @minitech, thank you for taking the time to reply to my post, it works great. Thank you also for the guidance re. the MySQL extensions. Kind regards

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