Unable to use a variable in mysql Select

Question

$select = " SELECT * FROM comments WHERE type=$row->title; ORDER BY id desc" .
         " LIMIT $low, $PerPage";
$final = mysql_query($select) or die('Error!');

$row->title; is previously created and it have value like Type1, Type2 or something else. When I start that script the result is "Error!". Could you tell me why? I have tried many ways to reslove the problem but without any result. This is one of them:

$mytype=$row->title;
$select = " SELECT * FROM comments WHERE type=$mytype; ORDER BY id desc" .
         " LIMIT $low, $PerPage";
$final = mysql_query($select) or die('Error!');

Chandu · Accepted Answer · 2010-12-28 21:23:19Z

3

Remove the ; after the $row-title i.e.:

$select = " SELECT * FROM comments WHERE type='$row->title'  ORDER BY id desc" .          " LIMIT $low, $PerPage";

answered Dec 28, 2010 at 21:23

Chandu

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1 Comment

Chandu Over a year ago

Added quotes. Thx dqhendricks almost forgot :)

TehShrike · Accepted Answer · 2010-12-28 22:27:40Z

0

You should echo out mysql_error() - MySQL will tell you what went wrong!

The answers by dqhendricks and Cybernate are correct - you should enclose your strings in single quotes.

However - you should also be escaping your text, or you will get more errors eventually:

$select = "SELECT * FROM comments WHERE type='" . mysql_real_escape_string($row->title) . "' ORDER BY id desc LIMIT $low, $PerPage";

answered Dec 28, 2010 at 22:27

TehShrike

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