PHP Error Message

jreynolds22 · **PHP Error Message -** 07-09-2012, 11:54 PM

I'm creating a web page using php, but I continue to get the following error message: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource on line 37 when trying to open the page. Below is the code that I have. Any help is greatly appericated.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>

<head>
<title>Weekday PM Pullouts</title>
<link rel="stylesheet" type="text/css" href="../styles.css">
<link rel="stylesheet" type="text/css" href="../print.css" media="print">
<script script type="text/javascript" src="../functions.js"></script>
</head>

<body>

<h1 class="hidden">WEEKDAY PM PULL-OUTS</h1>

<div id="noprint" class="header" align="center">
<ul>
<li><a href="#" onclick="show_alert()">All</a></li>
<li><a href="#" onclick="show_alert()">Channel 1</a></li>
<li><a href="#" onclick="show_alert()">Channel 4</a></li>
<li><a href="javascript:window.print()">Print</a></li>
</ul>
</div>

<?php
$con = mysql_connect($localhost,$user,$password);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("pullouts", $con);

$result = mysql_query ("SELECT * FROM pullouts WHERE pullouts.day='Weekday' AND pullouts.time > '12:00:00' AND pullouts.effective='2012/03/12' ORDER BY pullouts.time");

while($row = mysql_fetch_array($result))
{
echo $row['check'] . " " . $row['time'] . " - " . $row['details'] . " " . $row['day'] . " " . $row['channel'] . " " . $row['yard'] . " " . $row['effective'];
echo "<br />";
}
mysql_close($con);
?>

</body>
</html>

grace1004 · 07-10-2012, 05:29 AM

Is your database set up in 000webhost.com server? If so, db_name should be something like:
axxxxxxx_dbname, not "pullouts" as shown below.

mysql_select_db("pullouts", $con);

Doug Lochert · 07-10-2012, 04:23 PM

Quote:

$result = mysql_query ("SELECT * FROM pullouts WHERE pullouts.day='Weekday' AND pullouts.time > '12:00:00' AND pullouts.effective='2012/03/12' ORDER BY pullouts.time");

You have already referenced the db table pullouts in the SQL call, there is no need to continue referencing the pullouts table again for actual fields within it. [WHERE pullouts.day='Weekday'] should be [WHERE day='Weekday']

jreynolds22 · 07-11-2012, 01:07 AM

Quote:

Originally Posted by grace1004

Is your database set up in 000webhost.com server? If so, db_name should be something like:
axxxxxxx_dbname, not "pullouts" as shown below.

mysql_select_db("pullouts", $con);

Thanks, I tried that, buy still no luck.

jreynolds22 · 07-11-2012, 01:08 AM

Quote:

Originally Posted by Doug Lochert

You have already referenced the db table pullouts in the SQL call, there is no need to continue referencing the pullouts table again for actual fields within it. [WHERE pullouts.day='Weekday'] should be [WHERE day='Weekday']

Ok, trying to move from access to php and learn all the new tricks. Thanks for the knowledge.

Doug Lochert · 04:12 AM

Quote:

<?php
$con = mysql_connect($localhost, $user, $password);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("pullouts", $con);
?>

It's always a good idea to check DB calls for errors, similar to the error check in the mysql_connect() you used above... but there is an easier (and tidier) way.

Secondly, it appears that you are using a config file to store the DB login details. It is also a good idea to store the DB name in same way, providing a 2nd barrier of security if users access the source code. See below:

Code:

$con = @mysql_connect($localhost, $user, $password) or die (mysql_error());
$db = @mysql_select_db($dbname, $con) or die (mysql_error());

*define $dbname in same config file as $user,$password etc.

Note: Prepending calls with "@" suppresses the error message (best used when site goes live so users don't see the errors/urls etc) but is also ignored when mysql_error is appended to the same call (during development).