Code check - database query

martinb7 · Oct 26, 2003

Hi, can someone check this code for me, whenever I try to use it it comes up with this errors:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\inetpub\

http://wwwroot\gr2004\webrevolt\inbox.php

on line 18

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\inetpub\

http://wwwroot\gr2004\webrevolt\inbox.php

on line 41

::Code::

Code:

<?php 
// start the session 
session_start(); 
header(&quot;Cache-control: private&quot;); //IE 6 Fix 
?>

<html>
<head>
<title>Untitled Document</title>
<meta http-equiv=&quot;Content-Type&quot; content=&quot;text/html; charset=iso-8859-1&quot;>
</head>
<?php include (&quot;config1.php&quot;);

$conn = mysql_connect($host, $user, $pass); // DO NOT CHANGE THIS
mysql_select_db($db, $conn); // DO NOT CHANGE THIS
$query = mysql_query(&quot;SELECT * FROM messages WHERE to = '&quot;.$_SESSION['name'].&quot;'&quot;);

while ($r = mysql_fetch_array($query)){
	$new = $r[&quot;new&quot;];
	
	if($new > 0)
	{
		echo(&quot;<bgsound src=youhavemail.mp3>&quot;);
	}
	if($new == 0)
	{
		echo(&quot;<bgsound src=>&quot;);
	}
}
?>
<body>
<?php 
$user = $_SESSION['name'];

include (&quot;config1.php&quot;);

$conn = mysql_connect($host, $user, $pass); // DO NOT CHANGE THIS
mysql_select_db($db, $conn); // DO NOT CHANGE THIS
$query2 = mysql_query(&quot;SELECT * FROM messages WHERE to = 'MJB3000'&quot;);

while ($r = mysql_fetch_array($query2)){
	$title = $r[&quot;title&quot;];
	$new = $r[&quot;new&quot;];
	$nm += $new;
	
	echo(&quot;<a href=inbox.php>You have $nm message(s) in you Inbox</a>&quot;);
	echo(&quot;$title&quot;); 
}
	?>
<a href=&quot;login.php&quot;>Login </a>
</body>
</html>

any ideas??

Thanx

Martin

sleipnir214 · Oct 26, 2003

You're not trapping enough errors to be able to tell what's going wrong. The best I can tell you is that MySQL is barfing on your query.

See section 1.5 of faq434-2999

Want the best answers? Ask the best questions:

http://www.catb.org/~esr/faqs/smart-questions.html

TANSTAAFL!!

cs7536 · Oct 26, 2003

Try this, I used it on my machine.. I quickly created a table to match what i thought would correspond and I got the script to work by re-writing little changes.. hope it works for you.. test it without thehtml first then add the html.

<?php
// start the session
session_start();
header("Cache-control: private&quot

; //IE 6 Fix
?>

<?php include ("config1.php&quot

;

$conn = mysql_connect("$host", "$user", "$pass&quot

; // DO NOT CHANGE THIS
mysql_select_db("$db", $conn); // DO NOT CHANGE THIS
$query = mysql_query("SELECT * FROM messages WHERE to LIKE '".$_SESSION['name']."'&quot

;

while ($r = mysql_fetch_array($query)){
$new = $r['new'];

if($new > 0)
{
echo "<bgsound src=youhavemail.mp3>";
}
if($new == 0)
{
echo "<bgsound src=>";
}
}
?>

<?php
$user = $_SESSION['name'];

include ("config1.php&quot

;

$conn = mysql_connect("$host", "$user", "$pass&quot

; // DO NOT CHANGE THIS
mysql_select_db("$db", $conn); // DO NOT CHANGE THIS
$query2 = mysql_query("SELECT * FROM messages WHERE to LIKE 'MJB3000'&quot

;

while ($r = mysql_fetch_array($query2)){
$title = $r['title'];
$new = $r['new'];
$nm += $new;

echo "<a href=inbox.php>You have $nm message(s) in you Inbox</a>";
echo "$title";
}
?>
<a href="login.php">Login</a>

Nothing is hard when you realy want to learn it.

Max

sleipnir214 · Oct 26, 2003

cs7536:
The only problem with the posted script is that it, too, performs no error-catching. It will provide no more information about errors than the code martinb7 posted.

Want the best answers? Ask the best questions:

http://www.catb.org/~esr/faqs/smart-questions.html

TANSTAAFL!!

DRJ478 · Oct 27, 2003

Sleipnir214 is absolutely right.
In general, do error checking on these MySQL statements:

1. mysql_connect -> the server could be down, the password wrong etc.
2. mysql_select_db -> wrong db name or server down
3. mysql_query -> statement error by malformed SQL, server down

For #3 remember that a SQL query may execute correctly but not return any rows. If you e.g. search for something and nothing is there, it will return true, since the SQL command was successfully executed. You need to check for the presence of rows a different way.

flugh · Oct 28, 2003

I'm no php-wiz, so probably could be done more efficiently, but the number of rows thing could go like this:

Code:

$query2 = mysql_query(&quot;SELECT * FROM messages WHERE to LIKE 'MJB3000'&quot;);

Code:

if mysql_numrows($query2) > 0 {

Code:

    while ($r = mysql_fetch_array($query2)){
        $title = $r['title'];

...snip...

Also, to get good error info, add:

Code:

or die(&quot;Goodbye World: &quot; . mysql_error());

so you get this:

Code:

$conn = mysql_connect(&quot;$host&quot;, &quot;$user&quot;, &quot;$pass&quot;) or die(&quot;Goodbye World: &quot; . mysql_error());

Sorry if this is useless. Cold medicine has me slammed

----
JBR

ingresman · Oct 30, 2003

As a quick and dirty diagnostic, can you put the SQL into a variable i.e. $SQL = "select..." and use that in the mysql command rather than the hard coded value. Print the $SQL value before you call mysql command. Also print the the errorcode and errormessage (mysql_error and mysql_errno).

Hopefuly we can get to the botton of the problem.

BTW using like does mysql not want a % somewhere ??

kr

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Code check - database query

martinb7

Programmer

sleipnir214

Programmer

cs7536

Programmer

sleipnir214

Programmer

DRJ478

IS-IT--Management

flugh

Technical User

ingresman

IS-IT--Management

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