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[PoeFacedKilla]

PoeFacedKilla

PHP: Displaying From a DB

I am working on a private messaging system for my new site, but for some reason, i get an error when i try to make an inbox:
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/zelda38/public_html/mw383/Private/index.php on line 6

here's my pm SQL query:

CREATE TABLE `private_m` (
`id` INT( 11 ) NOT NULL AUTO_INCREMENT PRIMARY KEY ,
`name` TEXT NOT NULL ,
`from` TEXT NOT NULL ,
`title` TEXT NOT NULL ,
`message` TEXT NOT NULL
) ENGINE = MYISAM ;


here's the inbox script:

Code: [Select]

<? include $_SERVER['DOCUMENT_ROOT']."/includes/layout.top.php"; ?>
<ol>
<?
$query = sprintf("SELECT id, name, from, title, message FROM private_m ORDER BY `id` DESC WHERE name='$memusername'" );
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
?>
<li><a href="pmView.php?id=<? print $row['id']; ?>"><? print $row['title']; ?></a> from: <? print $row['from']; ?></li>
<?
}
?>
<? include $_SERVER['DOCUMENT_ROOT']."/includes/layout.bottom.php"; ?>


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Poe, The Independent Programmer

• 

[Benito]

Benito

Re: PHP: Displaying From a DB

You could try changing line 9:

Code: [Select]

while ($row = mysql_fetch_assoc($result)) {to

Code: [Select]

while ($row = mysql_fetch_array($result)) {
It doesn't look like your using an associative array so assoc shouldn't be used. Although I may be wrong its worth a try.

I ♥ Open Girlfriend (What!? I didn't add this.. must have been Solly!)

• 

[PoeFacedKilla]

PoeFacedKilla

Re: PHP: Displaying From a DB

i have already fixed it, it was because 'from' is a taken SQL variable.

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[Windy]

Windy

Re: PHP: Displaying From a DB

just so your aware its better to do something like

Code: [Select]

$result = mysql_query($query) or die(mysql_error());so if there's a problem with your sql it'll pick up on it and tell you what the actually problem is

[PoeFacedKilla]

PoeFacedKilla

Re: PHP: Displaying From a DB

just so your aware its better to do something like

Code: [Select]

$result = mysql_query($query) or die(mysql_error());so if there's a problem with your sql it'll pick up on it and tell you what the actually problem is

yeh i know that, i wrote that at like 5 in the morning, so i am going back through all of it and fixing the errors.

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