Notice: Undefined variable: VARIABLE ??

[yami007]

I just moved my code from Appserv to EasyPHP and it gave me this error, it was working fine on Appserv...what's with easyPHP ??

[floridaflatlander]

Where is the variable used?

 

On submits I use

 

if (issest($_POST(['stuff']))

 

or you can go $var = FALSE; or $var1 =$var2 = FALSE;

 

or = '';

 

PS You're getting it because you have more sensitive error reporting now.

[yami007]

it's in a WHILE

while( $data = mysql_fetch_array($result) )

 

i called it => $data['id']...and it won't work, it just keeps bringing that error.

 

Will this $var = false work ?

[Pikachu2000]

What is the actual error message, and the relevant code?

[yami007]

it's as simple as this:

<div id="cours">
<div class="title"><h3>Semestre 1</h3></div>
<div id="Sem1">
	<?php
		$query = "SELECT * FROM ce_cours ";//ORDER BY nom";
		$result = mysql_query($query) or die(mysql_error());
		$row = 0;
	?>
	<table border="0" width="95%" height="40%" cellpadding="0" cellspacing="0">
		<tr>
<?
while( $cours = mysql_fetch_array($result) ){
$row++;
?>
			<td valign="top">
				<h4><?php echo $cours['nom']; ?></h4>
				<?php 

					$query2 = "SELECT * FROM ce_cour_chap WHERE cour_id=".$cours['cid'];
					$result2 = mysql_query($query2);

					while( $chapitre = mysql_fetch_array($result2) ){
					?>
						<a href="<?php echo $chapitre['chid']; ?>"><?php echo $chapitre['nom']; ?></a><br />
					<?
					}
					?>
			</td>
<?php
				if ( $row%2==o ) {
					echo '</tr><tr>';
				}
}
?>
		</tr>
	</table>
</div>
</div>

 

I just cant tell what's wrong here !

[Pikachu2000]

And what is the actual error message?

[ManiacDan]

The word "variable" doesn't exist in your code. 

 

The only thing that can possibly help us (and you) is the actual error message.  Not a summary, not a paraphrase, the error itself. 

[Drummin]

Where is $cours['nom'] defined?

[mikosiko]

<?
while( $cours = mysql_fetch_array($result) ){
$row++;
?>

 

replace the short tags for the long ones

 

<php?
while( $cours = mysql_fetch_array($result) ){
$row++;
?>

[Pikachu2000]

One little thing I noticed is here:

if ( $row%2==o ) { 

 

The result of $row modulus 2 will never equal the letter o.

[yami007]

hey thanks Guys...the error was in <? as Mikosiko said...and the Pikachu2000 mentioned too...for the latter i think i mistook 0 with o...I sort of do that everytime