Help with syntax please..

[wudz]

Hi,

Thought I had a simple logic problem, but uncertain of my syntax/ logic.

The problem is that 'local_area 6' is always selected even though the test query town displayed is in 'local_area 3' which is correct.

Is '||' equal to OR which is what I require. Is there a shorter way of achiving the same result when the code is corrected...


Newbie John


<code>

$area = mysql_result ( $result, 0, "town" );
$TPL_area = $area;

if(($area ='Colne')||($area ='Nelson') || ($area ='Barrowford') || ($area ='Brierfield')) {
$local_area = 1;
}
if (($area ='Burnley') || ($area ='Padiham')||($area ='Fence')) {
$local_area = 2;
}
if (($area ='Hapton')|| ($area ='Accrington')|| ($area ='Oswaldtwistle')|| ($area ='Church')|| ($area ='Baxenden')) {
$local_area = 3;
}


if (($area ='Great Harwood')|| ($area ='Rishton')) {
$local_area = 4;

}
if (($area ='Blackburn')|| ($area ='Mellor')|| ($area ='Darwen')) {
$local_area = 5;

}
if (($area ='Chorley')|| ($area ='Bamber Bridge')|| ($area ='Leyland')) {
$local_area = 6;
}
$TPL_locationnum = $local_area;

</code>

 

[sleipnir214]

You're using the wrong operator.

In PHP, "=" is used exclusively for assignments. If you want to compare, you must use "==".



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[wudz]

Hi,
Just to clarify my last post.

No matter what town is held in variable $area it always displays $local_area as being 6....

Cheers
Newbie John

 

[wudz]

Cheers Sleipnir....crossed posts....I live and learn..Thanks again for all your help.

Cheers John