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PHP IF STATEMENT NOT WORKING

Posted on 2007-11-30
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330 Views
Last Modified: 2010-04-21
HI

I think i must be going mad but i cant see why the below code wont work!!!

ANy suggestions

--s--
<?php
 
// THIS WONT WORK
if($type != '' || $type != 'a')
   $search_url .= ',type%3D'.$type;
 
// THIS WONT WORK
if(!empty($type) || $type != 'a')
   $search_url .= ',type%3D'.$type;
 
// THIS WORK AROUND DOES
if($type == '' || $type == 'a'){}else $search_url .= ',type%3D'.$type;

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Question by:socross
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16 Comments
 
LVL 17

Expert Comment

by:nplib
ID: 20383927
if(($type != '') || ($type != 'a')) {
   $search_url .= ',type%3D'.$type;
}
0
 
LVL 1

Author Comment

by:socross
ID: 20384072
Already Tried that and its not working, this is driving me crazy!!

--s--

0
 
LVL 17

Expert Comment

by:nplib
ID: 20384093
where are you getting the value for $type,

I don't see it declared anywhere.
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LVL 20

Expert Comment

by:steelseth12
ID: 20384168
// THIS WONT WORK
if($type != '' && $type != 'a')
   $search_url .= ',type%3D'.$type;
 
// THIS WONT WORK
if(!empty($type) && $type != 'a')
   $search_url .= ',type%3D'.$type;
0
 
LVL 1

Author Comment

by:socross
ID: 20384188
nplib

$type is a posted varible.

steelseth12

Ok so if those wont work, what will and why dont they work??

--s--
0
 
LVL 20

Expert Comment

by:steelseth12
ID: 20384204
&& (AND) will work the ones i show above .... you had || (OR)
0
 
LVL 1

Author Comment

by:socross
ID: 20384277
As the value for type comes from a drop down. it will never be && i need it to work with ||

Why would it not?

--s--
0
 
LVL 20

Expert Comment

by:steelseth12
ID: 20384306
Its a double negative you have to use && ... it just will not work will || ... why do you have to use || ??????
0
 
LVL 1

Author Comment

by:socross
ID: 20384323
well i need to say

if (type does not equal '' or type does not equal 'a') add value to the varible
0
 
LVL 20

Expert Comment

by:steelseth12
ID: 20384359
if thats what you need to say then your if statement is returning the correct result.

If type does not equal to '' OR type not equal to a then it means that in the first part it can be any value except '' (INCLUDING a) the second part any value except a (including '') .... so if its either of the 2 then you are getting the correct result.
0
 
LVL 1

Author Comment

by:socross
ID: 20384433
Right ok, so under what circustances would it run the script contained in the if statment.

I only need it to run the script if type does not equal empty or not equal a

Could you write the script that will work for this?

--s--
0
 
LVL 20

Accepted Solution

by:
steelseth12 earned 125 total points
ID: 20384545
You need to use && to get the results you want .

if($type != '' && $type != 'a')
   $search_url .= ',type%3D'.$type;
0
 
LVL 17

Expert Comment

by:nplib
ID: 20384680
Man I go have lunch, and the problem gets solved.
Oh Well.
0
 
LVL 1

Author Comment

by:socross
ID: 20385160
just driven home and it all seemed to make sense, its amazing when you look at something so simple you can miss the really obvious stuff!!!

Thanks for persevering with me.

Thanks

--s--
0
 
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Author Closing Comment

by:socross
ID: 31411981
Mind caught up in the end
0
 
LVL 20

Expert Comment

by:steelseth12
ID: 20385180
i once try to explain it to a friend of mine that was been programming for over 10 years. After 3 hours 2 packs of cigarettes and waking up the wife we agreed not to bring it up again. :)
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