!= operator question

the code works if I remove the or portion but I need to check to make sure its not either 9 or 10.  I am not sure what I am doing wrong
if(($row['cnid'] != '9') || ($row['cnid'] != '10')) {

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iceman19330Asked:
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zappafan2k2Connect With a Mentor Commented:
Another way to write it would be
if (!($row['cnid'] == '9' || $row['cnid'] == 10)) {

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if that makes more sense.

It's all about the logic.
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zappafan2k2Commented:
Try && instead of ||
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Lukasz ChmielewskiCommented:
So to be complete
if(($row['cnid'] != '9') && ($row['cnid'] != '10'))

if the cnid is not 9 AND is not 10 - do something
Mind that you're comparing the string value '9' and '10'
To check if the $row['cnid'] is not the value and the type integer you would use !==
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Lukasz ChmielewskiCommented:
the logic is the same in result :P
if NOT (9 OR 10) equals if NOT 9 OR NOT 10
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Ray PaseurCommented:
The "or" condition is evaluated from left to right in this sequence.  So if you give it 10, the if() statement will be satisfied that 10 is not 9, and no further tests will be done.  That is why you want the && instead of the || here.
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iceman19330Author Commented:
thank you
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