Is it right to write this ?
if($site!=1 || $site!=4)
I want to execute a code if the site ID isn't 1 or 4. Two negations are possible with a OR? Because it doesn't work.
Thanks!
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2 Answers
When you using || if any of them satisfies it will be executed. When you dont want the both should be satisfied the you need to use &&. Should be -
if($site!=1 && $site!=4)
It will check the both conditions, i.e. $site is not equals to 1 or 4.
OR you can use in_array -
if(!in_array($site, array(1, 4)))
Currently it will check $site is 1 or not first, if it is not it will go inside the condition and ignore the rest of the conditions. So for 4 it will not be checked.
1 Comment
Ty Yt
Ok thanks, but in my head it was in two different cases : site1 or site4. And with && the two conditions must be executed. But it works thanks for your explanation.
Although you already have a correct answer, you can also use
if (!in_array($site,array(1,4)))
This simplifies it if there are a lot of numbers to check against. This is even simpler in PHP 5.4+
if (!in_array($site,[1,4]))
1 Comment
Sougata Bose
It is also a good trick. Help a lot if many values needed to be checked. :)