PHP Check if variable is contained in an array

Question

I have a numeric variable $i. This is a while loop, and $i increments after each iteration.

How can a conditional statement be made as to not be necessary to write such a long statement if($i == 1 || $i == 2 || $i == 25 [...])?

Thanks in advance!

Are the values arbitrary, or is their inclusion governed by a mathematical equation? — George Cummins
– George Cummins, Commented Jul 9, 2013 at 3:35
Adeneo's answer included very similar logic to Sumoanand's: both said to use in_array(). Sumoanand posted his first. — George Cummins
– George Cummins, Commented Jul 9, 2013 at 3:41

Sumoanand · Accepted Answer · 2013-07-09 03:35:21Z

3

Use php in_array.

$os = array("Mac", "NT", "Irix", "Linux");
if (in_array("Irix", $os)) {
    echo "Got Irix";
}

answered Jul 9, 2013 at 3:35

Sumoanand

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Sébastien VINCENT · Accepted Answer · 2013-07-09 03:47:45Z

0

Seems a bit obvious that you should take a deeper look at the documentation of PHP. There is standards functions that can do a lot and help you save a lot of time.

You say that you've got a string but show an example with an array. Assuming you really have a string in input you may try explode() function to explode it to an array of objects.

Then having a real Array in_array() function will do the job.

$resultString = "12 13 17 26";
$resultArray = explode(" ", $resultString);
if (in_array("13", $resultArray)) {
    echo "found";
} else {
    echo "not found";
}

answered Jul 9, 2013 at 3:47

Sébastien VINCENT

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