if array is in_array() in array?

Question

First time trying to find something in an array using an array as both a needle and a haystack. So, example of 2 arrays:

My Dynamically formed array:

Array ( 
    [0] => 
    [1] => zpp 
    [2] => enroll 
)

My Static comparison array:

Array ( 
    [0] => enroll 
)

And my in_array() if statement:

if (in_array($location_split, $this->_acceptable)) {
   echo 'found';
}
$location_split; // is my dynamic
$this->_acceptable // is my static

But from this found is not printed out, as I'd expect it to be? What exactly am I failing on here?

cHao · Accepted Answer · 2012-11-19 20:51:31Z

3

If i'm understanding you right, you want to see whether the first array's entries are present in the second array.

You might look at array_intersect, which would return you an array of stuff that's present in all the arrays you pass it.

$common = array_intersect($this->_acceptable, $location_split);
if (count($common)) {
    echo 'found';
}

If the count of that array is at least 1, then there's at least one element in common. If it's equal to your dynamic array's length, and the array's values are distinct, then they're all in there.

And, of course, the array will be able to tell you which values matched.

edited Nov 19, 2012 at 20:51

answered Nov 19, 2012 at 20:46

cHao

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2 Comments

chris Over a year ago

I want to see if any of the static array values exist in the dynamic one. But not all or nothing if that makes sense.

cHao Over a year ago

@chris: Then check whether the count from the array returned by array_intersect is at least 1. Added a code sample.

Madara's Ghost · Accepted Answer · 2012-11-19 20:43:38Z

1

Because there is no element in your array containing array('enroll') in it (only 'enroll').

Your best bet would be to use array_diff(), if the result is the same as the original array, no match is found.

answered Nov 19, 2012 at 20:43

Madara's Ghost

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Comments

Teena Thomas · Accepted Answer · 2012-11-19 20:53:35Z

0

You are interchanging the parameters for in_array (needle & haystack). It needs to be,

 if(in_array($this->_acceptable, $location_split))
{
  echo 'found';
 }

Edit: Try using array_intersect.

Demo

edited Nov 19, 2012 at 20:53

answered Nov 19, 2012 at 20:45

Teena Thomas

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1 Comment

chris Over a year ago

tried it that way to, thats what I first thought myself, but its the same either direction

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