How to compare key value in one array with key value in another array?

Question

Within a foreach iteration in PHP, I wish to check if the value of a key in one array is equal to the value of a different key in another array.

$array1 = array(word1=>"hello",word2=>"world");
$array2 = array(word1=>"hello",word2=>"peter");

I.e. Foreach $array1 as $element if $element [word1=>value] == $array2[word2=>value]

in_array doesn't seem to be specific enough, since it only looks for the value anywhere in the array. I want just to check for its existence as part of a specific key value pair.

if($arr[$key1] == $arr2[$key2]) Does you means something like this? — Rizier123
– Rizier123, Commented Jan 3, 2015 at 9:16
try using this:if($key1 != $key2 && $arr[$key1] == $arr2[$key2]).when keys are different and value is same. — Suchit kumar
– Suchit kumar, Commented Jan 3, 2015 at 9:19

Suchit kumar · Accepted Answer · 2015-01-03 10:18:32Z

1

try like this:if you only worried about value then remove the key inequality checking from if condition.

<?php 
$arr=array(1,3);
$arr2=array(3,4,1);
foreach($arr as $key1=>$val1){
    foreach($arr2 as $key2=>$val2){

        if($key1 != $key2 && $val1 == $val2){// if you do not want to check key inequality remove the first condition.
            echo "key1:".$key1.", key2:".$key2." ,value:".$val2."<br>";
        }
    }

}

edited Jan 3, 2015 at 10:18

answered Jan 3, 2015 at 9:59

Suchit kumar

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1 Comment

MrDave Over a year ago

I think you have identified the heart of my confusion...and that it is a nested foreach which will enable me to integrate through array2 to compare with integration of array1. I'll just check it works or whether there are other problems!

Utkarsh Dixit · Accepted Answer · 2015-01-03 09:55:39Z

0

Use the code below

 <?php
    $array1 = array("hello","world");
    $array2 = array("hello","peter");
    foreach( $array1 as $index => $p){
                if($p==$array2[$index]){
    echo "they have the same value<br>";
    }
else
{
echo "They have different value";
}
            }

Hope this helps you

edited Jan 3, 2015 at 9:55

answered Jan 3, 2015 at 9:37

Utkarsh Dixit

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