Check if specific array key exists in multidimensional array - PHP

Question

I have a multidimensional array e.g. (this can be many levels deep):

$array = Array ( 
    [21] => Array ( ) 
    [24] => Array ( 
        [22] => Array ( ) 
        [25] => Array ( 
            [26] => Array ( ) 
        ) 
    ) 
)

I am trying to loop through it to see if a certain key exists:

$keySearch = 22; // key searching for

function findKey($array, $keySearch) {
    foreach ($array as $item){
        if (isset($item[$keySearch]) && false === findKey($item[$keySearch], $item)){
            echo 'yes, it exists';
        }
    }
}

findKey($array, $keySearch);

But it finds nothing. Is there an error in the loop?

@Haddock-san "Note: array_key_exists() will search for the keys in the first dimension only. Nested keys in multidimensional arrays will not be found." (see official docs) — jave.web
– jave.web, Commented Sep 7 at 10:18

Tim · Accepted Answer · 2015-10-09 08:22:15Z

55

array_key_exists() is helpful.

Then something like this:

function multiKeyExists(array $arr, $key) {

    // is in base array?
    if (array_key_exists($key, $arr)) {
        return true;
    }

    // check arrays contained in this array
    foreach ($arr as $element) {
        if (is_array($element)) {
            if (multiKeyExists($element, $key)) {
                return true;
            }
        }

    }

    return false;
}

Working example: http://codepad.org/GU0qG5su

edited Oct 9, 2015 at 8:22

answered Oct 17, 2013 at 7:40

Tim

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Comments

mickmackusa · Accepted Answer · 2018-01-13 05:16:19Z

46

I played with your code to get it working :

function findKey($array, $keySearch)
{
    foreach ($array as $key => $item) {
        if ($key == $keySearch) {
            echo 'yes, it exists';
            return true;
        } elseif (is_array($item) && findKey($item, $keySearch)) {
            return true;
        }
    }
    return false;
}

edited Jan 13, 2018 at 5:16

mickmackusa♦

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answered Oct 17, 2013 at 7:52

Alexandre Nucera

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6 Comments

David Lawrence Over a year ago

Shouldnt: "if (isset( $array[$key] ) )" be: "if( is_array( $array[$key] ) )"? Obviously the $key is set in $array because the foreach is looping that array.

VolkerK Over a year ago

if (is_array($item)) findKey($item, $keySearch); - the return value of this call is lost.

mickmackusa Over a year ago

I've edited to condense the if block. If this was my code, I'd rather rather use a single if with ||. *using parentheses as logic requires.

Steven Yip Over a year ago

thanks for this...btw, from my testing, it's better to set the === operator for this to work properly.

tibelchior Over a year ago

To make it more complete, I would suggest also checking if $item isn't empty. This can avoid some problems down the road.

|

knot22 · Accepted Answer · 2017-11-28 14:00:15Z

14

Here is a one line solution:

echo strpos(json_encode($array), $key) > 0 ? "found" : "not found";

This converts the array to a string containing the JSON equivalent, then it uses that string as the haystack argument of the strpos() function and it uses $key as the needle argument ($key is the value to find in the JSON string).

It can be helpful to do this to see the converted string: echo json_encode($array);

Be sure to enclose the needle argument in single quotes then double quotes because the name portion of the name/value pair in the JSON string will appear with double quotes around it. For instance, if looking for 22 in the array below then $key = '"22"' will give the correct result of not found in this array:

$array =
Array ( 
        21 => Array ( ), 
        24 => 
        Array ( 
            522 => Array ( ),
            25 =>
                Array ( 
                26 => Array ( ) 
            )
        )
    );

However, if the single quotes are left off, as in $key = "22" then an incorrect result of found will result for the array above.

EDIT: A further improvement would be to search for $key = '"22":'; just incase a value of "22" exists in the array. ie. 27 => "22" In addition, this approach is not bullet proof. An incorrect found could result if any of the array's values contain the string '"22":'

edited Nov 28, 2017 at 14:00

answered Nov 28, 2017 at 13:06

knot22

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3 Comments

Nico Haase Over a year ago

I would refuse a solution that works in the simplified case, but not when "special circumstances" occurs. If the values of the array may not contain all arbitrary values cause that might disturb your algorithm, the algorithm is miscrafted

Jerry Over a year ago

Not agreed !! Because it will also return "found" if a value equals the searched key.

mickmackusa Over a year ago

This solution is only correct for the OP's basic, minimal sample data. It is VERY easy for values in the array to cause this technique to give false positive results. I do not endorse this "hacky" solution. This answer makes no attempt to differentiate between "keys" and "values" in the json payload.

minboost · Accepted Answer · 2013-10-17 07:51:20Z

function findKey($array, $keySearch)
{
    // check if it's even an array
    if (!is_array($array)) return false;

    // key exists
    if (array_key_exists($keySearch, $array)) return true;

    // key isn't in this array, go deeper
    foreach($array as $key => $val)
    {
        // return true if it's found
        if (findKey($val, $keySearch)) return true;
    }

    return false;
}

// test
$array = Array ( 
    21 => Array ( 24 => 'ok' ),
    24 => Array ( 
        22 => Array ( 29 => 'ok' ),
        25 => Array ( 
            26 => Array ( 32 => 'ok' ) 
        )
    )
);

$findKeys = Array(21, 22, 23, 24, 25, 26, 27, 28, 29, 30);
foreach ($findKeys as $key)
{
    echo (findKey($array, $key)) ? 'found ' : 'not found ';
    echo $key.'<br>';
}

Luke Snowden · Accepted Answer · 2017-03-16 11:51:18Z

3

returns false if doesn't exists, returns the first instance if does;

function searchArray( array $array, $search )
{
    while( $array ) {
        if( isset( $array[ $search ] ) ) return $array[ $search ];
            $segment = array_shift( $array );
            if( is_array( $segment ) ) {
                if( $return = searchArray( $segment, $search ) ) return $return;
            }
        }
    }
    return false;
}

answered Mar 16, 2017 at 11:51

Luke Snowden

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1 Comment

Nico Haase Over a year ago

What if "false" is the value stored under the requested key? How to distinguish that? ;)

Pradeep · Accepted Answer · 2018-05-02 16:09:25Z

2

For sure some errors, is this roughly what you are after? (Untested code):

$keySearch=22; // key seraching for

function findKey($array, $keySearch) 
{ 
    // check whether input is an array
    if(is_array($array)
    {
       foreach ($array as $item)
       {
         if (isset($item[$keySearch]) || findKey($item, $keysearch) === true)
          {
            echo 'yes, it exists';
            return true;
          }
       }
    }
}

edited May 2, 2018 at 16:09

Pradeep

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answered Oct 17, 2013 at 7:38

Jelle Ferwerda

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2 Comments

Nico Haase Over a year ago

This method is missing a clear return statement for the case that the key does not exist

mickmackusa Over a year ago

If the sought key is paired to a null value, a false negative result will be generated.

Mohammad Ali Fathi · Accepted Answer · 2020-12-06 12:55:22Z

0

Here is one solution that finds and return the value of the key in any dimension array..

function findValByKey($arr , $keySearch){
    $out = null;
    if (is_array($arr)){
        if (array_key_exists($keySearch, $arr)){
            $out = $arr[$keySearch];
        }else{
            foreach ($arr as $key => $value){
                if ($out = self::findValByKey($value, $keySearch)){
                    break;
                }
            }
        }
    }
    return $out;
}

answered Dec 6, 2020 at 12:55

Mohammad Ali Fathi

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Comments

Sanjay Trivedi · Accepted Answer · 2022-08-19 13:41:17Z

0

I did modified to return value of searched key:

function findKeyInArray($array, $keySearch, &$value)
{
    foreach ($array as $key => $item) {
        if ($key === $keySearch) {
            $value =  $item;
            break;
        } elseif (is_array($item)) {
            findKeyInArray($item, $keySearch,$value);
        }
    }
}
$timeZone = null;
findKeyInArray($request, 'timezone', $timeZone);

edited Aug 19, 2022 at 13:41

answered Aug 19, 2022 at 13:39

Sanjay Trivedi

11 bronze badge

1 Comment

ahuemmer Over a year ago

Welcome to SO! Please take a look at the other answers that were given before. Your approach is mentioned there already. In order to keep the site clear and make it easy to find answers, we try to avoid double answers.

alboforlizo · Accepted Answer · 2024-09-04 14:45:51Z

0

This prints "Found":

$a = ["a"];
$a["a"] = [];
$a["a"]["b"] = ["c"];

$exists = $a["a"]["b"] ?? false;

if (is_array($exists))
echo "Found";
else
echo "Not Found";

While this prints "Not Found":

$a = ["a"];
$a["a"] = [];
$a["a"]["b"] = ["c"];

$exists = $a["a"]["c"] ?? false;

if (is_array($exists))
echo "Found";
else
echo "Not Found";

answered Sep 4, 2024 at 14:45

alboforlizo

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