PHP in_array not working with current array structure

Question

I am using a custom method to return a query as an array.

This is being used to check if a discount code posted is in the DB.

The array ends up as example:

Array
(
[0] => stdClass Object
    (
        [code] => SS2015
    )

[1] => stdClass Object
    (
        [code] => SS2016
    )

)

So when I am trying to do:

if ( ! in_array($discount_code, $valid_codes)) {

}

Its not working. Is there a way I can still use the function for query to array I am using and check if its in the array?

No issues, I can make a plain array of the codes but just wanted to keep things consistent.

I think your array code is broken. you can't have the same key with different values in an assoc array. — Patrick
– Patrick, Commented May 27, 2015 at 10:22
Show the exact input from $discount_code and the exact values from $valid_codes — Rizier123
– Rizier123, Commented May 27, 2015 at 10:22
Fixed the array example using a print_r of the actual array. — Lovelock
– Lovelock, Commented May 27, 2015 at 10:27
Because it is object of array. That's why it is not searching. — Khushang Bhavnagarwala.
– Khushang Bhavnagarwala., Commented May 27, 2015 at 10:35

num8er · Accepted Answer · 2015-05-27 11:10:57Z

1

Read about json_encode (serialize data to json) and json_decode (return associative array from serialized json, if secondary param is true). Also array_column gets values by field name. so we have array of values in 1 dimensional array, then let's check with in_array.

function isInCodes($code, $codes) {
    $codes = json_encode($codes); // serialize array of objects to json
    $codes = json_decode($codes, true); // unserialize json to associative array
    $codes = array_column($codes, 'code'); // build 1 dimensional array of code fields
    return in_array($code, $codes); // check if exists
}

if(!isInCodes($discount_code, $valid_codes)) {
// do something
}

edited May 27, 2015 at 11:10

answered May 27, 2015 at 10:35

num8er

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2 Comments

Eugene Podskal Over a year ago

That is much better so I upvoted it. And it is a good idea not to just give a working solution, but also to describe why OP's attempt failed. And there are other markdown capabilities that can allow you to create far more visually appealing and readable posts (but it is obviously not a compulsory obligation to do so with every post). Good luck.

num8er Over a year ago

Thanks. I Appreciate Your recommendations.

axiac · Accepted Answer · 2015-05-27 10:38:10Z

Use array_filter() to identify the objects having property code equal with $discount_code:

$in_array = array_filter(
    $valid_codes,
    function ($item) use ($discount_code) {
        return $item->code == $discount_code;
    }
);

if (! count($in_array)) {
    // $discount_code is not in $valid_codes
}

If you need to do the same check many times, in different files, you can convert the above code snippet to a function:

function code_in_array($code, array $array)
{
    return count(
        array_filter(
            $array,
            function ($item) use ($code) {
                return $item->code == $code;
            }
        )
    ) != 0;
}


if (! code_in_array($discount_code, $valid_codes)) {
    // ...
}

Imtiaz Pabel · Accepted Answer · 2015-05-27 10:49:29Z

0

try this

function in_array_r($needle, $haystack, $strict = false) {
foreach ($haystack as $item) {
    if (($strict ? $item === $needle : $item == $needle) || (is_array($item) && in_array_r($needle, $item, $strict))) {
        return true;
    }
}

return false;
}

then

echo in_array_r("SS2015", $array) ? 'found' : 'not found';

answered May 27, 2015 at 10:49

Imtiaz Pabel

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Comments

splash58 · Accepted Answer · 2015-05-27 11:57:02Z

0

why not solve it as a school task - fast and easy:

for($i = 0; $i < count($valid_codes); $i++) if ($valid_codes[$]->code == $discount_code) break;
if ( ! ($i < count($valid_codes))) {  // not in array
}

answered May 27, 2015 at 11:57

splash58

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