How do I remove one array from another in php?

Question

I have the following arrays in PHP: print_r($employees) =

Array
(
    [0] => Array
        (
            [p_id] => T29083999
            [name] => Robert Plaxo
        )

    [1] => Array
        (
            [p_id] => T29083388
            [name] => Yvan Sergei
        )

    [2] => Array
        (
            [p_id] => T21083911
            [name] => Rhonda Saunders
        )

    [3] => Array
        (
            [p_id] => H02910382
            [name] => Miguel Mercado
        )
)

then this array: print_r($record) =

Array
(
    [0] => Array
        (
            [c_id] => 1
            [section] => 1061
            [term] => 201631
            [p_id] => T29083388
            [c_date] => 2016-04-01 09:14:00
        )

)

I want to remove the array element from $employees that matches the p_id of $record. Array $record may have multiple entries like the one shown. If so, any p_id in $record must be removed from $employees.

I tried:

foreach ($employees as $k => $e) {
    foreach ($record as $r) {
        if ($e['p_id']==$r['p_id']) {
            echo "found it!";
            // if I uncomment the next line, it crashes! (500 server error)
            //  unset $employees[$k];
        }           
    }   
}

I just want to remove any element from $employees that has any employee that matches any record in $record with that employee id.

Also tried using: unset $e; with same result (500 server error). Please help! — ShotgunPR
– ShotgunPR, Commented Apr 29, 2016 at 21:05
why? I just need the key element to delete from the $employees array. I do not need the key for the comparison array (AFAIK). — ShotgunPR
– ShotgunPR, Commented Apr 29, 2016 at 21:17

David Wyly · Accepted Answer · 2016-04-29 21:19:18Z

2

You were almost there; just needed parens around your unset()

I also took the liberty to change some of your variable names as single character variable names bother me.

$employees[] = [
    'p_id' => 'T29083999',
    'name' => 'Robert Plaxo',
];
$employees[] = [
    'p_id' => 'T29083388',
    'name' => 'Yvan Sergei',
];
$employees[] = [
    'p_id' => 'T21083911',
    'name' => 'Rhonda Saunders',
];
$employees[] = [
    'p_id' => 'H02910382',
    'name' => 'Miguel Mercado',
];

$records[] = [
        'c_id' => '1',
        'section' => '1061',
        'term' => '201631',
        'p_id' => 'T29083388',
        'c_date' => '2016-04-01 09:14:00',
];

foreach ($employees as $key => $employee) {
    foreach ($records as $record) {
        if ($employee['p_id'] == $record['p_id']) {
            echo "found it!";
            unset($employees[$key]);
        }
    }
}

echo "<pre>";
print_r($employees);

Outputs

found it!

Array
(
    [0] => Array
        (
            [p_id] => T29083999
            [name] => Robert Plaxo
        )

    [2] => Array
        (
            [p_id] => T21083911
            [name] => Rhonda Saunders
        )

    [3] => Array
        (
            [p_id] => H02910382
            [name] => Miguel Mercado
        )

)

edited Apr 29, 2016 at 21:19

answered Apr 29, 2016 at 21:12

David Wyly

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2 Comments

ShotgunPR Over a year ago

Thank you!!! Finally works! I got all messed up with the array, I have checked everything on the Internet and was going crazy with this!

David Wyly Over a year ago

@user200956 glad to have helped, feel free to upvote and/or accept the answer if it solved your problem

RomanPerekhrest · Accepted Answer · 2016-04-29 21:19:53Z

1

The short solution using array_column and array_filter functions. It will also fit your requirement "Array $record may have multiple entries":

$p_ids = array_column($record, "p_id");
$employees = array_filter($employees, function($v) use($p_ids){
    return !in_array($v["p_id"], $p_ids);     
});

print_r($employees);

The output:

Array
(
    [0] => Array
        (
            [p_id] => T29083999
            [name] => Robert Plaxo
        )

    [2] => Array
        (
            [p_id] => T21083911
            [name] => Rhonda Saunders
        )

    [3] => Array
        (
            [p_id] => H02910382
            [name] => Miguel Mercado
        )
)

answered Apr 29, 2016 at 21:19

RomanPerekhrest

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