Check if first two items in Laravel collection have same date; if not, select the item with the greater date

Question

I have a collection that contain two arrays :

array:2 [
  0 => array:3 [
 
    "cde" => "cd542"
    "startDate" => "2022-10-27"
    "endDate" => null
  ]
  1 => array:3 [
    "cde" => "cd547"
    "startDate" => "2022-10-27"
    "endDate" => null
  ]
]

1. take the first element if they have the same code & the same start date
2. if they have different startDate take the most recent one

i tried the ->contains method but is not working

Answer 1

This feels like an odd use case but this should get you on the right track:

$sameElements = [
    [
        "Code" => "AKAR02",
        "startDate" => "2022-10-27",
        "endDate" => null,
    ],
    [
        "Code" => "AKAR02",
        "startDate" => "2022-10-27",
        "endDate" => null,
    ]
];

$collection = collect($sameElements);
[$first, $second] = $collection->only(0, 1);

if (
    $first['code'] === $second['code']
    && $first['startDate'] === $second['endDate']
) {
    // take the first element if they have the same code & the same start date
} else {
    // do something else?
}

//if they have different startDate take the most recent one
$diffElements = [
    [
        "Code" => "AKAR02",
        "startDate" => "2022-10-27",
        "endDate" => null,
    ],
    [
        "Code" => "AKAR02",
        "startDate" => "2022-10-01",
        "endDate" => null,
    ]
];

$collection = collect($diffElements);
[$first, $second] = $collection->only(0, 1);

// if they have different startDate take the most recent one
if ($first['startDate'] != $second['startDate']) {
    // do something with the most recent one
    $mostRecent = $collection->sort(function ($a, $b) {
        return strtotime($a['startDate']) <=> strtotime($b['startDate']);
    })->first();
} else {
    // do something else
}