Question

我有一个对象数组

$a = [{"id":"20","invoice_id":"123"},{"id":"21","invoice_id":"123"},{"id":"22","invoice_id":"125"},{"id":"23","invoice_id":"125"},{"id":"24","invoice_id":"123"}];

这里我想创建一个新的abject数组，其中重复的对象不会存在（invoice_id），因为新数组将具有相同invoice_id的第一个对象。我是这样做的。

foreach ($a as $key => $value) {
        if(isset($new)) {
            foreach ($new as $k => $val) {
                if($val->id != $value->id) {
                    $new[] = $value;
                }
            }
        }else{
            $new[] = $value;
        }
    }

我的新阵列就像

$new = [{"id":"20","invoice_id":"123"},{"id":"22","invoice_id":"125"}]

但它没有提供所需的输出。该怎么办？

Answer 1

由于您将此标记为Laravel问题，请使用collections。

更少的代码（一行！）和性能损失不存在。

$a = json_decode('[{"id":"20","invoice_id":"123"},{"id":"21","invoice_id":"123"},{"id":"22","invoice_id":"125"},{"id":"23","invoice_id":"125"},{"id":"24","invoice_id":"123"}]');

$result = collect($a)->groupBy('invoice_id');

OP编辑后的问题：

$result = collect($a)->unique('invoice_id')->values()->toArray();

结果：

=> [
     {#826
       +"id": "20",
       +"invoice_id": "123",
     },
     {#824
       +"id": "22",
       +"invoice_id": "125",
     },
   ]

或使用->toJson()代替->toArray()

"[{"id":"20","invoice_id":"123"},{"id":"22","invoice_id":"125"}]"

Answer 2

请使用简单的逻辑尝试以下代码，

$temp = $new = array();
$b = json_decode($a, true);
foreach ($b as $key => $val) {
  if(!in_array($val['invoice_id'], $temp)) {
    $temp[$val['id']] = $val['invoice_id'];
    $new[] = array('id' => $val['id'], 'invoice_id' => $val['invoice_id']);
  }
}

print_r($new);

我只是创建一个临时数组，只存储唯一的invoice_id，以便在循环中进行比较。

它给出了以下结果，

Array
(
  [0] => Array
    (
        [id] => 20
        [invoice_id] => 123
    )

  [1] => Array
    (
        [id] => 22
        [invoice_id] => 125
    )

)

Answer 3

$result = [];
foreach ($a as $data) {
    $result[$data->id] = $data;
}

var_dump(array_values($results));

Answer 4

试试这个

$a = json_decode($a); 

$invoiceid = [];
$unique = [];
foreach ($a as $key => $value) { 
    if(!in_array($value->invoice_id,$invoiceid)) {
        $invoiceid[] = $value->invoice_id;
        $unique[] = $value;
    }

}
    echo $a = json_encode($unique);

检查数组中任何对象中是否存在键值（对象数组）

4 个答案: