Question

我有下面提到的数组，我的值总是3的倍数。

$xyz = array(array('name'=>'abc'),array("name"=>"snds"),array("name"=>""),array("number"=>"452"),array("number"=>"845120"),array("number" => "84514513200"),array("email" => "ddddf"),array("email" => "dkskns"),array("email" => "kjnksdnkds"));

但这不是我执行进一步操作的正确格式，所以我想要这个数组如下所述。

$abc = array(array("name"=>"abc","number"=>'452',"email" => "ddddf"),array("name" => "snds","number" =>"845120","email" => "dkskns"),array("name" => "", "number" => "84514513200","email" => "kjnksdnkds"));

注意：数组长度是动态的，但它总是3的倍数

Answer 1

一种可能性是使用modulo %运算符。

在foreach中，值是一个数组，您可以使用array_keys获取密钥，reset获取第一个数组元素的值。

$result = [];
$count = 0;
foreach ($xyz as $value) {
    if ($count%3 === 0) {
        $count = 0;
    }
    $result[$count][array_keys($value)[0]] = reset($value);
    $count++;
}

Demo

那会给你：

array(3) {
  [0]=>
  array(3) {
    ["name"]=>
    string(3) "abc"
    ["number"]=>
    string(3) "452"
    ["email"]=>
    string(5) "ddddf"
  }
  [1]=>
  array(3) {
    ["name"]=>
    string(4) "snds"
    ["number"]=>
    string(6) "845120"
    ["email"]=>
    string(6) "dkskns"
  }
  [2]=>
  array(3) {
    ["name"]=>
    string(0) ""
    ["number"]=>
    string(11) "84514513200"
    ["email"]=>
    string(10) "kjnksdnkds"
  }
}

Answer 2

这样做：

$result = array_map('array_merge', ...array_chunk($xyz, count($xyz) / 3));

改变php数组格式

2 个答案: