PHP在数组中找到相同的值并创建新的数组

Question

我有一个像这样的数组：

array 
  0 => 
    array (size=5)
      'id' => string '1' (length=1)
      'hostings_id' => string '4' (length=1)
      'email' => string 'info@test.de' (length=16)
  1 => 
    array (size=5)
      'id' => string '3' (length=1)
      'hostings_id' => string '4' (length=1)
      'email' => string 'kun@test.de' (length=26)
  2 => 
    array (size=5)
      'id' => string '5' (length=1)
      'hostings_id' => string '4' (length=1)
      'email' => string 'ni@test.de' (length=23)
  3 => 
    array (size=5)
      'id' => string '410' (length=3)
      'hostings_id' => string '5' (length=1)
      'email' => string 'spam@email.de' (length=13)
  4 => 
    array (size=5)
      'id' => string '148' (length=3)
      'hostings_id' => string '5' (length=1)
      'email' => string 'e@email.de' (length=23)
  5 => 
    array (size=5)
      'id' => string '165' (length=3)
      'hostings_id' => string '8' (length=1)
      'email' => string 'mobil@newmail.de' (length=15)
  6 => 
    ...

我的目标是将此数组放入一个新数组：

array 
  0 => 
    array (size=5)
      'id' => string '1' (length=1)
      'hostings_id' => string '4' (length=1)
      'email' => string 'info@test.de' (length=16)
    array (size=5)
      'id' => string '3' (length=1)
      'hostings_id' => string '4' (length=1)
      'email' => string 'kun@test.de' (length=26)
    array (size=5)
      'id' => string '5' (length=1)
      'hostings_id' => string '4' (length=1)
      'email' => string 'ni@test.de' (length=23)
  1 => 
    array (size=5)
      'id' => string '410' (length=3)
      'hostings_id' => string '5' (length=1)
      'email' => string 'spam@email.de' (length=13)
    array (size=5)
      'id' => string '148' (length=3)
      'hostings_id' => string '5' (length=1)
      'email' => string 'e@email.de' (length=23) 
   2 =>   
    array (size=5)
      'id' => string '165' (length=3)
      'hostings_id' => string '8' (length=1)
      'email' => string 'mobil@newmail.de' (length=15)
  3 => 

最后，我想要这样的JSON：

{
    "emails": {
        "0": {
            "id": "1",
            "hostings_id": "4",
            "email": "info@test.de",
        },
        "1": {
            "id": "3",
            "hostings_id": "4",
            "email": "kun@test.de",
        },
        "2": {
            "id": "5",
            "hostings_id": "4",
            "email": "ni@test.de",
        },
    }
}

{
    "emails": {
        "0": {
            "id": "410",
            "hostings_id": "5",
            "email": "spam@email.de",
        },
        "0": {
            "id": "148",
            "hostings_id": "5",
            "email": "e@email.de",
        },
    }
}
...

我想基于'hostings_id'创建JSON具有相同'hostings_id'值的所有内容都应放在单独的JSON中。

我已经尝试了很多foreach和array_count_values，但是没有得到我想要的结果。 非常感谢您的帮助。

问候

Answer 1

我认为您需要的是：

array_merge($array1, $array2);

假设我有以下数组：

$array1 = array('id'=>1,'name'=>'james');

$array2 = array('address'=>'usa','phone'=>'912092091');

// merge then and put all together

$array_final = array_merge($array1, $array2);

print_r($array_final);

检查以下链接以获取更多信息：

http://php.net/manual/it/function.array-merge.php

这是您的最终代码，包括具有相同hosting_id但位于不同json数组中的json：

<?php
$array1 = array('0'=>array('id'=>5,'hosting_id'=>'4','email'=>'example@sample.com'),'1'=>array('id'=>6,'hosting_id'=>'5','email'=>'example@sample.com'),'2'=>array('id'=>8,'hosting_id'=>'4','email'=>'example@sample.com'),'3'=>array('id'=>10,'hosting_id'=>'5','email'=>'example@sample.com'),'4'=>array('id'=>11,'hosting_id'=>'5','email'=>'example@sample.com'));
$array2 = array('0'=>array('id'=>78,'hosting_id'=>'4','email'=>'example@sample.com'),'1'=>array('id'=>96,'hosting_id'=>'5','email'=>'example@sample.com'),'2'=>array('id'=>78,'hosting_id'=>'4','email'=>'example@sample.com'),'3'=>array('id'=>110,'hosting_id'=>'5','email'=>'example@sample.com'),'4'=>array('id'=>111,'hosting_id'=>'5','email'=>'example@sample.com'));

// merge then and put all together

$array_final = array_merge($array1, $array2);

$out=array();
foreach($array_final as $x){
  $out[$x['hosting_id']]['hosting_id']=$x['hosting_id'];
  $out[$x['hosting_id']]['details'][]=array('id'=>$x['id'],'hosting_id'=>$x['hosting_id'],'email'=>$x['email']);
}

echo json_encode($out);

结果：

{
4: {
hosting_id: "4",
details: [
{
id: 5,
hosting_id: "4"
},
{
id: 8,
hosting_id: "4"
},
{
id: 78,
hosting_id: "4"
},
{
id: 78,
hosting_id: "4"
}
]
},
5: {
hosting_id: "5",
details: [
{
id: 6,
hosting_id: "5"
},
{
id: 10,
hosting_id: "5"
},
{
id: 11,
hosting_id: "5"
},
{
id: 96,
hosting_id: "5"
},
{
id: 110,
hosting_id: "5"
},
{
id: 111,
hosting_id: "5"
}
]
}
}

Answer 2

非常感谢您的快速回复！ 最后，我不需要array_merge（），但是您通过美妙的说明将我带回到了正确的道路。 有时候我觉得太复杂了。

就是说，对我而言，现在起作用的是一种魅力：

foreach($result as $x){
    $out[$x['hostings_id']][] = array(
        'email' => $x['email'],
        'login' => $x['login'],
        'password' => $x['passwort']
    );
}

$hosting = new Hosting();

foreach($out as $key => $value) {
    $result = $hosting->insertJsonIntoRowEmails($key, json_encode($value));
}

再次谢谢你！