php合并json数组

array 1:

[
{"PlayerID":"17794204","userName":"Vandiel","castleCount":"9","NotUpd":"1476253231000"},
{"PlayerID":"21532584","userName":"Mayland","castleCount":"1","NotUpd":"0"},
{"PlayerID":"21539896","userName":"Dana","castleCount":"9","NotUpd":"0"}
]

array 2:

[
{"PlayerID":"17794204","Trouble":"2"},
{"PlayerID":"21532584","Trouble":"0"},
{"PlayerID":"21539896","Trouble":"0"}
]

这两个都是使用 fetch_assoc_all() 从 mySQL 数据库中提取的

我尝试了 merge_array、merge_array_recursive、json_decode(xx,true) 以及我能想到的各种东西以及通过谷歌在其他地方想到的东西。我正在寻找一种将 array1、array2 合并为类似以下内容的方法：

[
{"PlayerID":"17794204","userName":"Vandiel","castleCount":"9","NotUpd":"1476253231000","Trouble":"2"},
{"PlayerID":"21532584","userName":"Mayland","castleCount":"1","NotUpd":"0","Trouble":"0"},
{"PlayerID":"21539896","userName":"Dana","castleCount":"9","NotUpd":"0","Trouble":"0"}
]

PlayerID 始终是唯一的。希望听到我可以做些什么来合并这两个数组（array1，array2）

（补充/编辑） 对于那些想知道 mySQL 是什么样子的人（我无法理解 JOIN 语句）：

$mSQL = 'SELECT nPlayer.PlayerID,userName,castleCount,IF(LastUpdate < (UNIX_TIMESTAMP() - 3720),LastUpdate*1000,0) NotUpd';
$mSQL .= ' FROM nPlayer';
$mSQL .= ' LEFT JOIN nMains ON nMains.mID = nPlayer.mID';
$mSQL .= ' WHERE nMains.Main = "'.$M.'"  AND nMains.Pass = "'.md5($P).'" AND nMains.Server = "'.$S.'"';
$mSQL .= ' ORDER BY nPlayer.PlayerID';

$mSQL = 'SELECT nCity.PlayerID,SUM(IF(Wartown > 0,1,0))+SUM(IF(support < 100,1,0))) Trouble';
$mSQL .= ' FROM nCity';
$mSQL .= ' INNER JOIN nPlayer ON nPlayer.PlayerID = nCity.PlayerID AND nPlayer.mID = nCity.mID';
$mSQL .= ' INNER JOIN nMains ON nMains.mID = nPlayer.mID';
$mSQL .= ' WHERE nMains.Main = "'.$M.'"  AND nMains.Pass = "'.md5($P).'" AND nMains.Server = "'.$S.'"';
$mSQL .= ' GROUP BY nCity.PlayerID';

详细说明

You can join the JSON array based on the key value that you obtain provided you have to give under which key you have to join the json_array().

我将考虑json_objects根据PHP代码如下。

<?php
$array1 = '[
{"PlayerID":"17794204","userName":"Vandiel","castleCount":"9","NotUpd":"1476253231000"},
{"PlayerID":"21532584","userName":"Mayland","castleCount":"1","NotUpd":"0"},
{"PlayerID":"21539896","userName":"Dana","castleCount":"9","NotUpd":"0"}
]';
$array2 = '[
{"PlayerID":"17794204","Trouble":"2"},
{"PlayerID":"21532584","Trouble":"0"},
{"PlayerID":"21539896","Trouble":"0"}
]';
?>

因此，为了合并 json 对象，我们必须首先使用json_decode()对于我们获得的两个数组。

$decode_one = json_decode($array1,TRUE);
$decode_two = json_decode($array2,TRUE);

因此输出为json_decoded()字符串将如下所示。

第一个解码字符串：

Array ( [0] => Array ( [PlayerID] => 17794204 [userName] => Vandiel [castleCount] => 9 [NotUpd] => 1476253231000 ) [1] => Array ( [PlayerID] => 21532584 [userName] => Mayland [castleCount] => 1 [NotUpd] => 0 ) [2] => Array ( [PlayerID] => 21539896 [userName] => Dana [castleCount] => 9 [NotUpd] => 0 ) ) 

第二个解码字符串：

Array ( [0] => Array ( [PlayerID] => 17794204 [Trouble] => 2 ) [1] => Array ( [PlayerID] => 21532584 [Trouble] => 0 ) [2] => Array ( [PlayerID] => 21539896 [Trouble] => 0 ) )

之后我们必须merge the two arrays我们根据key那是unique for us.

因此，代码的功能如下。

我已经考虑过PlayerID as the UNIQUE参数和组合了数组。

function merge_json_decoded_arrays($decode_one,$decode_two) {
    $data = array();
    $arrayAB = array_merge($decode_one,$decode_two);
    foreach ($arrayAB as $value) {
      $id = $value['PlayerID'];
      if (!isset($data[$id])) {
        $data[$id] = array();
      }
      $data[$id] = array_merge($data[$id],$value);
    }
    return $data;
  }

您需要从需要执行以下操作的代码中调用这样的函数array_merge()运营。

$merged_array = merge_json_decoded_arrays($decode_one,$decode_two);

最后，完整的代码与设置如下所示。

完整代码：

<?php
$array1 = '[
{"PlayerID":"17794204","userName":"Vandiel","castleCount":"9","NotUpd":"1476253231000"},
{"PlayerID":"21532584","userName":"Mayland","castleCount":"1","NotUpd":"0"},
{"PlayerID":"21539896","userName":"Dana","castleCount":"9","NotUpd":"0"}
]';
$array2 = '[
{"PlayerID":"17794204","Trouble":"2"},
{"PlayerID":"21532584","Trouble":"0"},
{"PlayerID":"21539896","Trouble":"0"}
]';

$decode_one = json_decode($array1,TRUE);
$decode_two = json_decode($array2,TRUE);

function merge_json_decoded_arrays($decode_one,$decode_two) {
    $data = array();
    $arrayAB = array_merge($decode_one,$decode_two);
    foreach ($arrayAB as $value) {
      $id = $value['PlayerID'];
      if (!isset($data[$id])) {
        $data[$id] = array();
      }
      $data[$id] = array_merge($data[$id],$value);
    }
    return $data;
  }
$merged_array = merge_json_decoded_arrays($decode_one,$decode_two);
?>

为了查看合并的数组，您需要print_r()数组并查看它。

数组输出代码：

print_r($merged_array);

Output:

Array ( [17794204] => Array ( [PlayerID] => 17794204 [userName] => Vandiel [castleCount] => 9 [NotUpd] => 1476253231000 [Trouble] => 2 ) [21532584] => Array ( [PlayerID] => 21532584 [userName] => Mayland [castleCount] => 1 [NotUpd] => 0 [Trouble] => 0 ) [21539896] => Array ( [PlayerID] => 21539896 [userName] => Dana [castleCount] => 9 [NotUpd] => 0 [Trouble] => 0 ) )

如果你需要它作为 JSON 输出，你必须json_encode()得到的array()并执行操作。

Note:它需要unique ID as the array key对于生成的每一行。

JSON 输出代码：

print_r(json_ecode($merged_array));

Output:

{"17794204":{"PlayerID":"17794204","userName":"Vandiel","castleCount":"9","NotUpd":"1476253231000","Trouble":"2"},"21532584":{"PlayerID":"21532584","userName":"Mayland","castleCount":"1","NotUpd":"0","Trouble":"0"},"21539896":{"PlayerID":"21539896","userName":"Dana","castleCount":"9","NotUpd":"0","Trouble":"0"}}

最快的执行方法将采取这种方式

您需要解码 json 字符串，然后必须将它们都循环通过foreach()然后与array()你需要加入它。

$decode_one = json_decode($array1,TRUE);
$decode_two = json_decode($array2,TRUE);
foreach ($decode_one as $key => $first_value) {
    foreach ($decode_two as $key_two => $second_value) {
        if($first_value['PlayerID']==$second_value['PlayerID'])
        { $decode_one[$key]['Trouble'] = $second_value['Trouble'];//Here if the key exists it will join the Trouble with the First decoded array }
        else {}
    }
}
$combined_output = json_encode($decode_one); //This will return the output in json format.

Output:

[{"PlayerID":"17794204","userName":"Vandiel","castleCount":"9","NotUpd":"1476253231000","Trouble":"2"},{"PlayerID":"21532584","userName":"Mayland","castleCount":"1","NotUpd":"0","Trouble":"0"},{"PlayerID":"21539896","userName":"Dana","castleCount":"9","NotUpd":"0","Trouble":"0"}]