php合并json数组

【问题标题】：php merge json arraysphp合并json数组
【发布时间】：2017-02-24 13:34:05
【问题描述】：

数组 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"}
]

数组 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) 以及各种我可以通过谷歌在我的头脑和其他地方想到的东西。我正在寻找一种将array1、array2 合并为类似的方法：

[
{"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）

（附加/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';

【问题讨论】：

  • 拉取数据时最好在查询中使用MYSQL JOIN操作
  • 我尝试了 mySQL JOIN，但无法解决这些问题，因为它来自 2 个不同的表，但一个查询需要 GROUP BY 而其他查询不需要。
  • 在上面编辑以包含 mySQL
  • 在第二个查询中，您可以尝试使用 nCity.PlayerID、nPlayer.userName、nPlayer.castleCount、.....
  • @sgkdnay。我已经提供了有关您所需输出的解释，也提供了代码。分享想法，如果您遇到任何障碍，请告诉我。

标签： php mysql arrays json


【解决方案1】：

详细说明

您可以根据获得的键值加入 JSON 数组，前提是您必须在哪个键下加入 json_array()。

我将根据PHP代码考虑json_objects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_objects，我们必须首先对我们获得的两个数组使用 json_decode()。

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

因此json_decoded() 字符串的输出将如下所示。

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，我们必须根据key获得merge the two arrays，即unique。

因此代码的功能如下。

我已将 PlayerID 视为 UNIQUE 参数，并合并了数组。

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array_merge() 操作的代码中调用这样的函数。

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

最后完整的代码如下所示。

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);
?>

要查看合并后的数组，您需要print_r() 并查看该数组。

数组输出代码：

print_r($merged_array);

输出：

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 ) )

如果您需要它作为 JSON 输出，您必须 json_encode() 获得的 array() 并执行操作。

注意：它把unique ID作为生成的每一行的array key。

JSON 输出代码：

print_r(json_ecode($merged_array));

输出：

{"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"}}

最快的执行方式会采用这种方式

您需要解码 json_strings，然后您必须通过 foreach() 将它们都 llop，然后与您需要加入的 array() 结合。

$decode_one = json_decode($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.

输出：

[{"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"}]

【讨论】：

  • 这种方法效果很好。直到有人提到加入，我只是把头撞到桌子上，并没有意识到它的简化。对于我计划使用其他 JSON 进行的额外工作，肯定会很有用
【解决方案2】：

假设json1和json2是你的两个JSON字符串，解决方案合并这两个JSON字符串是这样的：

  • 首先使用json_decode()函数解码这两个JSON字符串，得到$decodedArray1和$decodedArray2数组。
  • 运行两个嵌套的foreach 循环以将$decodedArray2 合并到$decodedArray1 数组中。
  • 最后，在$decodedArray1 数组上应用json_encode() 函数，得到生成的JSON 字符串。

代码如下：

$decodedArray1 = json_decode($json1, true);
$decodedArray2 = json_decode($json2, true);

foreach($decodedArray1 as $key => $array1){
    foreach($decodedArray2 as $array2){
        if($array1['PlayerID'] == $array2['PlayerID']){
            $decodedArray1[$key]['Trouble'] = $array2['Trouble'];
        }
    }
}

$resultantJson = json_encode($decodedArray1);

这是live demo

【讨论】：

    【解决方案3】：

    PHP 中没有任何功能可以满足您的需求。如果您查看常见的可疑框架，您会发现它们使用如下合并功能：

    /**
 * @param array $array
 * @param array $other
 *
 * @return array
 */
function _array_merge(array $array, array $other)
{
    foreach ($other as $key => $value) {
        if (isset($array[$key]) || array_key_exists($key, $array)) {
            if (is_int($key)) {
                $array[] = $value;
            } elseif (is_array($value) && is_array($array[$key])) {
                $array[$key] = _array_merge($array[$key], $value);
            } else {
                $array[$key] = $value;
            }
        } else {
            $array[$key] = $value;
        }
    }

    return $array;
}

    通过上面的方法，您可以json_decode() 每个元素然后合并它们，然后json_encode() 结果。

    为了完整起见，这里是上述合并功能的测试用例：

    <?php

class ArrayMergeTest extends \PHPUnit_Framework_TestCase
{
    public function testMerge(array $expected, array $a, array $b)
    {
        $this->assertSame($expected, _array_merge($a, $b));
    }

    /**
     * @return array
     */
    public function provideMerge()
    {
        return array(
            'should-preserve-integer-keys' => array(
                array(
                    2 => array('a', 'b', 'c'),
                    3 => array('d', 'e', 'f'),
                ),
                array(2 => array('a', 'b', 'c')),
                array(2 => array('d', 'e', 'f')),
            ),
            'should-merge-when-no-existing-key' => array(
                array(
                    'a' => array('b', 'c'),
                    'd' => array('e', 'f'),
                ),
                array('a' => array('b', 'c')),
                array('d' => array('e', 'f')),
            ),
            'should-overwrite-key-when-not-array' => array(
                array('a' => 'b'),
                array('a' => 'foo'),
                array('a' => 'b'),
            ),
            'should-merge-recursive' => array(
                array('a' => array(0 => 'b', 1 => 'c')),
                array('a' => array('b')),
                array('a' => array('c')),
            ),
            'should-replace-string-keys' => array(
                array('foo' => 'baz', 'bar' => 'bat'),
                array('foo' => 'bar', 'bar' => array()),
                array('foo' => 'baz', 'bar' => 'bat'),
            ),
            'should-merge-nested' => array(
                array('a' => array('b' => array('baz', 'bat'))),
                array('a' => array('b' => array('baz'))),
                array('a' => array('b' => array('bat'))),
            ),
            'should-merge-simple-keys' => array(
                array('a' => 'a_val', 'b' => 'b_val'),
                array('a' => 'a_val'),
                array('b' => 'b_val'),
            ),

            // Ensure documentation examples work

            'doc_example_1' => array(
                array(42 => 'x', 43 => 'y'),
                array(42 => 'x'),
                array(42 => 'y'),
            ),

            'doc_example_2_a' => array(
                array('x' => 'b'),
                array('x' => 'a'),
                array('x' => 'b'),
            ),
            'doc_example_2_b' => array(
                array('x' => 'b'),
                array('x' => array('a')),
                array('x' => 'b'),
            ),
            'doc_example_2_c' => array(
                array('x' => array('b')),
                array('x' => 'a'),
                array('x' => array('b')),
            ),
            'doc_example_2_d' => array(
                array('x' => array('a', 'b')),
                array('x' => array('a')),
                array('x' => array('b')),
            ),
            'merge-integer-and-string-keys' => array(
                array(
                    0 => 'foo',
                    3 => 'bar',
                    'baz' => 'baz',
                    4 => array(
                        'a',
                        1 => 'b',
                        'c',
                    ),
                    5 => 'baz',
                    6 => array(
                        'd' => 'd',
                    ),
                ),
                array(
                    'foo',
                    3 => 'bar',
                    'baz' => 'baz',
                    4 => array(
                        'a',
                        1 => 'b',
                        'c',
                    ),
                ),
                array(
                    'baz',
                    4 => array(
                        'd' => 'd',
                    ),
                ),
            ),
            'merge-arrays-recursively' => array(
                array(
                    'foo' => array(
                        0 => 'baz',
                        1 => 'baz',
                    ),
                ),
                array(
                    'foo' => array(
                        'baz',
                    ),
                ),
                array(
                    'foo' => array(
                        'baz',
                    ),
                ),
            ),
            'replace-string-keys' => array(
                array(
                    'foo' => 'baz',
                    'bar' => 'bat',
                ),
                array(
                    'foo' => 'bar',
                    'bar' => array(),
                ),
                array(
                    'foo' => 'baz',
                    'bar' => 'bat',
                ),
            ),
            'merge-with-null' => array(
                array(
                    'foo' => 'baz',
                    null => 'zad',
                    'cat' => 'bar',
                    'god' => null,
                ),
                array(
                    'foo' => null,
                    null => 'rod',
                    'cat' => 'bar',
                    'god' => 'rad',
                ),
                array(
                    'foo' => 'baz',
                    null => 'zad',
                    'god' => null,
                ),
            ),
        );
    }
}

    【讨论】：

      【解决方案4】：

      你能用foreach循环试试吗？

      示例 JSON 数据 1：

        $json1='[
  {"PlayerID":"17794204","userName":"Vandiel","castleCount":"9","NotUpd":"1476253231000"},
{"PlayerID":"21532584","userName":"Mayland","castleCount":"1","NotUpd":"0"},
{"PlayerID":"21539896","userName":"Dana","castleCount":"9","NotUpd":"0"}
]';

      示例 JSON 数据 2：

      $json2='[
 {"PlayerID":"17794204","Trouble":"2"},
 {"PlayerID":"21532584","Trouble":"0"},
 {"PlayerID":"21539896","Trouble":"0"}
]';

      首先解码您获取的 json 数据以便转换为数组

      $array1=json_decode($json1,true);
$array2=json_decode($json2,true);

      转换后使用以下 foreach 循环来检查两个数组中的“PlayerID”是否相等。如果相同，则与结果数组合并为以下脚本：

          $result = array();

    foreach($array1 as $data1)
    {
      foreach($array2 as $data2)
      {
      if($data1['PlayerID'] == $data2['PlayerID'])
      {
        $tmp = array("Trouble" => $data2['Trouble']);
        $tmp = array_merge($data1, $tmp);
        $result[] = $tmp;

      }
    }
  }

      输出将完全按照需要合并到数组后面：

                   array(3) {
             [0]=>
               array(5) {
               ["PlayerID"]=>
               string(8) "17794204"
               ["userName"]=>
                string(7) "Vandiel"
                ["castleCount"]=>
                string(1) "9"
                ["NotUpd"]=>
                 string(13) "1476253231000"
                ["Trouble"]=>
                string(1) "2"
                }
              [1]=>
                 array(5) {
                ["PlayerID"]=>
                string(8) "21532584"
                ["userName"]=>
                 string(7) "Mayland"
                 ["castleCount"]=>
                 string(1) "1"
                 ["NotUpd"]=>
                  string(1) "0"
                 ["Trouble"]=>
                  string(1) "0"
                 }
             [2]=>
               array(5) {
               ["PlayerID"]=>
               string(8) "21539896"
              ["userName"]=>
              string(4) "Dana"
              ["castleCount"]=>
              string(1) "9"
              ["NotUpd"]=>
              string(1) "0"
              ["Trouble"]=>
              string(1) "0"
             }
}

      【讨论】：

        【解决方案5】： 
        function mergeArrays(array $arr1, array $arr2, string $idKeyName) {
        $data = [];
        foreach ($arr1 as $value) {
            $id = $value[$idKeyName];
            $data[$id] = $value;
        }

        foreach ($arr2 as $value) {
            $id = $value[$idKeyName];
            if (isset($data[$id])) {
                $data[$id] = array_merge($data[$id], $value);
            } else {
                $data[$id] = $value;
            }
        }

        return array_values($data);
    }

        $arr1 = [["id" => "66395", "substanceId" => 182], ["id" => "66396", "substanceId" => 183]];
$arr2 = [["id" => "66395_new", "substanceId" => 182], ["id" => "66397", "substanceId" => 184]];
$result = mergeArrays($arr1, $arr2, 'substanceId');

# var_export($result)
array (
  0 => 
  array (
    'id' => '66395_new',
    'substanceId' => 182,
  ),
  1 => 
  array (
    'id' => '66396',
    'substanceId' => 183,
  ),
  2 => 
  array (
    'id' => '66397',
    'substanceId' => 184,
  ),
)

        【讨论】：

          【解决方案6】：

          这个 PHP 代码应该可以解决问题：

          $array1 = json_decode('[
    {"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 = json_decode('[
    {"PlayerID":"17794204","Trouble":"2"},
    {"PlayerID":"21532584","Trouble":"0"},
    {"PlayerID":"21539896","Trouble":"0"}
]');
$arrays_merged = array_merge($array1, $array2);

          编辑

          忘记了 json 数据的引号。对不起，我的错，已更正

          【讨论】：

          • 这也解决不了他的问题。请好好参考问题。他需要根据键合并数组。此解决方案将仅合并数组。
          猜你喜欢
          • 1970-01-01
          • 1970-01-01
          • 1970-01-01
          • 2013-12-25
          • 2020-02-29
          • 1970-01-01
          • 1970-01-01
          • 2023-03-16
          相关资源
          最近更新 更多
          热门标签
          Java Python linux javascript Mysql C# Docker 算法 前端 SpringBoot Redis Vue spring 设计模式 .net core .net kubernetes c++ 数据库 数据结构 大数据 js 机器学习 微服务 Android Go 程序员 面试 JVM ASP.net core 云原生 人工智能 后端 PHP git CSS golang k8s Nginx Django mybatis 深度学习 多线程 React 架构 devops 爬虫 云计算 Spring Boot LeetCode