使用php将Json嵌套数组到单个数组答案

【问题标题】：Json Nested arrays to single array using php使用php将Json嵌套数组到单个数组
【发布时间】：2018-05-10 11:38:02
【问题描述】：

我有一个嵌套的 JSON。我想在 php 中将其转换为简单的 json

var movies = [{
    "name": "Ice Age 3",
    "place" : "USA",
    "actors" : "cartoon",
    "details": [
        {"language": "English", "year": "2012"}, 
        {"language": "French", "year": "2011"}, 
        {"language": "German", "year": "2013"}
    ],
    "details2": [
        {"language2": "Spanish", "year2": "2015"}, 
        {"language2": "Arabic", "year2": "2016"}, 
        {"language2": "Hindi", "year2": "2017"}
    ]
}];

像这样……

  var movies = [
      {"name":"Ice Age 3","place" : "USA", "actors" : "cartoon", "details.language":"English", "details.year":"2012", "details2.language2":"English", "details2.year2":"2015"},
      {"name":"Ice Age 3","place" : "USA", "actors" : "cartoon", "details.language":"French", "details.year":"2011", "details2.language2":"French", "details2.year2":"2016"},
      {"name":"Ice Age 3","place" : "USA", "actors" : "cartoon", "details.language":"German", "details.year":"2013", "details2.language2":"German", "details2.year2":"2017"}
    ];

当我尝试这种方式时，我得到了一个扁平的 json。

function convert_flatten($array) { 
  if (!is_array($array)) { 
    return FALSE; 
  } 
  $result = array(); 
  foreach ($array as $key => $value) { 
    if (is_array($value)) { 
      $arrayList=convert_flatten($value);
      foreach ($arrayList as $listItem) {
        $result[] = $listItem; 
      }
    } 
   else { 
    $result[$key] = $value; 
   } 
  } 
  return $result; 
}

这实际上是一个表示 json。我正在寻找一个通用的答案。任何帮助将不胜感激

谢谢

【问题讨论】：

标签： php arrays json multidimensional-array

【解决方案1】：

类似于 Prashant 的回答...

$movies = '[{
    "name": "Ice Age 3",
    "details": [
    {"language": "English", "year": "2012"},
    {"language": "French", "year": "2011"},
    {"language": "German", "year": "2013"}
    ]
}]';

$movies = json_decode($movies, true);
$out = array();
foreach ( $movies as $movie )   {
    foreach ( $movie['details'] as $movieDetails ){
        $movieDetails['name'] = $movie['name'];
        $out[] = $movieDetails;

    }
}
echo json_encode($out);

输出...

[{"language":"English","year":"2012","name":"Ice Age 3"},
    {"language":"French","year":"2011","name":"Ice Age 3"},
    {"language":"German","year":"2013","name":"Ice Age 3"}]

与其试图将内容作为某种匿名 JSON 进行操作，此代码仅适用于呈现的数据。原始 JSON 中的每个元素一次处理一个（可能允许存在具有相同结构的多部电影），并且仅将每个详细信息数组元素提升到 $out 中的顶层（将电影名称添加到每次都这样）。

【讨论】：

我认为没有比这更干净的了。

【解决方案2】：

试试这个

     $(function () {
                    var movies = [{
                            "name": "Ice Age 3",
                            "details": [
                                {"language": "English", "year": "2012"},
                                {"language": "French", "year": "2011"},
                                {"language": "German", "year": "2013"}
                            ]
                        }];

                    var obj = JSON.parse(JSON.stringify(movies));
                    var obj2;

                    jsonObj = [];
                    for (var i = 0; i < obj.length; i++) {

                        item = {};
                        obj2 = JSON.parse(JSON.stringify(obj[i].details));
                        for (var j = 0; j < obj2.length; j++) {

                            item ["name"] = obj[i].name;
                            item ["language"] = obj2[j].language;
                            item ["year"] = obj2[j].year;

                            jsonObj.push(item);
                        }
                    }

                    var data = JSON.stringify(jsonObj);

                    alert(data);

                });

【讨论】：

试试这个答案在 StackOverflow 上的价值很低，因为它们在教育 OP 和成千上万的未来研究人员方面做得很差。请永远不要发布仅代码的答案。请花点时间解释一下您的答案是如何工作的，以及为什么它是一个好的建议。

【解决方案3】：

您不需要递归来执行此操作，这是未经测试的，但将为您的需要提供基础；

function flattenMovies($data)
{
    // Get the name of the movie
    $name = $data['name'];
    $return_val = array();
    // For each part of this...
    foreach ($data as $part)
    {
        // If it is an array...
        if (is_array($part))
        {
            // Create a new array for this row and add the movie name
            $add_parts = array("name" => $name);
            // For each of the parts parts...
            foreach ($part as $key => $value)
            {
                // Add this to the row array assoc by key = value
                $add_parts[$key] = $value;
            }
            // Add the row to the return array
            $return_val[] = $add_parts;
            // Blank this row so we know this does not have any "old" info in
            $add_parts = null;
        }
    }
    // Return the flattened array
    return $return_val;
}

您需要做的就是遍历数组并将各个部分添加到单个值数组中
正如我所说，您可能需要在 foreach 中添加另一个嵌套，未经测试

【讨论】：