用php解析非数组JSON

我正在用php+codeigniter开发一个后端，该后端连接到一个返回JSON的数据提供服务，如下所示：

{
"code": "36397_26320",
"type": "TEAMS",
"4522": {
    "id": "4522",
    "code": "324",
    "name": "IT24354"
},
"4524": {
    "id": "4524",
    "code": "1234",
    "name": "IT24234"
},
"4527": {
    "id": "4527",
    "code": "2134",
    "name": "IT2678"
},
"4529": {
    "id": "4529",
    "code": "653",
    "name": "IT3546",
    "info":{
         "type1":
         {
            "type":"1",
            "url":"www.someurl.com",
            "date":"some date"
         },
         "type2":
         {
            "type":"2",
            "url":"www.someurl.com",
            "date":"some date"
         }

      }
},
"4530": {
    "id": "4530",
    "code": "3456",
    "name": "IT8769"
},
"4534": {
    "id": "4534",
    "code": "6453",
    "name": "IT3456"
},
"4537": {
    "id": "4537",
    "code": "76856",
    "name": "IT2676"
},
"4540": {
    "id": "4540",
    "code": "5768",
    "name": "IT23454"
},
"16225": {
    "id": "16225",
    "code": "4675",
    "name": "IT90687"
}
}

我想获取这些数字标识符中的信息，这样JSON编码的输出将如下所示：

{
"items": [
    {
        "id": "4522",
        "code": "324",
        "name": "IT24354"
    },
    {
        "id": "4524",
        "code": "1234",
        "name": "IT24234"
    },
    {
        "id": "4527",
        "code": "2134",
        "name": "IT2678"
    },
    {
        "id": "4529",
        "code": "653",
        "name": "IT3546",
        "info":[
         {
            "type":"1",
            "url":"www.someurl.com",
            "date":"some date"
         },
         {
            "type":"2",
            "url":"www.someurl.com",
            "date":"some date"
         }

         ]
    },
    {
        "id": "4530",
        "code": "3456",
        "name": "IT8769"
    },
    {
        "id": "4534",
        "code": "6453",
        "name": "IT3456"
    },
    {
        "id": "4537",
        "code": "76856",
        "name": "IT2676"
    },
    {
        "id": "4540",
        "code": "5768",
        "name": "IT23454"
    },
    {
        "id": "16225",
        "code": "4675",
        "name": "IT90687"
    }
]
}

我的问题是，例如，我不知道项目是否有字段“info”，例如在项目4529中，或者在项目标识符级别，有两个名为code和type的字段没有进一步的信息

有什么简单的方法可以做这样的操作吗？或者说，我想要达到多少级别就执行一个foreach是唯一的方法吗？如何识别json中的键是否包含更多的键值对

---

谢谢大家!

获取您提供的第一个json对象，并将其放入本例的

$json

变量中：

<?php 

$json = '{
"code": "36397_26320",
"type": "TEAMS",
"4522": {
    "id": "4522",
    "code": "324",
    "name": "IT24354"
},
"4524": {
    "id": "4524",
    "code": "1234",
    "name": "IT24234"
},
"4527": {
    "id": "4527",
    "code": "2134",
    "name": "IT2678"
},
"4529": {
    "id": "4529",
    "code": "653",
    "name": "IT3546",
    "info":{
         "type1":
         {
            "type":"1",
            "url":"www.someurl.com",
            "date":"some date"
         },
         "type2":
         {
            "type":"2",
            "url":"www.someurl.com",
            "date":"some date"
         }

      }
},
"4530": {
    "id": "4530",
    "code": "3456",
    "name": "IT8769"
},
"4534": {
    "id": "4534",
    "code": "6453",
    "name": "IT3456"
},
"4537": {
    "id": "4537",
    "code": "76856",
    "name": "IT2676"
},
"4540": {
    "id": "4540",
    "code": "5768",
    "name": "IT23454"
},
"16225": {
    "id": "16225",
    "code": "4675",
    "name": "IT90687"
}
}';
?>

允许我们浏览该数据对象并查看是否设置了

info

：

foreach($data as $item){
    if(isset($item->info)){
        // info found...do what you need to...
        print $item->id . " <br />";print_r($item->info);
    }
}

---

非常感谢大家，这有点棘手，但我最后做了以下几点，因为它允许我在不知道键值是否是另一个json的情况下探索完整的json。现在它返回相同的输入json，但要将输出更改为数组，只需在递归时向返回值添加括号

$response[$key][] = $this->read_non_array_json(json_encode($value));

高度欢迎改进：

private function read_non_array_json($data)
{

    $json = json_decode($data);

    if(is_object($json))
    {
        if(isset($json->id))
        {
            //obtain values, use a global array maybe to save info or something...
                        $id = $json->id;
                        $code = $json->code;
                        $name = $json->name;
        }
        if(isset($json->info)
        {
              //more code...
        }

        foreach($json as $key => $value)
        {
            if(is_object($value))
            {
                $response[$key]= $this->read_non_array_json(json_encode($value));
            }
            else
            {
                $response[$key] = $value;
            }

        }

        return $response;

    }
    else
    {

        return $data;
    }
}

---

创建一个函数来遍历数组。使用

gettype（$var）

测试

object/array

，如果它是一个对象/数组，请再次调用该函数。这只需要几行代码。您可以使用

json\u decode

，将额外参数设置为

true

，将json转换为PHP数组，并使用

foreach

通过它进行交互。有关如何使用

json\u解码

，请参阅PHP手册。从那里，您可以很容易地使用例如

isset

来了解该项中是否存在

info

，

if（isset（$item['info']）{//bla}

或

is_array

来了解该项是否为数组。@ohgoodwhy正如您所注意的，该json中没有对象数组。是否有任何函数可以确定它是否有多个元素？谢谢@Darren我实际上最终使用isset函数来分析json中的值。@RobertoNovelo很高兴：）

$response[$key][] = $this->read_non_array_json(json_encode($value));

private function read_non_array_json($data)
{

    $json = json_decode($data);

    if(is_object($json))
    {
        if(isset($json->id))
        {
            //obtain values, use a global array maybe to save info or something...
                        $id = $json->id;
                        $code = $json->code;
                        $name = $json->name;
        }
        if(isset($json->info)
        {
              //more code...
        }

        foreach($json as $key => $value)
        {
            if(is_object($value))
            {
                $response[$key]= $this->read_non_array_json(json_encode($value));
            }
            else
            {
                $response[$key] = $value;
            }

        }

        return $response;

    }
    else
    {

        return $data;
    }
}