Question

我试图解析一个看起来像这样的JSON文件，

[
    {
        "id": "539eebdba276db40a4716726",
        "name": "Development Task",
        "idList": "539eebbb4e9a8d709704b254",
        "desc": "",
        "url": ""
    },
    {
        "id": "539eebe09b42c971d46b9ba1",
        "name": "Design Task",
        "idList": "539eebbe50dc4fa2a82474fc",
        "desc": "",
        "url": ""
    }
]

我试图从名为desc的数组中获取Development Task对象，系统需要是动态的，所以我不能只使用json_o[0](desc);

我尝试了不同的方法，例如多次预测数据，但我仍然无法想到解决方案，任何帮助都会很棒，欢呼。

Answer 1

试试这个：

$jsonData = '[
    {
        "id": "539eebdba276db40a4716726",
        "name": "Development Task",
        "idList": "539eebbb4e9a8d709704b254",
        "desc": "",
        "url": ""
    },
    {
        "id": "539eebe09b42c971d46b9ba1",
        "name": "Design Task",
        "idList": "539eebbe50dc4fa2a82474fc",
        "desc": "",
        "url": ""
    }
]';

$encoded = json_decode($jsonData);

foreach($encoded as $data) 
{
        if('Development Task' == $data->name) 
        {
            echo $data->desc;
        }
}

Answer 2

如果您搜索动态字段，我会选择：

$jsonData = '[
    {
        "id": "539eebdba276db40a4716726",
        "name": "Development Task",
        "idList": "539eebbb4e9a8d709704b254",
        "desc": "FirstDescription",
        "url": ""
    },
    {
        "id": "539eebe09b42c971d46b9ba1",
        "name": "Design Task",
        "idList": "539eebbe50dc4fa2a82474fc",
        "desc": "SecondDescription",
        "url": ""
    }
]';

$arrayFromJson = json_decode($jsonData, true);

function searchByKey($keyToSearchIn, $searchName, $array) {
   foreach ($array as $key => $val) {
       if ($val[$keyToSearchIn] == $searchName) {
           return $val['desc'];
       }
   }
   return null;
}

$return = searchByKey("name", "Development Task", $arrayFromJson);

所以

var_dump($return);

返回

string(16) "FirstDescription"

注意：如果您只需按名称搜索，则可以使用以下命令更改功能：

function searchByName( $searchName, $array) {
   foreach ($array as $key => $val) {
       if ($val['name'] == $searchName) {
           return $val['desc'];
       }
   }
   return null;
}

$return = searchByName("Development Task", $arrayFromJson);

希望它有所帮助。

Answer 3

代码中的注释解释了会发生什么：

    $jsonData = '[
    {
        "id": "539eebdba276db40a4716726",
        "name": "Development Task",
        "idList": "539eebbb4e9a8d709704b254",
        "desc": "moocow",
        "url": ""
    },
    {
        "id": "539eebe09b42c971d46b9ba1",
        "name": "Design Task",
        "idList": "539eebbe50dc4fa2a82474fc",
        "desc": "wowcow",
        "url": ""
    },
        {
        "id": "539eebe09rb42c971d46b9ba1",
        "name": "Development Task",
        "idList": "539eebbe50dc4fa2a82474fc",
        "desc": "sowccow",
        "url": ""
    }
]';

$encoded = json_decode($jsonData, true); //converts the json object to an array
foreach ($encoded as $arrayObject){
  if( in_array("Development Task", $arrayObject) ){
    print_r($arrayObject['desc'] . "\n");
  }
}
-> moocow sowccow

基于

我正在尝试从名为Development Task
的数组中获取desc对象

这应该做你想要的，你可以轻松地获取字符串并使其成为变量。 PHP.net上有很多内置函数可以调整，并且可能比滚动自己的代码好10倍。

过滤JSON对象并输出另一个

3 个答案: