Question

这是JSON

{
  "circuit_list": [
    {
      "_id": "58c0f378a986f808cdaf94cf", 
      "aggregation": {
        "dev_name": "ME2-D2-BOO", 
        "port": {
          "desc": "AKSES_SITE_SITE-TSEL_ME2-D2-BOO#1/2/5_200M_BOO082#CIPAKUBOO534", 
          "name": "1/2/5"
        }
      }, 
      "area": "AREA 2", 
      "site_id": "N/A", 
      "site_name": "N/A"
    }, 
    {
      "_id": "58c0f378a986f808cdaf94d0", 
      "aggregation": {
        "dev_name": "ME2-D2-BOO", 
        "port": {
          "desc": "AKSES_SITE_SITE-TSEL_ME2-D2-BOO#1/2/5_200M_BOO082#CIPAKUBOO534", 
          "name": "1/2/5"
        }
      }, 
      "area": "AREA 2", 
      "site_id": "N/A", 
      "site_name": "N/A"
    }
}

我已尝试使用此代码

$json = json_decode($url, true);

foreach($json as $value)
{
    $_id = $value->_id;

}

它没有用。请帮助，我需要获取值以在视图上显示它们。我做错了吗？这个json很难，因为我不了解结构。我通常用格式

解码json

[{"id":"1","name":"faisal"}]

像这样和我的foreach一起工作。

Answer 1

如果json_decode的第二个参数为true，则该函数将返回一个数组而不是一个对象。此外，您需要遍历对象的circuit_list属性。

$json = json_decode($url); // <- remove the parameter

foreach($json->circuit_list as $value) // <- loop over circuit_list
{
    $_id = $value->_id;
}

Answer 2

<?php
$json = json_decode($url,true);
foreach($json['circuit_list'] as $value) 
{
    $id = $value['_id'];
}

?>

如何用foreach解码这个json

2 个答案: