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无法分析json属性_Json_Api - Fatal编程技术网

无法分析json属性

无法分析json属性,json,api,Json,Api,我试图从下面的输出中获取username和id的值 { "listusersresponse": { "count": 1, "user": [ { "id": "f01e8ea4-2da5-11e4-a846-726c7bbb864f", "username": "admin", "firstname": "admin", "lastname": "test", "created":

我试图从下面的输出中获取username和id的值

{
  "listusersresponse": {
    "count": 1,
    "user": [
      {
        "id": "f01e8ea4-2da5-11e4-a846-726c7bbb864f",
        "username": "admin",
        "firstname": "admin",
        "lastname": "test",
        "created": "2014-08-26T20:52:24-0400",
        "state": "enabled",
        "account": "admin",
        "accounttype": 1,
        "domainid": "c091153a-2da5-11e4-a846-726c7bbb864f",
        "accountid": "f01e7c02-2da5-11e4-a846-726c7bbb864f",
        "iscallerchilddomain": false,
        "isdefault": true
      }
    ]
  }
}
这就是我尝试过的

$url = file_get_contents("http://URL/client/&response=json");

$data = json_decode($url);

var_dump($data);

foreach ($data as $value) {
  echo $value->count;
  }
因为这个有效,我得到1分

但是,当尝试访问“用户名”时

我得到以下错误:注意:试图获取“filename”中非对象的属性用户是一个数组

foreach ($data as $value) {
    echo $value->user[0]->username;
}
不确定您的json在多个用户中的外观如何,我猜一定是这样的:

foreach ($data as $value) {
    foreach ($value->user as $user) {
        echo $user->username;
    }
}
foreach ($data as $value) {
    foreach ($value->user as $user) {
        echo $user->username;
    }
}