Question

我正在尝试从以下输出中获取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;
  }

对于这项工作，我得到1

但是在尝试访问“用户名”

时

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

我收到以下错误：注意：尝试在“filename”中获取非对象的属性

Answer 1

用户是一个数组。

foreach ($data as $value) {
    echo $value->user[0]->username;
}

不确定你的json如何看待几个用户，它必须是这样的，我猜：

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

无法解析json属性

1 个答案: