Question

我有这个代码

$client = json_decode($client); 
echo "<pre>";
print_r($client);

在那里产生

Array
(
  [0] => stdClass Object
    (
        [id] => 1
        [name] => Jojohn@doe.com
        [email_verified_at] => 
        [password] => $2y$10$pAvJ9/K7ZPOqw10WhfmToumK0TY1XihY8M9uAEEs4GkHZr4LdGc4e
        [remember_token] => 
        [created_at] => 2020-07-29 21:08:02
        [updated_at] => 
        [userid] => 2
        [account_rep] => 3
    )

)

我的问题是如何获取我尝试过的name和account_rep的值

echo $client['0']['object']['name'];

但这不起作用，只会引发错误

Cannot use object of type stdClass as array

Answer 1

json_decode（$ variable），用于将JSON对象解码或转换为PHP对象。

因此您可以这样做，因为$client['0']是一个对象。

echo $client['0']->name;

但是我想说您应该通过将TRUE作为参数传递给json_decode，而不是将JSON对象转换为关联数组。当TRUE时，返回的对象将转换为关联数组。

$client = json_decode($client, true);

现在$ client是

Array
(
  [0] => Array
    (
        [id] => 1
        [name] => Jojohn@doe.com
        [email_verified_at] => 
        [password] => $2y$10$pAvJ9/K7ZPOqw10WhfmToumK0TY1XihY8M9uAEEs4GkHZr4LdGc4e
        [remember_token] => 
        [created_at] => 2020-07-29 21:08:02
        [updated_at] => 
        [userid] => 2
        [account_rep] => 3
    )

)

现在您可以轻松地

echo $client[0]['name'];

从json响应中获取特定值

1 个答案: