Question

这是代码：

    <?php
$userFriends = json_decode(file_get_contents("http://api.steampowered.com/ISteamUser/GetFriendList/v0001/?key=[api-key]&steamid=[steam-id]&relationship=all"), true);
foreach ($userFriends->data as $friend) {
    //$json = json_decode(file_get_contents("http://steamcommunity.com/market/priceoverview/?appid=730&market_hash_name=".rawurlencode($item)), true);
    echo $friend->friends;
}
?>

无法解析。哪里有问题？

Answer 1

首先，如果您希望使用->语法访问JSON数据，则必须从True中删除json_decode参数：

$userFriends = json_decode( file_get_contents( ... ), true );

然后，您的JSON具有以下结构：

{
    "friendslist": {
        "friends": [
            {
                "steamid": "0123456789",
                "relationship": "friend",
                "friend_since": 0
            },
            (...)
        ]

    }
}

所以，要访问朋友，你必须写下这样的东西：

foreach( $userFriends->friendslist->friends as $friend )
{
    echo $friend->steamid . PHP_EOL;
    echo $friend->relationship . PHP_EOL;
    echo $friend->friend_since . PHP_EOL;
}

Answer 2

您可以使用它来获取最后一个JSON错误，以防有一个错误。当然，只是回应错误而不是使用我的确切代码。

    switch (json_last_error()) {
        case JSON_ERROR_NONE:
            //echo ' - No errors';
        break;
        case JSON_ERROR_DEPTH:
            header($_SERVER["SERVER_PROTOCOL"].'500 Maximum stack depth exceeded');
            $this->response = array('error' => 'JSON Decode error - Maximum stack depth exceeded');
            return $this->return_response();
        break;
        case JSON_ERROR_STATE_MISMATCH:
            header($_SERVER["SERVER_PROTOCOL"].'500 Underflow or the modes mismatch');
            $this->response = array('error' => 'JSON Decode error - Underflow or the modes mismatch');
            return $this->return_response();
        break;
        case JSON_ERROR_CTRL_CHAR:
            header($_SERVER["SERVER_PROTOCOL"].'500 Unexpected control character found');
            $this->response = array('error' => 'JSON Decode error - Unexpected control character found');
            return $this->return_response();
        break;
        case JSON_ERROR_SYNTAX:
            header($_SERVER["SERVER_PROTOCOL"].'500 Syntax error, malformed JSON');
            $this->response = array('error' => 'JSON Decode error - Syntax error, malformed JSON');
            return $this->return_response();
        break;
        case JSON_ERROR_UTF8:
            header($_SERVER["SERVER_PROTOCOL"].'500 Malformed UTF-8 characters, possibly incorrectly encoded');
            $this->response = array('error' => 'JSON Decode error - Malformed UTF-8 characters, possibly incorrectly encoded');
            return $this->return_response();
        break;
        default:
            header($_SERVER["SERVER_PROTOCOL"].'500 Unknown JSON Decode error');
            $this->response = array('error' => 'JSON Decode error - Unknown error');
            return $this->return_response();
        break;
    }

Answer 3

可以将其作为数组读取

-(void) someAction{
  if(something){
  [[NSNotificationCenter defaultCenter] postNotificationName:@"syncFinish"
                                                        object:self];
}

为什么我无法在json中解码和解析用户朋友？

3 个答案: