Question

我正在使用Youtube Data API并尝试收到聊天消息，因为我必须提供df3 <- df3 %>% left_join(df2)和lifeChatId参数

我的代码

part

我收到错误

    $guzzle_client = new Client();
    $res = $guzzle_client->request('GET', 'https://www.googleapis.com/youtube/v3/liveChat/messages',
        [
           'liveChatId' => $broadcastsResponse['modelData']['snippet']['liveChatId'],
           'part' => 'id,snippet'
        ]
    );

但我确定我提供了两个必需的参数。这个var_dump写在guzzle请求之前

{
 "error": {
  "errors": [
   {
    "domain": "global",
    "reason": "required",
    "message": "Required parameter: liveChatId",
    "locationType": "parameter",
    "location": "liveChatId"
   },
   {
    "domain": "global",
    "reason": "required",
    "message": "Required parameter: part",
    "locationType": "parameter",
    "location": "part"
   }
  ],
  "code": 400,
  "message": "Required parameter: liveChatId"
 }
}

返回

var_dump([
    'liveChatId' => $broadcastsResponse['modelData']['snippet']['liveChatId'],
    'part' => 'id,snippet'
]);)

知道我为什么会收到这样的错误吗？

Answer 1

尝试使用query请求选项将它们作为查询字符串参数传递。

$guzzle_client = new Client();
$liveChatId = $broadcastsResponse['modelData']['snippet']['liveChatId'];

$res = $guzzle_client->request('GET', 'https://www.googleapis.com/youtube/v3/liveChat/messages', [
    'query' => ['liveChatId' => $liveChatId, 'part' => 'id,snippet']
]);

获取Youtube LiveChatMessages列表的Guzzle请求

1 个答案: