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javascript - YouTube API 查询返回 null

转载 作者:行者123 更新时间:2023-11-28 01:41:34 26 4
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YouTube 搜索查询“Kanye West Amazing”应该返回几个 YouTube 链接。但是,通过 api 进行的 youtube 查询返回 null(来自 console.log):

<!doctype html>
<html>
<head>
<title>Vidify</title>
</head>

<body>
<!--Add a button for the user to click to initiate auth sequence -->
<button id="authorize-button" style="visibility: hidden">Authorize</button>
<script type="text/javascript">

var clientId = '591751286145-ktngvf2s76uiuuevan6mo1fft0kchl8l.apps.googleusercontent.com';

var apiKey = 'AIzaSyB0mteDV5vDKFR-iAv4Fx4OC2gp1Yhqe9U';

var scopes = 'https://www.googleapis.com/auth/youtube';

function handleClientLoad() {
console.log('API key provided - authorizing client...');
// Step 2: Reference the API key
gapi.client.setApiKey(apiKey);
gapi.auth.init(checkAuth);
}

function checkAuth() {
console.log('entered checkAuth - authorizing...');
setTimeout(function() {
console.log("requesting auth");
gapi.auth.authorize({
client_id: clientId,
scope: scopes,
response_type:'token',
immediate: true
}, handleAuthResult);
}, 100);
}

function handleAuthResult(authResult) {
console.log('received authResult');
var authorizeButton = document.getElementById('authorize-button');
if (authResult && !authResult.error) {
console.log("auth. successful");
authorizeButton.style.visibility = 'hidden';
makeApiCall();
} else {
console.log('auth failed.');
authorizeButton.style.visibility = '';
authorizeButton.onclick = handleAuthClick;
}
}

function handleAuthClick(event) {
// Step 3: get authorization to use private data
console.log('retry');
gapi.auth.authorize({client_id: clientId, scope: scopes, immediate: false}, handleAuthResult);
return false;

}

// Load the API and make an API call. Display the results on the screen.
function makeApiCall() {
console.log("loaded client");
gapi.client.setApiKey(apiKey);
// Step 4: Load the Google+ API
gapi.client.load('youtube', 'v3', function() {
console.log('youtube API loaded...');
// Step 5: Assemble the API request
var qVar = "Kanye West Amazing"
// changed. added: type
var request = gapi.client.youtube.search.list({
type: 'video',
part: 'id',
q: qVar
});
// Step 6: Execute the API request
request.execute(function(resp) {

console.log(resp);
var vid = document.createElement('vid');
vid.value = resp.items[0].id.videoId;
console.log(vid.value);
var youtube_url = "http://www.youtube.com/watch?v=";
youtube_url += vid.value;
youtube_url += "&rel=0";
window.open(youtube_url);
});
});
}
</script>

<script src="https://apis.google.com/js/client.js?onload=makeApiCall"></script>

</body>
</html>

所以我实际上正在尝试从 Spotify 桌面应用程序中运行它。 Spotify 应用程序系统允许我从“spotify 浏览器”中加载 index.html。以下是我现在遇到的错误:

Uncaught TypeError: Cannot call method 'setApiKey' of undefined

index.html:63 makeApiCall index.html:63 (anonymous function)

client.js?onload=makeApiCall:6 fa client.js?onload=makeApiCall:1 b

client.js?onload=makeApiCall:6 v.(anonymous function)

client.js?onload=makeApiCall:6 H.(anonymous function)

client.js?onload=makeApiCall:6 (anonymous function) cb=gapi.loaded_0:1

Uncaught TypeError: Cannot read property 'prototype' of undefinedcb=gapi.loaded_0:6

_.J cb=gapi.loaded_0:6 (anonymous function) cb=gapi.loaded_1:6 (anonymous function) client.js?onload=makeApiCall:4 Z

client.js?onload=makeApiCall:6 ua client.js?onload=makeApiCall:4 E

client.js?onload=makeApiCall:5 b client.js?onload=makeApiCall:6

v.(anonymous function) client.js?onload=makeApiCall:6 H.(anonymousfunction) client.js?onload=makeApiCall:6 (anonymous function)cb=gapi.loaded_1:1

最佳答案

看来您的嵌套可能不正确;您正在gapi.client.load函数的回调之外运行request.execute调用,因此当它尝试运行时,实际的请求本身不会被创建(gapi客户端异步加载API)。如果您尝试这样做:

  function makeApiCall() {
// Step 4: Load the Google+ API
gapi.client.load('youtube', 'v3', function() {
console.log('youtube API loaded...');
// Step 5: Assemble the API request
var qVar = "Kanye West Amazing"
// changed. added: type
var request = gapi.client.youtube.search.list({
type: 'video',
part: 'id',
q: qVar
});
// Step 6: Execute the API request
request.execute(function(resp) {
document.getElementById('vid').value = resp.items[1].id.videoId;
console.log('saved video id successfully');
});
});
}

然后它应该等待并仅在 API 代码加载后才执行请求。当然,这还假设您有一个 id 为“vid”的元素(您的示例代码没有)。

就范围而言,您不需要它们来执行此特定调用,因为它不需要进行身份验证(即您可以在未经授权的情况下运行 MakeApiCall() )。但是,如果您稍后要做其他需要授权的事情,它会正常工作,因此您可以轻松地将其保留。

关于javascript - YouTube API 查询返回 null,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20864312/

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