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Javascript Youtube API在DW中工作,但在JSFIDLE中发布错误_Javascript_Youtube_Youtube Api_Youtube Javascript Api_Youtube Data Api - Fatal编程技术网

Javascript Youtube API在DW中工作,但在JSFIDLE中发布错误

Javascript Youtube API在DW中工作,但在JSFIDLE中发布错误,javascript,youtube,youtube-api,youtube-javascript-api,youtube-data-api,Javascript,Youtube,Youtube Api,Youtube Javascript Api,Youtube Data Api,我已经和Youtube API玩了几天了,这只是我的第二个项目。我刚刚完成了FlickrAPI,所以我想我应该试试youtube 代码在dreamweaver中工作,但在JSFIDLE中不工作。我想这与提交有关,但在过去的几个小时里我一直在玩弄它,肯定错过了什么 非常感谢您的任何意见 <script src="https://apis.google.com/js/client.js?onload=loaded" type="text/javascript"></script&g

我已经和Youtube API玩了几天了,这只是我的第二个项目。我刚刚完成了FlickrAPI,所以我想我应该试试youtube

代码在dreamweaver中工作,但在JSFIDLE中不工作。我想这与提交有关,但在过去的几个小时里我一直在玩弄它,肯定错过了什么

非常感谢您的任何意见

<script src="https://apis.google.com/js/client.js?onload=loaded" type="text/javascript"></script>
    </head>
    <body>

        <body>
    <h1 id="header"> Search YouTube </h1>
    <section id="center">
        <form id="theform">
            <p><input type="text" name="term" id="term" value="dog"></p>
            <p><input type="submit" id="submit" value="Search"></p>
        </form>
    </section>
    <div id='here'> Here: </div>
    <pre id="response"></pre>
    </body>


var apiKey = 'myKey';

function showResponse(response) {
    var responseString = JSON.stringify(response, '', 2);
    document.getElementById('response').innerHTML += responseString;
}

function loaded(event) {
    gapi.client.load('youtube', 'v3', gapi.client.setApiKey(apiKey));
}   


$("#theform").on("submit", function(e) {
    e.preventDefault();
    var request = gapi.client.youtube.search.list({
        part: 'snippet',
        q: document.getElementById('term').value,
        maxResults: 1
    });
});

    request.execute(onSearchResponse);
}

function onSearchResponse(response) {
    showResponse(response);
}