【发布时间】:2017-10-13 22:01:03
【问题描述】:
我试图找到解决这个问题的方法,我目前不知所措。我正在尝试使用 twitch API 收集有关流的信息。
当我传入一个用户名数组时,我能够让 .getJSON 请求工作,如果用户离线或无效,则返回 null。
$(document).ready(function() {
var streamers = ["ESL_SC2","OgamingSC2","cretetion","freecodecamp","storbeck","habathcx","RobotCaleb","noobs2ninjas","meteos","c9sneaky","JoshOG"];
var url = "https://api.twitch.tv/kraken/streams/";
var cb = "?client_id=5j0r5b7qb7kro03fvka3o8kbq262wwm&callback=?";
getInitialStreams(streamers, url, cb); //load hard-coded streams
});
function getInitialStreams(streamers, url, cb) {
//loop through the streamers array to add the initial streams
for (var i = 0; i < streamers.length; i++) {
(function(i) {
$.getJSON(url + streamers[i] + cb, function(data) {
//If data is returned, the streamer is online
if (data.stream != null) {
console.log("online");
} else {
console.log("offline");
}
});
})(i);
}
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form class="form-inline">
<input id="addStreamerForm" class="form-control" type="text" placeholder="Add a Streamer">
<button id="addStreamerBtn" class="btn" type="submit">Add</button>
</form>
但是,当我尝试通过单击函数调用 API(最终通过表单提取任何输入)时,.getJSON 失败。如下所示。
$(document).ready(function() {
var url = "https://api.twitch.tv/kraken/streams/";
var cb = "?client_id=5j0r5b7qb7kro03fvka3o8kbq262wwm&callback=?";
$("#addStreamerBtn").click(function(e) {
addStreamer("Ben", url, cb);
});
});
function addStreamer(streamer, url, cb) {
console.log("I'm inside the function");
$.getJSON(url + streamer + cb, function(data) {
console.log("WE DID IT");
});
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form class="form-inline">
<input id="addStreamerForm" class="form-control" type="text" placeholder="Add a Streamer">
<button id="addStreamerBtn" class="btn" type="submit">Add</button>
</form>
我不明白为什么该代码不适用于第二个 sn-p。任何指导将不胜感激。
【问题讨论】:
标签: javascript jquery html json api