After yesterday I solve the hoisting problem. Here is the post.This morning I get this 404 not found error message in console. And it's just like the page before. Still got many
Solution 1:
This is because your displayAll()
function was executed before $.ajax
promises returned. And allStreamInfo[j]["logo"] = data.logo;
code like this in the success callback
function must be undefined
. Do read How do I return the response from an asynchronous call?. And you will know $.ajax
's basic priciple. Here is my code:
$(document).ready(function () {
loadStreamInfo();
});
var allStreamInfo = [{"user" : "ogaminglol"},{"user" : "faceittv"},{"user" :"twitch"},{"user" :"hearthstonesea"},{"user" :"zondalol"},{"user" :"aegabriel"}];
functionloadStreamInfo() {
for(var i = 0; i < 6; i++) {
(function (j) {
$.ajax({
url:("https://wind-bow.gomix.me/twitch-api/streams/" + allStreamInfo[j].user),
async: false,
jsonp: "callback",
dataType: "jsonp",
success: function (data) {
if(data.stream == null){
allStreamInfo[j]["status"] = "offline";
} else {
allStreamInfo[j]["status"] = "online";
}
$.ajax({
url:("https://wind-bow.gomix.me/twitch-api/channels/" + allStreamInfo[j].user),
async: false,
jsonp : "callback",
dataType : "jsonp",
success: function (data) {
allStreamInfo[j]["logo"] = data.logo;
allStreamInfo[j]["gameName"] = data.game;
allStreamInfo[j]["views"] = data.views;
allStreamInfo[j]["followers"] = data.followers;
allStreamInfo[j]["url"] = data.url;
displayAll(j);
}
});
}
});
})(i)
}
}
functiondisplayAll(i) {
var outString = "";
outString += "<div class='item'>";
outString += "<img src='" + allStreamInfo[i].logo + "' alt='logo'>";
outString += "<a href='" +allStreamInfo[i].url + "'><span id='gameName'>" + allStreamInfo[i].gameName +"</span></a>";
outString += "<span id='state'>" + allStreamInfo[i].status+"</span>";
outString += "<span id='views-block'>Views:<span id='view'>" + allStreamInfo[i].views + "</span></span>";
outString += "<span id='follow-block'>Followers:<span id='followed'>" + allStreamInfo[i].followers +"</span></span>";
outString += "</div>";
$("#result").append(outString);
}
body {
padding: 40px;
}
.toggle-button {
width: 400px;
color: white;
height: 100px;
text-align: center;
margin: 0 auto;
}
.all {
background-color: #6699CC;
width: 30%;
height: 70px;
line-height: 70px;
border-right: 2px solid grey;
display: inline;
float: left;
cursor: pointer;
}
.online {
cursor: pointer;
line-height: 70px;
background-color: cadetblue;
border-right: 2px solid grey;
width: 30%;
height: 70px;
display: inline;
float: left;
}
.offline {
cursor: pointer;
background-color: darkorange;
line-height: 70px;
width: 30%;
height: 70px;
display: inline;
float: left;
}
#result {
margin-top: 30px;
}
.item {
width: 500px;
height: 70px;
margin: 5px auto;
background-color: #666699;
border-left: 4px solid red;
color: whitesmoke;
/*border: 2px solid red;*/
}
a {
text-decoration: none;
}
img {
width: 50px;
height: 50px;
margin-top: 10px;
margin-left: 20px;
margin-right: 21px;
}
span#gameName,span#views-block,span#state ,span#follow-block {
position: relative;
bottom: 18px;
}
span#gameName,span#state,span#views-block{
margin-right: 21px;
}
<scriptsrc="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script><divclass="toggle-button"><divclass="all"onclick="displayAll()">All</div><divclass="online"onclick="displayOnline()">Online</div><divclass="offline"onclick="displayOffline()">Offline</div></div><divid="result"></div>
You must put the display()
function in success callback
. The only downside is you can't control which (function(j){...})(i)
is executed first. That's because asynchronous
.
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