jQuery ajax pass success into outside variable

Question

I have a problem using jquery ajax. I have already a js player that works fine without ajax. I.E.

/jwplayer.js

window.onload = function () {

function etc etc etc

jwplayer('player').setup({
playlist: [{
file: video_url,
}],
width: "640",
height: "380",
autostart: "true",
stretching: "exactfit",
volume: "100",
});
}

PHP page

<script type="text/javascript" src="/jwplayer.js"></script>
<script src="http://code.jquery.com/jquery-1.11.3.min.js"></script>

<div id='player'></div>

<script type="text/javascript">
var video_url = some_website_dot_com/file/.m3u8
</script>

This works just fine. But i cant add an ajax function to var video_url. This is the script that i'm trying to make it work

<script type='text/javascript'>
var video_url = function () {
$.ajax({
    type: 'get',
    url: "some_website_dot_com/file/.m3u8",
    dataType: "html",
    success: function (data) {
        var result = data.match(/(http\:\/\/\S+m3u8)/);
        return result[1];
    }
});
}();
</script>

@jgabb was the code that i found it here, http://stackoverflow.com/questions/16805306/jquery-return-ajax-result-into-outside-variable However it doesnt work neither without `()` — Avel, Oct 23 '15 at 19:50
It's an [iffe](http://stackoverflow.com/questions/8228281/what-is-the-function-construct-in-javascript). — Shanoor, Oct 23 '15 at 19:51

score 2 · Answer 1 · edited May 23 '17 at 11:58

2

You're trying to return something from an inner function, why not just do this?

var video_url;
$.ajax({
    type: 'get',
    url: "some_website_dot_com/file/.m3u8",
    dataType: "html",
    success: function (data) {
        var result = data.match(/(http\:\/\/\S+m3u8)/);
        video_url = result[1];
    }
});

Update: to get your player created and working when you get the url:

function setupVideo(data) {
    var result = data.match(/(http\:\/\/\S+m3u8)/);
    var video_url = result[1];
    jwplayer('player').setup({
        playlist: [{
            file: video_url,
        }],
        width: "640",
        height: "380",
        autostart: "true",
        stretching: "exactfit",
        volume: "100",
    });
}

$.ajax({
    type: 'get',
    url: "some_website_dot_com/file/.m3u8",
    dataType: "html",
    success: setupVideo
});

A good read on AJAX and specifically the A for asynchronous: How do I return the response from an asynchronous call?

edited May 23 '17 at 11:58

Community

1
1

answered Oct 23 '15 at 19:50

Shanoor

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Even i'm not sure if this is a proper way to use ajax, but i tried before, and it doesnt work. – Avel Oct 23 '15 at 19:58
It works for getting `video_url` but I guess you're not familiar with async? I update my answer. – Shanoor Oct 23 '15 at 20:03
I updated my question posting my jwplayer.js file exactly as it is. – Avel Oct 23 '15 at 20:19
But it's true, i'm not familiar with async – Avel Oct 23 '15 at 20:21
I have found a simple way: `var page = $.ajax({ type: 'get', url: "some_website_dot_com/file/.m3u8", dataType: "html", async: false, success: function (data) { return data; } }).responseText; var video = page.match(/(http\:\/\/\S+m3u8)/); var video_url = video[1];` – Avel Oct 23 '15 at 23:13

jQuery ajax pass success into outside variable

1 Answers1