返回从父函数调用Ajax的值 [英] Return a value from an ajax call to parent function

问题描述

我有一个功能，我需要返回一个网址，我通过Ajax调用我得到。

I have a function where I need to return a url that I am getting via an ajax call.

var heatmap = new google.maps.ImageMapType({
    getTileUrl: function(coord, zoom) {
        var tileURL;
        $.get('getimage.php', { zoom: zoom, x: coord.x, y: coord.y }, function(data) {
            if(data.status) { tileURL=data.image; }
        }, "json");
        return "tileURL";
    },
       tileSize: new google.maps.Size(256, 256),
       opacity:0.55,
       isPng: true
});

显然，Ajax调用是异步的，所以我理解为什么上面的code将返回tileURL为未定义。我知道，一般人用的回调函数来解决这个问题，但我不知道如何让一个回调返回一个值给父功能。我正在使用谷歌地图API，所以我真的没有任何灵活地改变是如何工作的。

Obviously, the ajax call is asynchronous so I understand why the above code will return tileURL as undefined. I know in general people use callback functions to solve this issue, but I don't know how to get a call back to RETURN a value to the parent function. I'm working with the google maps API, so I don't really have any flexibility to change how that works.

推荐答案

由于Ajax请求是异步的，你唯一的选择是使用一个回调。如果你能返回一个值父（实际上是呼唤）功能，则该请求将不会是异步的;它会阻塞调用函数。此外，也没有办法从在该函数中的封闭件内获得一个参考到调用函数。 （也就是说，你不能返回来一个较高的功能在调用栈）。

Because the Ajax request is asynchronous, your only option is to use a callback. If you could return a value to the "parent" (really, "calling") function, then the request wouldn't be asynchronous; it would block the calling function. Also, there is no way to get a reference to the calling function from within a closure in that function. (i.e., you can't return to a function higher up in the call stack).

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