如何从一个异步回调函数的返回值? [英] How to return value from an asynchronous callback function?
问题描述
这个问题是在SO问了很多次。但我仍然无法得到的东西。
我想从回调一定的价值。看看下面澄清脚本。
函数foo(地址){ //谷歌地图的东西
地理coder.geo code({'地址':地址},功能(结果状态){
结果[0] .geometry.location; //我想返回该值
}) }
富(); //结果应该是结果[0] .geometry.location;值
如果我试图返回值刚开不确定。我跟着从这么一些想法,但
仍然失败。
这些是:
函数foo(地址){
VAR返回值;
地理coder.geo code({'地址':地址},功能(结果状态){
返回值=结果[0] .geometry.location;
})
返回返回值;
}
富(); //仍然不确定
这是不可能的,因为你不能从一个异步调用同步方法中返回。
在这种情况下,你需要通过回调来富将接收返回值
函数foo(地址,FN){
地理coder.geo code({'地址':地址},功能(结果状态){
FN(结果[0] .geometry.location);
});
}富(地址功能(位置){
警报(位置); //这是你在哪里得到的返回值
});
问题是,如果一个内部函数调用是异步的,那么所有的功能包裹这一呼吁也必须按顺序异步回归的回应。
如果你有很多的回调可以考虑冒险尝试和使用承诺图书馆像:Q
This question is asked many times in SO. But still I can't get stuff.
I want to get some value from callback. Look at the script below for clarification.
function foo(address){
// google map stuff
geocoder.geocode( { 'address': address}, function(results, status) {
results[0].geometry.location; // I want to return this value
})
}
foo(); //result should be results[0].geometry.location; value
If I try to return that value just getting "undefined". I followed some ideas from SO, but still fails.
Those are:
function foo(address){
var returnvalue;
geocoder.geocode( { 'address': address}, function(results, status) {
returnvalue = results[0].geometry.location;
})
return returnvalue;
}
foo(); //still undefined
This is impossible as you cannot return from an asynchronous call inside a synchronous method.
In this case you need to pass a callback to foo that will receive the return value
function foo(address, fn){
geocoder.geocode( { 'address': address}, function(results, status) {
fn(results[0].geometry.location);
});
}
foo("address", function(location){
alert(location); // this is where you get the return value
});
The thing is, if an inner function call is asynchronous, then all the functions 'wrapping' this call must also be asynchronous in order to 'return' a response.
If you have a lot of callbacks you might consider taking the plunge and use a promise library like Q.
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