我在javascript中遇到回调函数问题。我想做的是:在for上循环并调用传递i
作为参数的函数。考虑到这一点,我仅在上一个交互完成后才循环到下一个交互。我不知道这是否有问题,但是在我将i
作为参数发送的函数中,我还有另一个回调函数。这是我的代码:
for(i=0; i<10; i++) {
aux(i, function(success) {
/*
* this should be made interaction by interaction
* but what happens is: while I'm still running my first interaction
* (i=0), the code loops for i=1, i=2, etc. before the response of
* the previous interaction
*/
if(!success)
doSomething();
else
doSomethingElse();
});
}
function aux(i, success) {
... //here I make my logic with "i" sent as parameter
getReturnFromAjax(function(response) {
if(response)
return success(true);
else
return success(false);
});
});
function getReturnFromAjax(callback) {
...
$.ajax({
url: myUrl,
type: "POST",
success: function (response) {
return callback(response);
}
});
}
最佳答案
jQuery's Deferred
可能有点棘手。您要做的就是将您的诺言链式地堆叠起来。例如:
var
// create a deferred object
dfd = $.Deferred(),
// get the promise
promise = dfd.promise(),
// the loop variable
i
;
for(i = 0; i < 10; i += 1) {
// use `then` and use the new promise for next itteration
promise = promise.then(
// prepare the function to be called, but don't execute it!
// (see docs for .bind)
aux.bind(null, i, function(success) {
success ? doSomethingElse() : doSomething();
})
);
}
// resolve the deferred object
dfd.resolve();
为了使它起作用,
aux
还必须返回一个诺言,但是$.ajax
已经做到了,因此只需传递它,一切都应该起作用:在
aux
中:function aux(i, callback) {
console.log('executing for `aux` with', i);
// return the ajax-promise
return getReturnFromAjax(function(response) {
callback(Boolean(response));
});
}
在
getReturnFromAjax
中:function getReturnFromAjax(callback) {
// return the ajax-promise
return $.ajax({
url: '%your-url%',
type: '%method%',
success: function (response) {
callback(response);
}
});
}
演示:http://jsbin.com/pilebofi/2/
关于javascript - 仅在上一次交互完成后才增加(回调),我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/23630683/