更新时间:2022-03-23 19:38:56
还有一个更好的方法比使用同步Ajax调用做到这一点。 jQuery的AJAX返回一个延迟,所以你可以只使用管道链接,以确保每一个AJAX调用下一个运行之前完成。这里有一个工作的例子有更深入的例子中,你可以上的jsfiddle 的播放。
There's a much better way to do this than using synchronous ajax calls. Jquery ajax returns a deferred so you can just use pipe chaining to make sure that each ajax call finishes before the next runs. Here's a working example with a more in depth example you can play with on jsfiddle.
// How to force async functions to execute sequentially
// by using deferred pipe chaining.
// The master deferred.
var dfd = $.Deferred(), // Master deferred
dfdNext = dfd; // Next deferred in the chain
x = 0, // Loop index
values = [],
// Simulates $.ajax, but with predictable behaviour.
// You only need to understand that higher 'value' param
// will finish earlier.
simulateAjax = function (value) {
var dfdAjax = $.Deferred();
setTimeout(
function () {
dfdAjax.resolve(value);
},
1000 - (value * 100)
);
return dfdAjax.promise();
},
// This would be a user function that makes an ajax request.
// In normal code you'd be using $.ajax instead of simulateAjax.
requestAjax = function (value) {
return simulateAjax(value);
};
// Start the pipe chain. You should be able to do
// this anywhere in the program, even
// at the end,and it should still give the same results.
dfd.resolve();
// Deferred pipe chaining.
// What you want to note here is that an new
// ajax call will not start until the previous
// ajax call is completely finished.
for (x = 1; x <= 4; x++) {
values.push(x);
dfdNext = dfdNext.pipe(function () {
var value = values.shift();
return requestAjax(value).
done(function(response) {
// Process the response here.
});
});
}
有些人评论说,他们不知道什么是code一样。为了了解它,你首先需要了解JavaScript的承诺。我是pretty的肯定承诺很快成为本地JavaScript语言特性,所以应该给你一个很好的学习积极性。
Some people have commented they have no clue what the code does. In order to understand it, you first need to understand javascript promises. I am pretty sure promises are soon to be a native javascript language feature, so that should give you a good incentive to learn.