我有一个回调数组,因此每个回调都会在作业完成时执行另一个回调。例如:
var queue = [
function (done) {
console.log('Executing first job');
setTimeout(done, 1000); // actually an AJAX call here
// I can't call abort() for the request
},
function (done) {
console.log('Executing second job');
setTimeout(done, 1000);
// also AJAX call here
// it is made from a third-party library that I can't change
// this means that I can't call abort() on it
},
function (done) {
console.log('Executing third job');
setTimeout(done, 1000);
},
function (done) {
console.log('Executing fourth job');
setTimeout(done, 1000);
} // and so forth
];
var proceed = true;
(function dequeue() {
var fn = queue.shift();
proceed && fn instanceof Function && fn(dequeue);
})();
对我来说很好,除了为了加快一切,我最好一次启动四个回调,同时仍然可以通过从其他地方更改proceed
标志来停止进一步执行。我该怎么做?
我在这个项目中使用了最新版本的jQuery,所以如果库中有任何东西可以帮助完成这个任务,我会使用它。一切都在浏览器中发生。
答案 0 :(得分:1)
var queue = [];
// let's think we have a thousand of jobs
for (var i = 0; i < 1000; i++) {
(function(jobNumber) {
queue.push(function (done) {
console.log('Started job '+ jobNumber);
setTimeout(function() {
console.log('Finished job '+ jobNumber);
done();
}, 1000);
});
})(i);
}
var proceed = true;
setTimeout(function() {
console.log('Full stop called after 10 seconds');
proceed = false;
}, 10000);
queue.depth = 0;
queue.maxDepth = 4;
(function dequeue() {
while (queue.depth < queue.maxDepth) {
queue.depth += 1;
var fn = queue.shift();
proceed && fn instanceof Function && fn(function() {
queue.depth -= 1;
dequeue();
});
}
})();