JavaScript:4个asyncronys函数在继续之前依次等待对方完成?
我有四个功能:JavaScript:4个asyncronys函数在继续之前依次等待对方完成?,javascript,jquery,asynchronous,google-chrome-extension,callback,Javascript,Jquery,Asynchronous,Google Chrome Extension,Callback,我有四个功能: // NOTE: localDataStore is a function to stringify/parse an item from localStorage function removeCookie() { chrome.cookies.remove({"name":"v1guid","url":"http://website.com"},function (cookie){ console.log("cookie removed"); }); } func
// NOTE: localDataStore is a function to stringify/parse an item from localStorage
function removeCookie() {
chrome.cookies.remove({"name":"v1guid","url":"http://website.com"},function (cookie){
console.log("cookie removed");
});
}
function getCookie() {
chrome.cookies.get({"url": "http://website.com", "name": "v1guid"}, function(cookie) {
console.log(cookie);
if(cookie !== null) {
console.log("getting cookie");
localDataStore.set("cookie", cookie);
console.log(localDataStore.get("cookie"));
}
});
}
function setCookie() {
var cookiedata = localDataStore.get("cookie");
chrome.cookies.set({
"url": "http://website.com",
"name": cookiedata.name,
"value": cookiedata.value,
"domain": cookiedata.domain,
"path": cookiedata.path,
"secure": cookiedata.secure,
"httpOnly": cookiedata.httpOnly,
"storeId": cookiedata.storeId}, function(cookie) {
console.log("cookietest");
console.log(JSON.stringify(cookie));
});
}
function minePosts() {
// Do a bunch of stuff
}
我试图让它们在继续执行下一个函数之前等待彼此执行:
getCookie();
removeCookie();
minePosts();
setCookie();
我知道这可以通过回调来实现,但是相互嵌套的4个回调使得我的代码很难阅读,并且使我的代码模块化程度大大降低——我可能写错了回调,我不确定。我删除了这个帖子的回调,希望有人能帮助我正确地完成它。我如何才能有效地让这4个函数按顺序运行并依次等待对方?我还使用JQuery并阅读了有关命令$.Deferred()的内容,但我不确定这是否是在中使用它的正确做法
有人能帮我吗?我在互联网上搜索,找不到任何人有和我一样的问题,我试图获得4个以上的函数来互相等待。谢谢 您可以使用回调或$.Deferred()来延迟函数。前一种方法更简单,另一方面后一种方法更健壮,因为您可以使用
deferred.reject()
来处理错误状态
1)回调
function removeCookie(callback) {
chrome.cookies.remove({"name":"v1guid","url":"http://website.com"},function (cookie){
console.log("cookie removed");
callback();
});
}
等等
2)$.Deferred()
function removeCookie() {
var deferred = $.Deferred();
chrome.cookies.remove({"name":"v1guid","url":"http://website.com"},function (cookie){
console.log("cookie removed");
deferred.resolve();
});
return deferred.promise();
}
等等
或者按照riovel(jQuery 1.8+)的建议进行链接
这里有一个。如果异步函数返回承诺,则只能使用
$.Deferred
。否则,您需要使用显式回调。那么我应该让它们都返回承诺吗?什么是最好的方式和最合适的方式来做这件事?谢谢!这正是我要找的!结果发现我没有正确地进行回拨,这有助于我澄清问题。我很高兴我帮助了你:)
function removeCookie() {
var deferred = $.Deferred();
chrome.cookies.remove({"name":"v1guid","url":"http://website.com"},function (cookie){
console.log("cookie removed");
deferred.resolve();
});
return deferred.promise();
}
getCookie().then(function(){
removeCookie().then(function(){
minePosts().then(function(){
setCookie().then(function(){
// do something
});
});
});
});
getCookie()
.then(removeCookie)
.then(minePosts)
.then(setCookie)
.then(function(){
// do something
});
function one() {
console.log('one');
deferred = jQuery.Deferred();
setTimeout(function() {
console.log('one done');
deferred.resolve();
}, 1000);
return deferred;
}
function two() {
console.log('two');
deferred = jQuery.Deferred();
setTimeout(function() {
console.log('two done');
deferred.resolve();
}, 1000);
return deferred;
}
function three() {
console.log('three');
deferred = jQuery.Deferred();
setTimeout(function() {
console.log('three done');
deferred.resolve();
}, 1000);
return deferred;
}
function four() {
console.log('four');
deferred = jQuery.Deferred();
setTimeout(function() {
console.log('four done');
deferred.resolve();
}, 1000);
return deferred;
}
one()
.then(two)
.then(three)
.then(four);