一次设置多个超时 运行
mutliple set timeouts running at once
所以,我有一个 for 循环,我想 运行 一次设置多个超时,这样它就可以连续多次检查是否是某个时间。我不能一次检查所有多次,检查需要在不同的时间后重复。它只会继续重复循环的最后一次迭代,即纬度中的最后一个字典。
我的方法稍微简化了一点:
lat = list of dictionaries with some values that differentiate them + times
for(i = 0; i< lat.length; i++){
if(lat[i][differentiate value]) = {
function checktime(){
if(currenttime != lat[i]){
setTimeout(checktime,30000)
}
else{
console.log("yay the time is right!")
}
}
else if(lat[i][differentiate value]) = {
function checktime(){
if(currenttime != lat[i]){
setTimeout(checktime,50000)
}
else{
console.log("yay the time is right!")
}
}
}
我将如何制作这个(它用于通知应用程序)
原代码:
(后面的每个值看起来像 [[75,null,[7,28,2021]], null,[[9,52,"p"]],"e&r"] 带有通知数量,预设日期,这些提醒的日期,按时间,这将是第一个 if,以及 attimes 的数组。为了测试,我有 2 个 laters,有 2 个不同的 attimes):
chrome.storage.sync.get('later', function(data){
if (data.later !== null){
var lat = data.later;
for(i = 0; i< lat.length; i++){
var currentDict = lat[i];
if(currentDict['notis'][1] !== null){
console.log("cheese")
var by = currentDict['notis'][1];
console.log(by)
const d = new Date();
var hr = d.getHours();
var min = d.getMinutes();
var sec = d.getSeconds();
var da = currentDict['notis'][0][2][1];
var mo = currentDict['notis'][0][2][0];
var ye = currentDict['notis'][0][2][2]
var h = by[0];
var m = by[1];
var ampm = by[2];
if(ampm == "p"){
h = h + 12;
}
var byMS = h*3600000 + m*60000;
var currentMS = hr*3600000 + min*60000 + sec*1000;
//check if right date then check if time is lesss than
function checkdate(){
var day = d.getDate();
var month = d.getMonth() + 1;
var year = d.getFullYear();
if(da == day && mo == month && ye == year){
var amt = 0;
function checktime(){
if(byMS >= currentMS){
//noti and delete
var int = setInterval(function(){
chrome.notifications.create({
title: currentDict['name'],
message: "do da " + currentDict['name'],
iconUrl: "logo.png",
type: "basic"
})
amt++
console.log(amt)
const dddd = new Date();
console.log(dddd.getMinutes() + " / " + dddd.getSeconds())
if(amt >= currentDict['notis'][0][0]){
clearInterval(int)
console.log("done")
//ju finish taht
console.log(lat)
lat.splice(lat.indexOf(currentDict),1)
console.log(lat)
chrome.storage.sync.set({'later': lat})
}
}, (byMS-currentMS)/currentDict['notis'][0][0])
}
else{
setTimeout(checktime,30000)
}
}
checktime();
}
else{
setTimeout(checkdate,66400000)
}
}
checkdate();
}
else if(currentDict['notis'][2] !== null){
console.log("cheese")
var at = currentDict['notis'][2];
console.log(at)
var arrayat = [];
for(j = 0; j<= at.length-1; j++){
var atcycle = at[j];
console.log(atcycle)
const ddd = new Date();
var hr = ddd.getHours();
var min = ddd.getMinutes();
var da = currentDict['notis'][0][2][1];
var mo = currentDict['notis'][0][2][0];
var ye = currentDict['notis'][0][2][2]
var hrat = atcycle[0];
var minat = atcycle[1];
var ampm = atcycle[2];
if(ampm == "p"){
hrat = hrat + 12;
}
console.log(hrat + "/" + minat + "/" + ampm)
if(hr <= hrat){
var temparray = [];
temparray.push(hrat)
temparray.push(minat)
arrayat.push(temparray);
console.log(arrayat)
}
else if(hr == hrat){
if(min<minat){
var temparray = [];
temparray.push(hrat)
temparray.push(minat)
arrayat.push(temparray);
console.log(arrayat)}
}
}
console.log(arrayat.length)
function checkdate(){
console.log(arrayat.length)
console.log("date")
const d = new Date();
var day = d.getDate();
var month = d.getMonth() + 1;
var year = d.getFullYear();
if(da == day && mo == month && ye == year){
function check(){
console.log(arrayat.length)
console.log("check")
for(l=0; l<arrayat.length; l++){
console.log(arrayat.length)
const dd = new Date();
var hr = dd.getHours();
var min = dd.getMinutes();
console.log(arrayat[l][1])
console.log(min)
if(arrayat[l][0] == hr && arrayat[l][1] == min ){ //at one of the times
console.log(arrayat)
arrayat.splice(l,1)
console.log(arrayat)
if(arrayat.length == 0){
lat.splice(lat.indexOf(currentDict),1)
chrome.storage.sync.set({'later': lat})
console.log(lat)
}
chrome.notifications.create({
title: currentDict['name'],
message: "do da " + currentDict['name'],
iconUrl: "logo.png",
type: "basic"
})
//add noti with name and delete it
console.log(arrayat)
check();
}
}
console.log(arrayat.length)
if(arrayat.length !== 0){
console.log("and repeat")
setTimeout(check,15000);//SETINTERVAL INSTEAD? ANDCLEAR
}
}
check();
}
else{
setTimeout(checkdate,66400000)
}
}
checkdate();
}
}
}
})
}
这是错误的做法。你知道时间,所以你知道他们在时间上有多远。如果时间到了,您无需一遍又一遍地检查。而是算出距离相关时间还有多少毫秒,然后为该时间设置超时。
你基本上需要将不同的函数传递给每个 setTimeout()
:
function checktime1(){
if(currenttime != lat[i]){
setTimeout(checktime1,30000)
}
else{
console.log("yay the time is right!")
}
}
function checktime2(){
if(currenttime != lat[i]){
setTimeout(checktime2,50000)
}
else{
console.log("yay the time is right!")
}
}
但是,这显然是不可扩展的。如果你有很多超时,你会写很多重复的函数。更糟糕的是,如果您有动态数量的超时,这将无法解决您的问题。
幸运的是,javascript 让我们可以在另一个函数中声明函数:
function checktime(timeout){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout)
}
else{
console.log("yay the time is right!")
}
}
loop();
}
这允许您根据需要动态创建函数:
checktime(30000);
checktime(50000);
如果您需要将参数传递给每个超时(例如,如果您需要显示自定义警报),您可以将它传递给您的外部函数,它将在闭包中捕获:
function checktime(timeout, message){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout)
}
else{
console.log(message)
}
}
loop();
}
那么你可以这样做:
checktime(30000, "Yay, the time is right for timer 1");
checktime(50000, "Yay, the time is right for timer 2");
同样,如果您需要为每个定时器执行自定义逻辑,您也可以将其作为参数(回调)传递:
function checktime(timeout, callback){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout);
}
else{
console.log("yay the time is right!");
callback(); // <----------------- execute custom logic
}
}
loop();
}
你可以像这样使用:
checktime(30000, () -> console.log("TIMER 1"));
checktime(50000, () -> alert("TIMER 2"));
所以,我有一个 for 循环,我想 运行 一次设置多个超时,这样它就可以连续多次检查是否是某个时间。我不能一次检查所有多次,检查需要在不同的时间后重复。它只会继续重复循环的最后一次迭代,即纬度中的最后一个字典。
我的方法稍微简化了一点:
lat = list of dictionaries with some values that differentiate them + times
for(i = 0; i< lat.length; i++){
if(lat[i][differentiate value]) = {
function checktime(){
if(currenttime != lat[i]){
setTimeout(checktime,30000)
}
else{
console.log("yay the time is right!")
}
}
else if(lat[i][differentiate value]) = {
function checktime(){
if(currenttime != lat[i]){
setTimeout(checktime,50000)
}
else{
console.log("yay the time is right!")
}
}
}
我将如何制作这个(它用于通知应用程序)
原代码: (后面的每个值看起来像 [[75,null,[7,28,2021]], null,[[9,52,"p"]],"e&r"] 带有通知数量,预设日期,这些提醒的日期,按时间,这将是第一个 if,以及 attimes 的数组。为了测试,我有 2 个 laters,有 2 个不同的 attimes):
chrome.storage.sync.get('later', function(data){
if (data.later !== null){
var lat = data.later;
for(i = 0; i< lat.length; i++){
var currentDict = lat[i];
if(currentDict['notis'][1] !== null){
console.log("cheese")
var by = currentDict['notis'][1];
console.log(by)
const d = new Date();
var hr = d.getHours();
var min = d.getMinutes();
var sec = d.getSeconds();
var da = currentDict['notis'][0][2][1];
var mo = currentDict['notis'][0][2][0];
var ye = currentDict['notis'][0][2][2]
var h = by[0];
var m = by[1];
var ampm = by[2];
if(ampm == "p"){
h = h + 12;
}
var byMS = h*3600000 + m*60000;
var currentMS = hr*3600000 + min*60000 + sec*1000;
//check if right date then check if time is lesss than
function checkdate(){
var day = d.getDate();
var month = d.getMonth() + 1;
var year = d.getFullYear();
if(da == day && mo == month && ye == year){
var amt = 0;
function checktime(){
if(byMS >= currentMS){
//noti and delete
var int = setInterval(function(){
chrome.notifications.create({
title: currentDict['name'],
message: "do da " + currentDict['name'],
iconUrl: "logo.png",
type: "basic"
})
amt++
console.log(amt)
const dddd = new Date();
console.log(dddd.getMinutes() + " / " + dddd.getSeconds())
if(amt >= currentDict['notis'][0][0]){
clearInterval(int)
console.log("done")
//ju finish taht
console.log(lat)
lat.splice(lat.indexOf(currentDict),1)
console.log(lat)
chrome.storage.sync.set({'later': lat})
}
}, (byMS-currentMS)/currentDict['notis'][0][0])
}
else{
setTimeout(checktime,30000)
}
}
checktime();
}
else{
setTimeout(checkdate,66400000)
}
}
checkdate();
}
else if(currentDict['notis'][2] !== null){
console.log("cheese")
var at = currentDict['notis'][2];
console.log(at)
var arrayat = [];
for(j = 0; j<= at.length-1; j++){
var atcycle = at[j];
console.log(atcycle)
const ddd = new Date();
var hr = ddd.getHours();
var min = ddd.getMinutes();
var da = currentDict['notis'][0][2][1];
var mo = currentDict['notis'][0][2][0];
var ye = currentDict['notis'][0][2][2]
var hrat = atcycle[0];
var minat = atcycle[1];
var ampm = atcycle[2];
if(ampm == "p"){
hrat = hrat + 12;
}
console.log(hrat + "/" + minat + "/" + ampm)
if(hr <= hrat){
var temparray = [];
temparray.push(hrat)
temparray.push(minat)
arrayat.push(temparray);
console.log(arrayat)
}
else if(hr == hrat){
if(min<minat){
var temparray = [];
temparray.push(hrat)
temparray.push(minat)
arrayat.push(temparray);
console.log(arrayat)}
}
}
console.log(arrayat.length)
function checkdate(){
console.log(arrayat.length)
console.log("date")
const d = new Date();
var day = d.getDate();
var month = d.getMonth() + 1;
var year = d.getFullYear();
if(da == day && mo == month && ye == year){
function check(){
console.log(arrayat.length)
console.log("check")
for(l=0; l<arrayat.length; l++){
console.log(arrayat.length)
const dd = new Date();
var hr = dd.getHours();
var min = dd.getMinutes();
console.log(arrayat[l][1])
console.log(min)
if(arrayat[l][0] == hr && arrayat[l][1] == min ){ //at one of the times
console.log(arrayat)
arrayat.splice(l,1)
console.log(arrayat)
if(arrayat.length == 0){
lat.splice(lat.indexOf(currentDict),1)
chrome.storage.sync.set({'later': lat})
console.log(lat)
}
chrome.notifications.create({
title: currentDict['name'],
message: "do da " + currentDict['name'],
iconUrl: "logo.png",
type: "basic"
})
//add noti with name and delete it
console.log(arrayat)
check();
}
}
console.log(arrayat.length)
if(arrayat.length !== 0){
console.log("and repeat")
setTimeout(check,15000);//SETINTERVAL INSTEAD? ANDCLEAR
}
}
check();
}
else{
setTimeout(checkdate,66400000)
}
}
checkdate();
}
}
}
})
}
这是错误的做法。你知道时间,所以你知道他们在时间上有多远。如果时间到了,您无需一遍又一遍地检查。而是算出距离相关时间还有多少毫秒,然后为该时间设置超时。
你基本上需要将不同的函数传递给每个 setTimeout()
:
function checktime1(){
if(currenttime != lat[i]){
setTimeout(checktime1,30000)
}
else{
console.log("yay the time is right!")
}
}
function checktime2(){
if(currenttime != lat[i]){
setTimeout(checktime2,50000)
}
else{
console.log("yay the time is right!")
}
}
但是,这显然是不可扩展的。如果你有很多超时,你会写很多重复的函数。更糟糕的是,如果您有动态数量的超时,这将无法解决您的问题。
幸运的是,javascript 让我们可以在另一个函数中声明函数:
function checktime(timeout){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout)
}
else{
console.log("yay the time is right!")
}
}
loop();
}
这允许您根据需要动态创建函数:
checktime(30000);
checktime(50000);
如果您需要将参数传递给每个超时(例如,如果您需要显示自定义警报),您可以将它传递给您的外部函数,它将在闭包中捕获:
function checktime(timeout, message){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout)
}
else{
console.log(message)
}
}
loop();
}
那么你可以这样做:
checktime(30000, "Yay, the time is right for timer 1");
checktime(50000, "Yay, the time is right for timer 2");
同样,如果您需要为每个定时器执行自定义逻辑,您也可以将其作为参数(回调)传递:
function checktime(timeout, callback){
function loop () {
if(currenttime != lat[i]){
setTimeout(loop,timeout);
}
else{
console.log("yay the time is right!");
callback(); // <----------------- execute custom logic
}
}
loop();
}
你可以像这样使用:
checktime(30000, () -> console.log("TIMER 1"));
checktime(50000, () -> alert("TIMER 2"));