var p1 = new Promise(function(resolve, reject) {
setTimeout(() => resolve("first"), 5000);
});
var p2 = new Promise(function(resolve, reject) {
setTimeout(() => resolve("second"), 2000);
});
var p3 = new Promise(function(resolve, reject) {
setTimeout(() => resolve("third"), 1000);
});
console.log("last to print");
p1.then(()=>p2).then(()=>p3).then(()=> console.log("last to be printed"))
As I was reading about promises, I know that I can print promises synchronous (in this case print: first, second, third, last to print) when I use async /await. Now I have also been reading that the same thing can be achieved using .then chaining and async/await is nothing 'special'. When I try to chain my promises, however, nothing happens except for the console.log of "last to be printed". Any insight would be great! Thanks!!
Edit to question:
var p1 = new Promise(function (resolve, reject) {
setTimeout(() => console.log("first"), 5000);
resolve("first resolved")
});
var p2 = new Promise(function (resolve, reject) {
setTimeout(() => console.log("second"), 2000);
resolve("second resolved")
});
var p3 = new Promise(function (resolve, reject) {
setTimeout(() => console.log("third"), 0);
resolve("third resolved")
});
console.log("starting");
p1.then((val) => {
console.log("(1)", val)
return p2
}).then((val) => {
console.log("(2)", val)
return p3
}).then((val) => {
console.log("(3)", val)
})
Loggs:
starting
(1) first resolved
(2) second resolved
(3) third resolved
third
second
first
1: if executor function passed to new Promise is executed immediately, before the new promise is returned, then why are here promises resolved ()synchronously) first and after the setTimeouts (asynchronously) gets executed?
Return value vs. resolve promise:
var sync = function () {
return new Promise(function(resolve, reject){
setTimeout(()=> {
console.log("start")
resolve("hello") //--works
// return "hello" //--> doesnt do anything
}, 3000);
})
}
sync().then((val)=> console.log("val", val))
The executor function you pass to new Promise is executed immediately, before the new promise is returned. So when you do:
var p1 = new Promise(function(resolve, reject) {
setTimeout(() => resolve("first"), 5000);
});
...by the time the promise is assigned to p1, the setTimeout has already been called and scheduled the callback for five seconds later. That callback happens whether you listen for the resolution of the promise or not, and it happens whether you listen for resolution via the await keyword or the then method.
So your code starts three setTimeouts immediately, and then starts waiting for the first promise's resolution, and only then waiting for the second promise's resolution (it'll already be resolved, so that's almost immediate), and then waiting for the third (same again).
To have your code execute those setTimeout calls only sequentially when the previous timeout has completed, you have to not create the new promise until the previous promise resolves (using shorter timeouts to avoid lots of waiting):
console.log("starting");
new Promise(function(resolve, reject) {
setTimeout(() => resolve("first"), 1000);
})
.then(result => {
console.log("(1) got " + result);
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("second"), 500);
});
})
.then(result => {
console.log("(2) got " + result);
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("third"), 100);
});
})
.then(result => {
console.log("(3) got " + result);
console.log("last to print");
});
Remember that a promise doesn't do anything, and doesn't change the nature of the code in the promise executor. All a promise does is provide a means of observing the result of something (with really handy combinable semantics).
Let's factor out the common parts of those three promises into a function:
function delay(ms, ...args) {
return new Promise(resolve => {
setTimeout(resolve, ms, ...args);
});
}
Then the code becomes a bit clearer:
function delay(ms, ...args) {
return new Promise(resolve => {
setTimeout(resolve, ms, ...args);
});
}
console.log("starting");
delay(1000, "first")
.then(result => {
console.log("(1) got " + result);
return delay(500, "second");
})
.then(result => {
console.log("(2) got " + result);
return delay(100, "third");
})
.then(result => {
console.log("(3) got " + result);
console.log("last to print");
});
Now, let's put that in an async function and use await:
function delay(ms, ...args) {
return new Promise(resolve => {
setTimeout(resolve, ms, ...args);
});
}
(async() => {
console.log("starting");
console.log("(1) got " + await delay(1000, "first"));
console.log("(2) got " + await delay(500, "second"));
console.log("(3) got " + await delay(100, "third"));
console.log("last to print");
})();
Promises make that syntax possible, by standardizing how we observe asynchronous processes.
Re your edit:
1: if executor function passed to new Promise is executed immediately, before the new promise is returned, then why are here promises resolved ()synchronously) first and after the setTimeouts (asynchronously) gets executed?
There are two parts to that question:
A) "...why are here promises resolved ()synchronously) first..."
B) "...why are here promises resolved...after the setTimeouts (asynchronously) gets executed"
The answer to (A) is: Although you resolve them synchronously, then always calls its callback asynchronously. It's one of the guarantees promises provide. You're resolving p1 (in that edit) before the executor function returns. But the way you're observing the resolutions ensures that you observe the resolutions in order, because you don't start observing p2 until p1 has resolved, and then you don't start observing p3 until p2 is resolved.
The answer to (B) is: They don't, you're resolving them synchronously, and then observing those resolutions asynchronously, and since they're already resolved that happens very quickly; later, the timer callbacks run. Let's look at how you create p1 in that edit:
var p1 = new Promise(function (resolve, reject) {
setTimeout(() => console.log("first"), 5000);
resolve("first resolved")
});
What happens there is:
new Promise gets called
It calls the executor function
The executor function calls setTimeout to schedule a callback
You immediately resolve the promise with "first resolved"
new Promise returns and the resolved promise is assigned to p1
Later, the timeout occurs and you output "first" to the console
Then later you do:
p1.then((val) => {
console.log("(1)", val)
return p2
})
// ...
Since then always calls its callback asynchronously, that happens asynchronously — but very soon, because the promise is already resolved.
So when you run that code, you see all three promises resolve before the first setTimeout callback occurs — because the promises aren't waiting for the setTimeout callback to occur.
You may be wondering why you see your final then callback run before you see "third" in the console, since both the promise resolutions and the console.log("third") are happening asynchronously but very soon (since it's a setTimeout(..., 0) and the promises are all pre-resolved): The answer is that promise resolutions are microtasks and setTimeout calls are macrotasks (or just "tasks"). All of the microtasks a task schedules are run as soon as that task finishes (and any microtasks that they schedule are then executed as well), before the next task is taken from the task queue. So the task running your script does this:
Schedules a task for the setTimeout callback
Schedules a microtask to call p1's then callback
When the task ends, its microtasks are processed:
The first then handler is run, scheduling a microtask to run the second then handler
The second then handler runs and schedules a micro task to call the third then handler
Etc. until all the then handlers have run
The next task is picked up from the task queue. It's probably the setTimeout callback for p3, so it gets run and "third" appears in the console
Return value vs. resolve promise:
The part you've put in the question doesn't make sense to me, but your comment on this does:
I read that returning a value or resolving a promise is same...
What you've probably read is that returning a value from then or catch is the same as returning a resolved promise from then or catch. That's because then and catch create and return new promises when they're called, and if their callbacks return a simple (non-promise) value, they resolve the promise they create with that value; if the callback returns a promise, they resolve or reject the promise they created based on whether that promise resolves or rejects.
So for instance:
.then(() => {
return 42;
})
and
.then(() => {
return new Promise(resolve => resolve(42));
})
have the same end result (but the second one is less efficient).
Within a then or catch callback:
Returning a non-promise resolves the promise then/catch created with that value
Throwing an error (throw ...) rejects that promise with the value you throw
Returning a promise makes then/catch's promise resolve or reject based on the promise the callback returns
You cannot make asynchronous code execute synchronously.
Even async / await are just syntax that gives you a synchronous-style control flow inside a promise.
When I try to chain my promises, however, nothing happens except for the console.log of "last to be printed". Any insight would be great!
The other functions don't generate any output. That has nothing to do with them being in promises.
You start three timers (all at the same time), then log 'last to print', then chain some promises so that 'last to be printed' will print when all three promises resolve (5 seconds after you start them all going).
If you want the timers to run sequentially, then you have to initiate them only when the previous one has finished, and if you want to see what they resolve with then you have to write code that actually looks at that.
function p1() {
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("first"), 5000);
});
}
function p2() {
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("second"), 2000);
});
}
function p3() {
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("third"), 1000);
});
}
function log(value) {
console.log("Previous promise resolved with " + value);
}
p1()
.then(log)
.then(p2)
.then(log)
.then(p3)
.then(log)
.then(() => console.log("last to be printed"));
Async/await is, arguably, neater:
function p1() {
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("first"), 5000);
});
}
function p2() {
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("second"), 2000);
});
}
function p3() {
return new Promise(function(resolve, reject) {
setTimeout(() => resolve("third"), 1000);
});
}
function log(value) {
console.log("Previous promise resolved with " + value);
}
(async function() {
log(await p1());
log(await p2());
log(await p3());
console.log("last to be printed");
}());
If you need to call an await but the function that contains that await doesn't have to be async, because you need, for example, a "number" and not a "Promise number ", you can do de next:
var ex: number = new Number(async resolve => {
var f = await funcionExample();
resolve(f);
}).valueOf();
Related
in one of youtube tutorial videos about promises I found following code:
let cleanRoom = function() {
return new Promise((resolve, reject) => {
resolve();
});
};
let removeGarbage = function() {
return new Promise((resolve, reject) => {
resolve();
});
};
let winIcecream = function() {
return new Promise((resolve, reject) => {
resolve();
});
};
cleanRoom().then(function(){
return removeGarbage();
}).then(function() {
return winIcecream();
}).then(function() {
console.log('finished');
})
Why the promises aren't chained like that the every .then word is after the previous promise? I mean, why for example .then is not immediately after removeGarbage() but it is after the cleanRoom().then(), how is it happening that the winIcecream() will run after resolving the removeGarbage promise?
Also do I need to type return declaring every promise like in the code above? If yes, why do I need to do so?
Your initial questions may be answered by rewriting things to variable assignments.
I'm using arrow function syntax to implicitly return the new expression here; if you're using regular functions, then yes, you do have to return the new chained promise if you want to run them in sequence.
const roomCleanedPromise = cleanRoom();
const roomCleanedAndGarbageTakenOutPromise = roomCleanedPromise.then(() => removeGarbage());
const roomCleanedAndGarbageTakenOutAndIcecreamWonPromise = roomCleanedAndGarbageTakenOutPromise.then(() => winIcecream());
const finishedPromise = roomCleanedAndGarbageTakenOutAndIcecreamWonPromise.then(() => console.log('finished'));
However things are easier written using the more modern async/await syntax - the YouTube tutorial you mention is a little outdated, perhaps.
async function cleanRoom() {
console.log('Cleaning room.');
// this could do other async things or just take a while
return {room: 'clean'}; // just to demonstrate a return value
}
async function removeGarbage() {
console.log('Removing garbage.');
// this could do other async things or just take a while
return {garbage: 'removed'};
}
// third function elided for brevity
async function doAllTheThings() {
const roomStatus = await cleanRoom();
const garbageStatus = await removeGarbage();
console.log('finished');
}
The purpose of using a fulfillment handler (the functions passed to then in your example) is to wait for the promise to be fulfilled and then, at that point, do something else.
The goal of that code (apparently) is to wait until the cleanRoom promise is fulfilled, then start the removeGarbage process, and then when that is fulfilled, start the winIcecream process. It's also worth noting that if cleanRoom's promise was rejected instead of being fulfilled, removeGarbage wouldn't happen at all, because it's in a fulfillment handler, not a rejection handler.
If you did this instead:
cleanRoom().then(function() { /*...*/ });
removeGarbage().then(function() { /*...*/ });
winIcecream().then(function() { /*...*/ });
...all three processes would be started immediately and run in parallel (to the extent whatever async process they're modelling can run in parallel with other JavaScript code). There'd be no coordination between them at all.
...how is it happening that the winIcecream() will run after resolving the removeGarbage promise...
then, catch, and finally create and return new promises. Those promises are fulfilled or rejected based on what happens to the promise they were called on and what happens in or is returned by their handler. So for example:
doThis()
.then(function() { return doThat(); })
.then(function() { console.log("done"); });
Let's call the promise from doThis() "Promise A". Calling then on it creates a new promise ("Promise B"), that will either be rejected (if Promise A is rejected) or will call its handler if Promise A is fulfilled. Promise B is resolved to whatever that handler returns. In the code above, suppose Promise A is fulfilled and doThat() returns a promise ("Promise C"). Now, Promise B is resolved to Promise C — whatever happens to Promise C is what will happen to Promise B. If Promise C is fulfilled, Promise B is fulfilled, and the second handler with the console.log is called.
The MDN article on using promises may be helpful.
Your suggestion / thinking I mean, why for example .then is not immediately after removeGarbage() but it is after the cleanRoom().then() is wrong.
Each promise ( .then ) is not execute immediately.
You can take a look at your example ( little edited by me )
const cleanRoom = () => new Promise((resolve, reject) => {
resolve();
});
const removeGarbage = () => new Promise((resolve, reject) => {
resolve();
});
const winIcecream = () => new Promise((resolve, reject) => {
resolve();
})
cleanRoom().then(function(){
console.log('second');
return removeGarbage();
}).then(function() {
return winIcecream();
}).then(function() {
console.log('finished');
})
console.log('first');
You should read more how the event loop works.
I've been playing with Promises and I put together the following code that isn't acting like I'm expecting it to. You can run it in a Fiddle here:
https://fiddle.sencha.com/#view/editor&fiddle/1pmk
function log(txt) {
var lapsed = new Date().getTime() - START;
console.log(lapsed, txt);
}
START = new Date().getTime();
var promise = new Promise(function promiseExecutor(resolve, reject) {
log('in Promise 1 Executor');
setTimeout(function onSetTimeout() {
log('in Promise 1 Timeout');
resolve(123);
}, 500);
});
var promise2 = new Promise(function promise2Executor(resolve, reject) {
log('in Promise 2 Executor');
setTimeout(function onSetTimeout() {
log('in Promise 2 Timeout');
resolve(123);
}, 1000);
});
// sequence
promise
.then(promise2)
.then(function onPromiseDone(p) {
// TODO: why this executes before promise2 resolve?
log('ALL Promise Done! ' + p);
});
If you watch the Console, you'll notice that the "ALL Promise Done" message is fired immediately after the FIRST promise is resolved, and then half a second later the 2nd promise is resolved...
How is it possible that the final then(onPromiseDone) is fired without resolve()?
Now, I understand that the Promise executor function starts running immediately, and this here is not a good example for "sequencing". I am able to achieve the correct behavior by wrapping my promises in "deferred" functions as shown here:
https://fiddle.sencha.com/#view/editor&fiddle/1pml
Still, this doesn't answer my question for the original fiddle, which I'll repeat:
How is it possible that the final then(onPromiseDone) is fired without resolve()?
Thanks!
.then receives two arguments: a function that triggers when the Promise successfully resolves and a function that's invoked when the Promise is rejected (can be ommited).
In your example, you're passing a new promisse as an argument.
In order to achieve your goal (chain), I suggest something like that:
var promise = new Promise(function (resolve, reject) {
log('in Promise 1 Executor');
setTimeout(function onSetTimeout() {
log('in Promise 1 Timeout');
resolve(123);
}, 2000);
});
var callback = function(val) {
log('received from the first promisse: ' + val)
return new Promise(function (resolve, reject) {
log('in Promise 2 Executor');
setTimeout(function onSetTimeout() {
log('in Promise 2 Timeout');
resolve(456);
}, 2000);
});
}
// Promise.all([promise, promise2]) // parallelize
// sequence
promise
.then(callback)
.then(function(val) {
log('received from the second promisse: ' + val)
log('ALL Promise Done! ');
}
);
This way, you run your promisse and set the callback function as .then parameter. So, just when the first setTimeout finishes, it will call resolve and invoke callback function, that will create a new Promise and then set the new setTimeout.
Will output:
0 "in Promise 1 Executor"
2002 "in Promise 1 Timeout"
2002 "received from the first promisse: 123"
2002 "in Promise 2 Executor"
4002 "in Promise 2 Timeout"
4003 "received from the second promisse: 456"
4003 "ALL Promise Done! "
Working fiddle: https://jsfiddle.net/mrlew/L8b8fow9/
Edited after #Bergi's comment. Thanks.
I'm trying to learn how to use promises, but am having trouble comprehending the chaining. I assume that with this code, both promises will run. Then when I call test.then() it should know that test has resolved and pass the resolve data to then().
Once that function finishes, it goes onto the next then(), repeating the same process with the test2 promise.
However, I can only get it to print out the first promise results, not the second. Any ideas what is missing here?
var test = new Promise(function(resolve, reject){
resolve('done1');
});
var test2 = new Promise(function(resolve, reject){
resolve('done2');
});
test
.then(function(data) {
console.log(data);
})
.then(test2)
.then(function(data) {
console.log(data);
});
Your first .then call is returning undefined, whereas any subsequent .then is expecting a returned promise. So you'd need to change your code to:
var test = new Promise(function(resolve, reject){
resolve('done1');
});
var test2 = new Promise(function(resolve, reject){
resolve('done2');
});
test
.then(function(data) {
console.log(data);
return test2;
})
.then(resultOfTest2 => doSomething)
.then(function(data) {
console.log(data);
});
You need to return next promise from the then callback:
test.then(function(data) {
console.log(data);
return test2;
}).then(function(data) {
console.log(data);
});
Summary:
The basic concept of promise chaining with promises is that every then / catch method on a fulfilled promise returns another promise. It works in the following manner:
When a promise is resolved the callback passed in the then method is called. The then method wraps the value which is returned in its callback in a resolved promise and returns this resolved promise.
When a promise is rejected the callback passed in the catch method is called. The catch method wraps the value which is returned in its callback in a rejected promise and returns this rejected promise.
Example:
Before fully understanding the concept of chaining multiple then methods it is important to know what exactly the return values of then and catch are. Take the following example:
let prom1 = new Promise((res, rej) => {
res('res');
});
const resolvedProm1 = prom1.then((val) => {return val});
// setTimeout needed for the promise to actually be resolved
setTimeout(() => console.log(resolvedProm1));
let prom2 = new Promise((res, rej) => {
rej('rej');
});
const resolvedProm2 = prom2.catch((err) => {throw err});
// setTimeout needed for the promise to actually be rejected
setTimeout(() => console.log(resolvedProm2));
We can observe the status of the promises in the chrome devtools:
What basically happens is that in a then or catch callback is the following:
Any value returned in a then or catch callback is wrapped in Promise.resolve() and a new resolved promise is returned.
Any error thrown in a then or catch callback is wrapped in Promise.reject() and a new rejected promise is returned.
Because we are getting returned a rejected or resolved promise object we can repeat the cycle and call the then or catch method on it again. For example:
const prom = new Promise((res, rej) => {
if (Math.random() > 0.5) {
res('success');
} else {
rej('error');
}
});
prom.then((val) => {
return val;
}).then((val) => {
return val
}).then((val) => {
console.log(val)
}).catch((err) => {
console.log('err');
})
This calling of then and catch methods which are executed in their respective order is called promise chaining. It is a very useful technique to make working with asynchronous code easier, especially if multiple asynchronous operations need to be performed which are dependend on each others data.
you need to return the other promise(test2) in the first promise (test1) to allow for chaining:
var test = new Promise(function(resolve, reject){
resolve('done1');
});
var test2 = new Promise(function(resolve, reject){
resolve('done2');
});
test
.then(function(data) {
console.log(data);
return test2;
});
You may also want to try -
let test = new Promise(function(resolve, reject){
resolve('done1');
});
let test2 = new Promise(function(resolve, reject){
resolve('done2');
});
try {
let logOne = test();
let logTwo = test2();
console.log(logOne);
console.log(logTwo);
} catch(error) {
console.error(error);
}
In this way, you can also properly handle any promise dependencies. For example if test one relied on test two's data your could -
try {
let logOne = test();
let logTwo = test2(logOne);
console.log(logOne);
console.log(logTwo);
} catch(error) {
console.error(error);
}
I have the following promise-returning function:
function createJourney() {
return new Promise((resolve, reject) => {
// code ...
doOperation((err, data) {
// code ...
return reject('We need to exit now!')
});
// ---> Why is code stil executing here? <---
})
}
Why is the code executing below the reject? When rejecting, I want to stop the execution of the createJourney function.
I am using Bluebird promise.
Imagine
return new Promise((resolve, reject) => {
doThing1()
doOperation(..);
doThing2()
})
there is no reason why doThing2() should not be executed depending on what happens inside doOperation()
doOperation will probably start an asynchronous operation and doThing2 will be called long before reject is called in your example.
How could I fire off promises one after the other?
waitFor(t), is a function that returns a promise that resolves after t time. What I wish to be able to do with that is:
waitFor(1000) Then when finished, console.log('Finished wait of 1000 millis') then
waitFor(2000) Then when finished, console.log('Finished wait of 2000 millis') then
waitFor(3000) Then when finished, console.log('Finished wait of 3000 millis')
Here is what I tried:
waitFor(1000).then(function(resolve, reject) {
console.log(resolve);
}).then(waitFor(2000).then(function(resolve, reject) {
console.log(resolve);
})).then(waitFor(3000).then(function(resolve, reject) {
console.log(resolve);
}));
Unfortunately this console.logs the statements each 1 second after another, which means that the promises where all called at once.
I managed to fix this with callbacks like so, yet that makes everything very ugly:
waitFor(1000).then(function(resolve, reject) {
console.log(resolve+' # '+(new Date().getSeconds()));
waitFor(2000).then(function(resolve, reject) {
console.log(resolve+' # '+(new Date().getSeconds()));
waitFor(3000).then(function(resolve, reject) {
console.log(resolve+' # '+(new Date().getSeconds()));
});
});
});
So how should I do this with promises that makes it work, yet isn't using ugly callback hell?
Undesired result: http://jsfiddle.net/nxjd563r/1/
Desired result: http://jsfiddle.net/4xxps2cg/
I found your solution.
You need to have each then to return a new promise, so that the next then can react once the previous one has been resolved.
waitFor(1000).then(function(result) {
$('#result').append(result+' # '+(new Date().getSeconds())+'<br>');
return waitFor(2000);
}).then(function(result) {
$('#result').append(result+' # '+(new Date().getSeconds())+'<br>');
return waitFor(3000);
}).then(function(result) {
$('#result').append(result+' # '+(new Date().getSeconds())+'<br>');
});
jsfiddle http://jsfiddle.net/4xxps2cg/2/
You can put promises into array and use reduce to chain them, starting with one extra resolved promise.
function waitPromise(time) {
//console.log(time);
return new Promise( (resolve,reject) => {
setTimeout( () => {resolve('resolved');}, time);
});
}
function log(data) {
return new Promise( (resolve,reject) => {
console.log( data +' # '+(new Date().getSeconds()));
resolve();
});
}
var ps = [];
for (var i=0;i<3;i++) {
let time = (i+1) * 1000;
ps.push( () => waitPromise(time) );
ps.push( log );
}
console.log( 'started' +' # '+(new Date().getSeconds()));
var p = Promise.resolve();
ps.reduce( (p,c) => {return p.then(c)}, p);
The format for calling and waiting is a bit off, your then should be a function that returns a promise, since now you're passing a function call instead of a function, its running that request instantly instead of waiting to call the function as a result of the promise.
This should do it:
function waitFor(timeout) {
return new Promise(function(resolve, reject) {
setTimeout(function() {
resolve(`Finished waiting ${timeout} milliseconds`);
}, timeout);
});
}
waitFor(1000).then(function(resolve, reject) {
$('#result').append(resolve+' # '+new Date().getSeconds()+'<br/>');
}).then(function(){
return waitFor(2000)
}).then(function(resolve, reject) {
$('#result').append(resolve+' # '+new Date().getSeconds()+'<br/>');
}).then(function() {
return waitFor(2000)
}).then(function(resolve, reject) {
$('#result').append(resolve+' # '+new Date().getSeconds()+'<br/>');
})
waitFor() appear to be called immediately at .then() ; try returning waitFor() from within .then() anonymous function.
Could alternatively create array of duration values , use Array.prototype.shift() to call waitFor with each duration value in succession , or pass parameter timeout to waitFor ; if same process called at each .then() , include process at .then() chained to Promise in waitFor ; call same function waitFor at .then() chained to initial waitFor() call
var t = [1000, 2000, 3000];
function waitFor(timeout) {
return new Promise(function (resolve, reject) {
setTimeout(function () {
resolve("`Finished waiting ${timeout} milliseconds`");
}, timeout && t.shift() || t.shift());
}).then(function (data) {
$('#result').append(data + ' # ' + new Date().getSeconds() + '<br/>');
})
}
waitFor().then(waitFor).then(waitFor)
//.then(function() {return waitFor(5000)})