Passing data between deferred functions

Question

I'm using some deferred functions with the .done so, I'm having something like that:

askTime(number).done(formatTime).done(function(html){
   times += html;
});

But although formatTime returns data, the html var has the data returned by askTime.

I don't know where if the problem. formatTime receives a data variable which is returned by askTime but if I put:

askTime(number).done(formatTime(data)).done(function(html)

It says that data isn't defined.

Try askTime(number).done(function(data) {formatTime(data);}).done(function(html) — hungryMind, Dec 02 '11 at 13:42
Inconsistency in Q, But although formatTime returns data & formatTime receives a data variable. Code for askTIme, formatTime will help. — hungryMind, Dec 02 '11 at 13:44
@hungryMind both functions works since both were part of the same function I splitted to be able to use them elsewhere. — Antonio Laguna, Dec 02 '11 at 13:48
@hungryMind also, your first try (although) seemed logicall to me, didn't work :( — Antonio Laguna, Dec 02 '11 at 13:51

score 2 · Accepted Answer · answered Dec 02 '11 at 13:42

2

To chain deferred methods, you need to invoke .pipe().

answered Dec 02 '11 at 13:42

jAndy

I'm having a look but unable to understand correctly. Examples given are a bit hard or unclear (for me at least) – Antonio Laguna Dec 02 '11 at 13:47
@AntonioLaguna Just try ".pipe()" instead of the first ".done()" in your code. – Pointy Dec 02 '11 at 14:00
I picked up the final code from here http://stackoverflow.com/questions/5921283/understanding-jquery-deferred-pipe and works and seems to understandable to me but this is not exactly what comes in the docs. Could you give me more light please? – Antonio Laguna Dec 02 '11 at 14:02
Thanks again @jAndy you seem to be a master of the deferred! – Antonio Laguna Dec 02 '11 at 14:04

1 Answers1