jquery using $.deferred

Question

I'm using jQuery $.when function to synchronize two $.getJSON asynchronous calls I have. Merging the data from both calls now works fine, however I want to return the contents of the merge , which is an array of jQuery objects.

I'm trying to achieve this using a $.when function wrapped around another function which contains the code where I synchronize my asynchronous calls. My code looks like this:

 $.fn.getResultFromMergedJsonCalls = function(params){ $.when( $.fn.jsonCall1(params), $.fn.jsonCAll2(params) ).then(function(){ var mergedData = mergeResults(resultOfJsonCAll1,resultOfJsonCAll2) return mergedData; }); }; )

How can I return the contents of the mergedData array to a calling function , i know the nature of the asynchronous calls is causing this behavior so i just need a pattern that solves the problem. Thanks.

You can't. It's asynchronous. You have to process the result inside the callback you pass to .then, or pass it to a function called from inside the callback. — Felix Kling
– Felix Kling, Commented Dec 16, 2011 at 14:32

dgvid · Accepted Answer · 2011-12-16 14:35:14Z

Pass a callback function to getResultsFromMergedJsonCalls, then invoke the callback from the then handler. For example:

$.fn.getResultFromMergedJsonCalls = function(params, success){ $.when( $.fn.jsonCall1(params), $.fn.jsonCAll2(params) ).then(function(){ if (success) { var mergedData = mergeResults(resultOfJsonCAll1,resultOfJsonCAll2); success(mergedData); });

};

The caller of getResultsFromMergedJsonCalls of course must implement a suitable function to pass as that new, callback argument.

SLaks · Accepted Answer · 2011-12-16 14:31:35Z

That's asynchronous; you can't return it.

Instead, you need to pass the value back using a callback.

Collectives™ on Stack Overflow

jquery using $.deferred

2 Answers 2

Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Related