jQuery AJAX listener

Question

If I have object myApi with execute function

var api = new myApi(); api.execute();

Inside execute I have (*that is myApi instance)

 function execute() { $.ajax({ type: this.getRequestMethod(), data: this.getDataParams(), complete: function(xmlHttp){ that.setResult(jQuery.parseJSON(xmlHttp.responseText)); that.setHttpStatus(xmlHttp.status); }, url: this.getUrl(), beforeSend: setHeader }); }

How can I make callback/listener so I can do this

var api = new myApi(); api.execute(); var result = api.getResult(); var statusCode = api.getStatusCode(); switch(statusCode) {...};

if I leave it just this way, these bottom two lines are executed before ajax call is finished (complete isn't called yet) so I have undefined variables.

jAndy · Accepted Answer · 2012-03-11 22:11:09Z

You can't do it that way, unless you would force the AJAX requests to be syncronous (which probably is a bad idea). You need need to attach somekind of callback method, you can also use some jQuery Deferred magic.

Therefore, return the jqXHR object which encapsulates a Deferred:

function execute() { return $.ajax({ type: this.getRequestMethod(), data: this.getDataParams(), complete: function(xmlHttp){ that.setResult(jQuery.parseJSON(xmlHttp.responseText)); that.setHttpStatus(xmlHttp.status); }, url: this.getUrl(), beforeSend: setHeader }); }

and then use it like

var api = new myApi(); var req = api.execute(); req.done(function( data ) { }); req.fail(function( xhr ) { var statusCode = xhr.status; // etc. });

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