JQuery. $.post request .done() .fail() avoid code duplication

Question

I have a post request like

  $.post("test", {
    ajax: "true",
    action: ""
  }).done(function(data){
    if (data == "ok"){
        //xxx
    } else if (data == "err"){
        //yyy
    }
  }).fail(function(){
    //yyy
  });

How to avoid code duplication in the post request if code in .done() method (comment 'yyy') the same in fail method (comment 'yyy') ??

Solution 1

You can use always callback method, and the request will always go in that block. As you know when data contains error and not, this method will work for server-side errors. And you can catch client-side errors by defining the final else block.

$.post("test", {
    ajax: "true",
    action: ""
}).always(function(data){
    if (data == "ok"){
        //xxx
    } else if (data == "err"){
        //handle server-side errors
    } else {
        //handle client-side errors like 404 errors
    }
});

Solution 2

The most obvious and simple solution would be to simply have a failure callback like so:

function ajaxFailed() {
    // yyy
}

$.post("test", {
    ajax: "true",
    action: ""
}).done(function(data){
    if (data == "ok"){
        //xxx
    } else if (data == "err"){
        ajaxFailed();
    }
}).fail(ajaxFailed);

Solution 3

Have them call the same function, e.g.

function onErr() { 
    //yyy
}
$.post("test", {
    ajax: "true",
    action: ""
}).done(function(data){
    if (data == "ok"){
        //xxx
    } else if (data == "err"){
        onErr();
    }
}).fail(onErr);

Comments

[John Dvorak]

I don't think you'll get data=="err" when the request fails.

[Rudi Visser]

I'm pretty sure the data=="err" is for when the request fails at the server side, not client. Exactly like @JanDvorak said.

[halilb]

you're right. data == 'err' will work for server side errors. I added the last else block to catch client-side errors.