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') ??

Answer 1

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);

Answer 2

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
    }
});

Answer 3

Have them call the same function, eg

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

Answer 4

An alternative would be to change the protocol slightly, and make use of the HTTP status codes to indicate success or failure:

if($sqlError){
  header("HTTP/1.1 503 Service unavailable");
}

...

.done(function(){
   //xxx
}).fail(function(){
   //yyy
});