How to generate a PHP response for $.ajax

Question

A form is being submitted using the ajax code below. I am unsure what response to generate using PHP, so that $.ajax can call the appropriate callbacks done() and fail()

request = $.ajax({
        url: "php_process.php",
        type: "post",
        data: serializedData
    });

    // callback handler that will be called on success
    request.done(function (response, textStatus, jqXHR){
        // log a message to the console
        console.log("Hooray, it worked!");

    });

    // callback handler that will be called on failure
    request.fail(function (jqXHR, textStatus, errorThrown){
        // log the error to the console
        console.error(
            "The following error occured: "+
            textStatus, errorThrown
        );
    });`

`.fail` will be called when your script responds with an HTTP error code, `.done` otherwise. — CBroe, Aug 13 '14 at 23:59
The failure is called based upon a request fail, so there should be no concern about what the PHP does. The PHP never succeeds in running at all. — XaxD, Aug 14 '14 at 00:00
send something back that you can validate in the `done` handler, can be text, json or whatever data format you prefer — charlietfl, Aug 14 '14 at 00:02
How can I return a json back from php so that i can do `response.some_key` — dr_rk, Aug 14 '14 at 15:32
@dr_rk Updated jsfiddle link at post http://jsfiddle.net/guest271314/L3jbvnex/1/ — guest271314, Aug 20 '14 at 19:34

score 0 · Accepted Answer · edited May 23 '17 at 12:20

See .done() and .fail() responses at result area , network tab at console

$(function() {
    var urls = ["/echo/jsons/", "/echo/json/"];
    var request = function(url) {

        return $.ajax({
        url: url,
        type: "POST",
        data: {json : JSON.stringify({"abc":[123]}) }            
        });
    };           
    // callback handler that will be called on success
 $.each(urls, function(k, v) {
     $.when(request(v))
    .done(function (response, textStatus, jqXHR){
        // log a message to the console
        console.log("Hooray, it worked!", response);
        $("body").prepend("DONE: <br>" 
                          + Object.keys(response) + ":" 
                          +  response[Object.keys(response)] 
                          + "<br><br>")
    })    
    // callback handler that will be called on failure
   .fail(function (jqXHR, textStatus, errorThrown){
        // log the error to the console
        console.log("The following error occured: "
                    + textStatus, errorThrown);
       $("textarea")
       .before("FAIL: <br>")
       .val(jqXHR.getAllResponseHeaders() +"\n" 
           + jqXHR.status +"\n"+ textStatus 
           +"\n"+ errorThrown +"\n" +  jqXHR.responseText)
    });  
});
});

jsfiddle http://jsfiddle.net/guest271314/L3jbvnex/1/

See

PHP: How to send HTTP response code?

http://php.net/manual/en/function.http-response-code.php

http://php.net/manual/en/function.header.php

score -1 · Answer 2 · answered Aug 14 '14 at 00:00

-1

fail() is only called if there is an error in getting the response from server. Else the done() is called. So fail() doesn't depend on the response from the PHP.

answered Aug 14 '14 at 00:00

webNeat

2,768
1
20
22

Of course it does – because PHP could issue an HTTP error status code as well. – CBroe Aug 14 '14 at 13:02

How to generate a PHP response for $.ajax

2 Answers2