Submitting form with ajax not working

Question

My ajax isn't working out. I have a form that I'm submitting through ajax to my php file, to update my database. It's updating the database correctly, however there must be something wrong with my ajax. The problem is it always calls my .fail function even when successful and it's not reseting the values of the input fields. Please, take a look:

$(".myForm").submit(function(e) {
e.preventDefault();
var url = "myurl.php"; 
$.ajax({
       type: "POST",
       url: url,
       data: $(this).serialize()
       }).done(function() {
         alert("Success");
         $('#email').value='';
       }).fail(function() {
            alert("Fail!");
        });
     });

It's giving me a fail alert, but my database is updating. What did I do wrong?

EDIT: My php is returning a result and I edited my code like this. My ajax looks like this now, but I'm still getting same error

$(".myForm").submit(function(e) {
e.preventDefault();
var url = "http://fullyawaken.com/website/database/updatedatabase.php"; 
$.ajax({
       type: "POST",
       url: url,
       data: $(this).serialize(),
       statusCode: {
        200: function(){
            alert("Success!");
            $('#email').val("");
                    }
             }
        }).fail(function() {
            $('#email').val("");
            alert("Fail!");
        });
     });

A bug in jquery which makes status 201 Created an error. Use statusCode: instead of done: — Eyeball, Nov 29 '15 at 21:14
And are you returning anything from the server, or does it just time out and fail — adeneo, Nov 29 '15 at 21:14
sorry url is supposed to be the url; miss-typed; that's not the issue. Correcting it now — Teodor Mavrodiev, Nov 29 '15 at 21:18
Use your post like this: http://stackoverflow.com/a/30663438/2040431 — Eyeball, Nov 29 '15 at 21:19
Because in jQuery, it would be `$('#email').val("");`, there is no `value` property — adeneo, Nov 29 '15 at 21:22
And again, are you returning a response from the server, otherwise it will always fail — adeneo, Nov 29 '15 at 21:23
You're absolutely right. My bad about that. I don't know what you mean, but I don't think so. This is my php code. if (empty($_POST["email"])) { echo "Please, add an email. "; } else { $query = query("INSERT INTO emails (email) VALUES (?)", $_POST["email"]); } — Teodor Mavrodiev, Nov 29 '15 at 21:28
You have to echo something back in the `else` clause, just something like "I did it" is enough, but it should return a response. — adeneo, Nov 29 '15 at 21:29
Inside the `fail` handler, do `console.log(arguments);`, and at least you'll see what the error is in the console. — adeneo, Nov 29 '15 at 21:49
If `$query = query("INSERT INTO emails (email) VALUES (?)", $_POST["email"])` is your real code, that isn't how prepared statements work. Unsure which API you're using to connect with, so check both http://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php or http://php.net/pdo.prepared-statements - check for errors also. choose the error method for the API you're using and look at your console. You may need to post your HTML form also. Next to impossible to answer this. — Funk Forty Niner, Nov 29 '15 at 21:58
*"but I'm still getting same error"* - which is what exactly? — Funk Forty Niner, Nov 29 '15 at 21:59
the error is that it's giving me my fail alert, while still updating the database. — Teodor Mavrodiev, Nov 29 '15 at 22:18
@Fred-ii- also this query is just calling a function I've defined in another file to streamline the database inquires. It works fine, since the database itself is getting updated. I just don;t understand why it's updating and yet giving me the fail response. — Teodor Mavrodiev, Nov 29 '15 at 22:19
@TeodorMavrodiev, the code is 201 for post, if your backend is implemented correctly. — Eyeball, Nov 30 '15 at 08:04

score 1 · Answer 1 · answered Nov 29 '15 at 21:27

1

try to use this syntax

$.ajax({
   type: 'post',
   url: url,
   data: $(this).serialize(),
   success: function (data) {
   console.log(data)
   },
   error: function (err) {
   console.log(err);
   },

});

answered Nov 29 '15 at 21:27

gmrash

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Submitting form with ajax not working

1 Answers1