.serialize does not pass parameters correctly

Question

I am sending a serialized form with $.post to a Controller PHP:

$.post('/Controller.php',
    {
        action: 'register',
      data: $('#regForm').serialize()                       
    },
  function(msg){

        if(parseInt(msg.status)==1)
        {
            window.location=msg.txt;
        }
        else if(parseInt(msg.status)==0)
        {
            error(1,msg.txt);
        }

        hideshow('loading',0);
    }, 
    "json"      
);

I would now expect to be able to access the Form fields by $_POST['fieldname'] but INSTEAD I have a Get like parameter string in $_POST['data'] -.- What do I do wrong ?

score 1 · Accepted Answer · answered Feb 05 '14 at 15:51

It's sending the data just as you told it to. .serialize() creates a "query string", which you added into $_POST['data'].

Instead of what you have, try this:

var formData = $('#regForm').serialize();
var extraFields = {action: 'register'};

$.post('/Controller.php', $.param(extraFields)+'&'+formData, function(msg){
    if(parseInt(msg.status)==1)
    {
        window.location=msg.txt;
    }
    else if(parseInt(msg.status)==0)
    {
        error(1,msg.txt);
    }

    hideshow('loading',0);
}, 'json');

Now you should be able to get $_POST['action'] and $_POST['fieldname'].

This does exactly what I need. Thank you! – user2942586 Feb 05 '14 at 16:24 — user2942586, Feb 05 '14 at 16:24

score 0 · Answer 2 · answered Feb 05 '14 at 15:50

the $.post format is like this:

jQuery.post( url [, data ] [, success(data, textStatus, jqXHR) ] [, dataType ] )

so your first parameter is url correct, the second is data. that is why you are getting $_POST['data'] and $_POST['action'] as post parameters, since indeed you send two url parameters 'action' and 'data'. you could do like this and send the form directly as data:

$.post('/Controller.php',
       $('#regForm').serialize(),
  function(msg){

        if(parseInt(msg.status)==1)
        {
            window.location=msg.txt;
        }
        else if(parseInt(msg.status)==0)
        {
            error(1,msg.txt);
        }

        hideshow('loading',0);
    }, 
    "json"      
);

then you'll get what you ask for, $_POST['fieldname']

I assume he needed the `$_POST['action']` value too. – gen_Eric Feb 05 '14 at 15:52 — gen_Eric, Feb 05 '14 at 15:52
in that case your answer is more complete. – Volkan Ulukut Feb 05 '14 at 15:55 — Volkan Ulukut, Feb 05 '14 at 15:55

score 0 · Answer 3 · edited May 23 '17 at 12:21

0

Make sure that your input fields in your form have the name attribute assigned otherwise jQuery will not send the data correctly.

The other option is to use serializeArray. The difference between the two has been discussed here.

$.post('/Controller.php',
    {
        action: 'register',
      data: $('#regForm').serializeArray()                       
    },
  function(msg){
        if(parseInt(msg.status)==1)
        {
            window.location=msg.txt;
        }
        else if(parseInt(msg.status)==0)
        {
            error(1,msg.txt);
        }

        hideshow('loading',0);
    }      
)

edited May 23 '17 at 12:21

Community

1
1

answered Feb 05 '14 at 15:56

Fraggy

569
7
17

This won't work quite as you expect it to. In PHP, you will get `$_POST['data'][0]['name']` and `$_POST['data'][0]['value']`. You'd probably want to do something like: `var data = $('#regForm').serializeArray(); data.push({name:'action',value:'register'}); $.post('/Controller.php', data, function(){});`. – gen_Eric Feb 05 '14 at 16:02

.serialize does not pass parameters correctly

3 Answers3