Use variable in jquery object

Don't redeclare the staylogged variable multiple times. A variable should be declared only once within a given scope. Also when building your form_data variable try to write valid javascript:

var staylogged = 'false';
if($('input[name="staylogged"]').is(':checked')) {
    staylogged = 'yes';
}

var form_data = {
    username: $('#username').val(),
    password: $('#password').val(),
    staylogged: staylogged
};

or a shorthand using the ? : operator:

var staylogged = $('input[name="staylogged"]').is(':checked') ? 'yes' : 'false';
var form_data = {
    username: $('#username').val(),
    password: $('#password').val(),
    staylogged: staylogged
};

Your code inserts the value of staylogged where you've put it in the code. So your code basically reads like this:

var form_data =  {
    username: $("#username").val(),
    password: $("#password").val(),
    true
};

Which obviously doesn't make sense. You need to provide a key:

var form_data = {
    username: $("#username").val(),
    password: $("#password").val(),
    staylogged: staylogged
};

Object literals need key-value pairs. The name of the key isn't inferred from a variable name.

 var form_data = {
     username: $("#username").val(),
     password: $("#password").val(),
     staylogged: staylogged
 };

This will result in form_date.staylogged being yes or false

Also i would recommend you set staylogged to true and false, it's easier to validate.

if($("input[name='staylogged']").is(":checked")){
    var staylogged = true;
} else {
    var staylogged = false;
}

Did you just want a field named staylogged that has the same value as the local one? If so, then you still just need to assign it like normal:

var form_data = {
    username: $("#username").val(),
    password: $("#password").val(),
    staylogged: staylogged
};