CODEWITHSUNDEEP
javascriptjquery
Matteo Pagliazzi
Matteo Pagliazzi

Reputation: 5270

Use variable in jquery object

i have this code 

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

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

but the var staylogged in form_data doesn't work as expected where is the error?

i don't know how to declare the var staylogged in form_data

Upvotes: 0

Views: 198

Answers (5)

Darin Dimitrov
Darin Dimitrov

Reputation: 1039508

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
};

Upvotes: 8

jtfairbank
jtfairbank

Reputation: 2307

Your staylogged variable is out of scope. Try declaring var staylogged outside of your if statement. You also need to give it a key.

var staylogged = ""; // give it a default value
if($("input[name='staylogged']").is(":checked")){
    var staylogged = "yes";
} else {
    var staylogged = "false";
}

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

Upvotes: 1

Joe Enos
Joe Enos

Reputation: 40431

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
};

Upvotes: 1

amosrivera
amosrivera

Reputation: 26554

 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;
}

Upvotes: 3

lonesomeday
lonesomeday

Reputation: 238115

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.

Upvotes: 3

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