Adding input field value

Question

I am trying to add the values from an input field with the values of a variable inside a jquery. When doing so I get [object Object] displaying as the result. My goal is to get the value from the input field so I can then added to a variable and ultimately display on result on another textbox with id called person. The texbox I am trying to get the value from is count. How can I achieve this?

$.ajax({
    type: 'POST',
    url: 'person.php',
    data: $(this).serialize(),
    dataType: 'json',
    success: function (data) {
    var num = $('#count').val(data);
    $('#person').val(num+ newUniqueAI);
    }
});

What is data? Why are you passing it to .val if you want to retrieve the value? Please read the jQuery documentation: api.jquery.com/val. It contains information about how .val should be called. — Felix Kling
– Felix Kling, Commented Dec 4, 2013 at 8:40
So, do you want to set the value of #count and also use the new value for #person? What is the value of data? What is newUniqueAI and its value? — Felix Kling
– Felix Kling, Commented Dec 4, 2013 at 8:43

Bas van Dijk · Accepted Answer · 2013-12-04 08:40:51Z

3

You are using:

$('#count').val(data);

This sets the value of #count to the value of data and returns a jQuery object. You want to get the value of #count. Therefore you should use:

$('#count').val();

If you do not specify a parameter the val function returns the value. More information about val() could be read here: http://api.jquery.com/val/

answered Dec 4, 2013 at 8:40

Bas van Dijk

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Comments

Rituraj ratan · Accepted Answer · 2013-12-04 08:51:10Z

1

$('#count').val(data);// first set the value     

var num = $('#count').val(); //get by val()
$('#person').val(parseInt(num)+ parseInt(newUniqueAI)); // set by val(something)

reference .val()

edited Dec 4, 2013 at 8:51

answered Dec 4, 2013 at 8:41

Rituraj ratan

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3 Comments

Code_Ed_Student Over a year ago

+1 Great but now (num+ newUniqueAI) will not add for example if num=5 and newUniqueAI =3 I get result 53. How come?

Felix Kling Over a year ago

@Code_Ed_Student: One of the values (or both) is a string. Convert strings to numbers first, e.g. with the unary + operator: +num.

Felix Kling Over a year ago

If you use parseInt, don't forget to pass the radix as second argument.

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