0
\$\begingroup\$

I am trying to write something that will copy the current <input>s value and enter it into any <input> that start with the same name.

<input> names will follow this pattern: price-0, price-1, price-2, upc-0, upc-1, upc-2.

So if a user enters a value in <input name="price-0"> and hits copy the value should be transferred over to all input whos name start with price

This is the code I've written:

$(document).on('click', '.--copy', function () {
    var input_name = $(this).closest('div').find('input').attr('name').split('-')[0];
    $('input[name^=' + input_name + ']').val($(this).closest('div').find('input').val());
});

A fiddle to make everyones life easier: http://jsfiddle.net/6jGLD/

I feel like there are too many selectors being called upon and the code is somewhat difficult to read.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ It would probably be easier to give each related input an identical class. Other than that, I would do $input = $(this).closest('div').find('input') at the beginning of your function so you don't have to traverse the dom to find the input twice. \$\endgroup\$ – Jason P Oct 25 '13 at 0:32
2
\$\begingroup\$

One minor improvement I can think of is to create a variable to hold the input element since the element is used twice like

$(document).on('click', '.--copy', function () {
    var $input = $(this).closest('div').find('input'), input_name = $input.attr('name').split('-')[0];
    $('input[name^=' + input_name + ']').val($input.val());
});

Demo: Fiddle

\$\endgroup\$
0
\$\begingroup\$

You can just grab the value of the first input put it in a variable and put that variable as the value of the second input for example

var value = $('input.firstInput').val();
            $('input.secondInput').val(value);

here is a jsfiddle http://jsfiddle.net/suyRd/

\$\endgroup\$
1
  • \$\begingroup\$ The HTML should not be changed in the fiddle \$\endgroup\$ – Zaki Aziz Oct 25 '13 at 5:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.