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Question

Question

Passing values into table dropdown

asked on September 30, 2015 • Show version history

I'm looking to pass the values of two fields into a dropdown in a table. 

Here is the working code i have for passing those values into a dropdown outside of the table.

When replacing '#Field25' with #Field21(1)' nothing happens and when replacing it with '.droppy1' the class i gave to the dropdown field in the table it just adds another label under the field rather than add an item to the drop down list.

  //Set dropdown
    $('.test1').on('blur', 'input', fill0);
  	function fill0() {   
      //clear dropdown
      $('#Field25').empty();

      //append values
        var numbers = [$('#q5 input').val(), $('#q17 input').val()];
        var option = '';
        for (var i=0;i<numbers.length;i++){
           option += '<option value="'+ numbers[i] + '">' + numbers[i] + '</option>';
        }
        $('#Field25').append(option);
    };  

Any help would be greatly appreciated.

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Answer

SELECTED ANSWER
replied on October 1, 2015

Hi there Casey,

You might have to select the drop down in the table a little differently by using something like:

$(".droppy1 select").append(option);

This should select the drop down's 'select' HTML tag from inside the table.

Cheers! Dan.

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Replies

replied on October 1, 2015

Thanks a ton Daniel!

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