Answers

• allan Posts: 61,972Questions: 1Answers: 10,160 Site admin

  February 2016 Answer ✓

  The issue is that you are calling field().update() once for every option. It expects the full list of options to be passed in, which explains why you are only ever seeing the final item in the array.

  What you need to do is create an array of objects with label and value and then pass that resulting array into field().update() just once (i.e. the array has all the options in it).

  Allan
• daniegloy Posts: 35Questions: 12Answers: 5

  February 2016

  Am i at least close?

               var YdataArray = $.parseJSON(Ydata);
                  var x,text,select;
                  var setText = new Array();
                        for (x in YdataArray) {
                              text = YdataArray[x]['RegYear'];
                              console.log(text);
  
                               setText[x] = text;
                         }
                         select = '';
  
                        for (x in setText) {
                          select += '{'+setText[x]+'},';
                        }
                        editorNewForm.field('YEAR').update([select]);
                })
              })
  

  now its only creating one select option with both values, what am i missing?
• daniegloy Posts: 35Questions: 12Answers: 5

  February 2016 Answer ✓

  Fixed Thank You. Found the solution here --- http://datatables.net/forums/discussion/9899
• allan Posts: 61,972Questions: 1Answers: 10,160 Site admin

  February 2016 Answer ✓

  Great stuff - good to hear you've got it working now :-)

  Allan