Function not defined with addition of ajax logic

Question

I'm not sure what's going on here, but the following code works perfectly by returning and logging my data:

function myFunction() {
  var itemArray = JSON.parse('<?php echo json_encode($items);?>');

  //console.log(itemArray);
  var promo_code = document.getElementById("promo_code").value;
  $.ajax({
       type:'POST',
       url:'validateCode',
       data:{promo_code:promo_code},
        _token: '{{ csrf_token() }}',
        success:function(data){
          console.log(data);           

        }
    });

  }

But as soon as I add function logic into my success block it says "myFunction() is not defined"

function myFunction() {
  var itemArray = JSON.parse('<?php echo json_encode($items);?>');

  //console.log(itemArray);
  var promo_code = document.getElementById("promo_code").value;
  $.ajax({
       type:'POST',
       url:'validateCode',
       data:{promo_code:promo_code},
        _token: '{{ csrf_token() }}',
        success:function(data){
          console.log(data);


          function(data){
            let results = [];
            $.each(itemArray, function(index1, value1) {
                let result = false;
                $.each(data.promo_codet_id.code.rule_type, function(index2,value2) {
                    if(value1.frame_number === value2.frame){
                        result = true;
                        break;
                    }
                });
                results.push(result);
            });
            console.log(results);
          }

        }
    });

  }

Emanuele Scarabattoli · Accepted Answer · 2019-07-31 12:53:05Z

1

You should not re-declare the function again inside the function, just put this:

success:function(data) {
  console.log(data);
  let results = [];
  $.each(itemArray, function (index1, value1) {
    let result = false;
    $.each(data.promo_codet_id.code.rule_type, function (index2, value2) {
      if (value1.frame_number === value2.frame) {
        result = true;
        break;
      }
    });
    results.push(result);
  });
  console.log(results);
}

answered Jul 31, 2019 at 12:53

Emanuele Scarabattoli

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Comments

Clarity · Accepted Answer · 2019-07-31 13:04:10Z

0

function(data){ is missing the function name, also you're not calling it after declaring.

Additionally if you want to break out from $.each, you need to return false:

function getResults(data) {
  let results = [];
  $.each(itemArray, function(index1, value1) {
    let result = false;
    $.each(data.promo_codet_id.code.rule_type, function(index2, value2) {
      if (value1.frame_number === value2.frame) {
        result = true;
        return false;
      }
    });
    results.push(result);
  });
  return results;
}

const results = getResults();
console.log(results);

answered Jul 31, 2019 at 13:04

Clarity

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