Grab the return value and get out of forEach in JavaScript?

Question

How can I modify this code so that I can grab the field.DependencyFieldEvaluated value and get out of the function as soon I get this value?

function discoverDependentFields(fields) {
                    fields.forEach(function (field) {
                        if (field.DependencyField) {
                            var foundFields = fields.filter(function (fieldToFind) { return fieldToFind.Name === field.DependencyField; });
                            if (foundFields.length === 1) {
                               return field.DependencyFieldEvaluated = foundFields[0];

                            }
                        }
                    });
                }

you can't break out of forEach(), try using `every` instead http://stackoverflow.com/questions/6260756/how-to-stop-javascript-foreach — gurvinder372, Feb 17 '16 at 06:46
firstly it complains that not all path returns a value. secondly it continues on the loop even when it reaches the return line! — Maryam, Feb 17 '16 at 06:47
@Maryam - the .forEach() function is *supposed to* continue the loop - it ignores the return value from your function. If you use .every() instead you can break out early but *only by returning false* - that is to say that you can't directly return a value to the code outside the loop, you have to assign the value to a variable declared outside. If what you really want to do is return that value from the `discoverDependentFields()` function then the easiest way is just to use a traditional `for` loop. — nnnnnn, Feb 17 '16 at 06:57
I tried for loop. The problem is that in the nested function then it cannot access fields[i]. any solution to this? — Maryam, Feb 17 '16 at 07:14
Please share a sample json you are using to call the function. — zakaiter, Feb 17 '16 at 07:16

score 27 · Accepted Answer · answered Feb 17 '16 at 07:12

Use a good old vanilla for loop:

function discoverDependentFields(fields) {
  for (var fieldIndex = 0; fieldIndex < fields.length; fieldIndex ++) {
    var field = fields[fieldIndex];

    if (field.DependencyField) {
      var foundFields = fields.filter(function(fieldToFind) {
        return fieldToFind.Name === field.DependencyField;
      });
      if (foundFields.length === 1) {
        return foundFields[0];
      }
    }
  }
}

Well, if you want to stay fancy, use filter again:

function discoverDependentFields(fields) {
  return fields.filter(function(field) {
    if (field.DependencyField) {
      var foundFields = fields.filter(function(fieldToFind) {
        return fieldToFind.Name === field.DependencyField;
      });
      if (foundFields.length === 1) {
        return foundFields[0];
      }
    }
  })[0];
}

this solution worked! thanks:) – Maryam Feb 17 '16 at 07:29 — Maryam, Feb 17 '16 at 07:29

score 10 · Answer 2 · edited Mar 05 '17 at 23:36

10

instead of fields.forEach you can use fields.map. here is an example:

var source=[1,2,3,4,5];

var destination=source.map(function(item){
 if(item==3)
     return 'OUTPUT';
}).filter(function(item){return item;})[0];

console.log(destination); // prints 'OUTPUT'

edited Mar 05 '17 at 23:36

Barkermn01

6,781
33
83

answered Mar 05 '17 at 23:15

behzad besharati

5,873
3
18
22

score -11 · Answer 3 · answered Feb 17 '16 at 07:02

-11

Use return statement where you want to break-out like below.

[1, 2, 3, 4, 5, 6].forEach(function(value){
  if(value > 2) return; 
  console.log(value)
});

answered Feb 17 '16 at 07:02

Ravi Tiwari

946
8
17

Break in what respect. The callback that you pass in forEach executes for each item of the data and hence can't have a single return value. However, if you want not to process the callback for specific condition, you can use return. If you could provide the problem at hand, not the code, I guess I would be able to help, not sure though :) – Ravi Tiwari Feb 17 '16 at 07:17
A return statement inside a .forEach() callback behaves like a `continue`, not like a `break`. But in any case if you look at the code in the question you'll see the OP is already trying to use `return`. – nnnnnn Feb 17 '16 at 10:03

Grab the return value and get out of forEach in JavaScript?

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