What's the best way to break from nested loops in JavaScript?

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What's the best way to break from nested loops in Javascript?

//Write the links to the page.
for (var x = 0; x < Args.length; x++)
{
   for (var Heading in Navigation.Headings)
   {
      for (var Item in Navigation.Headings[Heading])
      {
         if (Args[x] == Navigation.Headings[Heading][Item].Name)
         {
            document.write("<a href=\"" 
               + Navigation.Headings[Heading][Item].URL + "\">" 
               + Navigation.Headings[Heading][Item].Name + "</a> : ");
            break; // <---HERE, I need to break out of two loops.
         }
      }
   }
}
18 Answers

Just like Perl,

loop1:
    for (var i in set1) {
loop2:
        for (var j in set2) {
loop3:
            for (var k in set3) {
                break loop2;  // breaks out of loop3 and loop2
            }
        }
    }

as defined in EMCA-262 section 12.12. [MDN Docs]

Unlike C, these labels can only be used for continue and break, as Javascript does not have goto.

Wrap that up in a function and then just return.

I'm a little late to the party but the following is a language-agnostic approach which doesn't use GOTO/labels or function wrapping:

for (var x = Set1.length; x > 0; x--)
{
   for (var y = Set2.length; y > 0; y--)
   {
      for (var z = Set3.length; z > 0; z--)
      {
          z = y = -1; // terminates second loop
          // z = y = x = -1; // terminate first loop
      }
   }
}

On the upside it flows naturally which should please the non-GOTO crowd. On the downside, the inner loop needs to complete the current iteration before terminating so it might not be applicable in some scenarios.

Quite simple:

var a = [1, 2, 3];
var b = [4, 5, 6];
var breakCheck1 = false;

for (var i in a) {
    for (var j in b) {
        breakCheck1 = true;
        break;
    }
    if (breakCheck1) break;
}

Here are five ways to break out of nested loops in JavaScript:

1) Set parent(s) loop to the end

for (i = 0; i < 5; i++)
{
    for (j = 0; j < 5; j++)
    {
        if (j === 2)
        {
            i = 5;
            break;
        }
    }
}

2) Use label

exit_loops:
for (i = 0; i < 5; i++)
{
    for (j = 0; j < 5; j++)
    {
        if (j === 2)
            break exit_loops;
    }
}

3) Use variable

var exit_loops = false;
for (i = 0; i < 5; i++)
{
    for (j = 0; j < 5; j++)
    {
        if (j === 2)
        {
            exit_loops = true;
            break;
        }
    }
    if (exit_loops)
        break;
}

4) Use self executing function

(function()
{
    for (i = 0; i < 5; i++)
    {
        for (j = 0; j < 5; j++)
        {
             if (j === 2)
                 return;
        }
    }
})();

5) Use regular function

function nested_loops()
{
    for (i = 0; i < 5; i++)
    {
        for (j = 0; j < 5; j++)
        {
             if (j === 2)
                 return;
        }
    }
}
nested_loops();
var str = "";
for (var x = 0; x < 3; x++) {
    (function() {  // here's an anonymous function
        for (var y = 0; y < 3; y++) {
            for (var z = 0; z < 3; z++) {
                // you have access to 'x' because of closures
                str += "x=" + x + "  y=" + y + "  z=" + z + "<br />";
                if (x == z && z == 2) {
                    return;
                }
            }
        }
    })();  // here, you execute your anonymous function
}

How's that? :)

How about pushing loops to their end limits

    for(var a=0; a<data_a.length; a++){
       for(var b=0; b<data_b.length; b++){
           for(var c=0; c<data_c.length; c++){
              for(var d=0; d<data_d.length; d++){
                 a =  data_a.length;
                 b =  data_b.length;
                 c =  data_b.length;
                 d =  data_d.length;
            }
         }
       }
     }

Already mentioned previously by swilliams, but with an example below (Javascript):

// Function wrapping inner for loop
function CriteriaMatch(record, criteria) {
  for (var k in criteria) {
    if (!(k in record))
      return false;

    if (record[k] != criteria[k])
      return false;
  }

  return true;
}

// Outer for loop implementing continue if inner for loop returns false
var result = [];

for (var i = 0; i < _table.length; i++) {
  var r = _table[i];

  if (!CriteriaMatch(r[i], criteria))
    continue;

  result.add(r);
}

There are many excellent solutions above. IMO, if your break conditions are exceptions, you can use try-catch:

try{  
    for (var i in set1) {
        for (var j in set2) {
            for (var k in set3) {
                throw error;
            }
        }
    }
}catch (error) {

}

Hmmm hi to the 10 years old party ?

Why not put some condition in your for ?

var condition = true
for (var i = 0 ; i < Args.length && condition ; i++) {
    for (var j = 0 ; j < Args[i].length && condition ; j++) {
        if (Args[i].obj[j] == "[condition]") {
            condition = false
        }
    }
}

Like this you stop when you want

In my case, using Typescript, we can use some() which go through the array and stop when condition is met So my code become like this :

Args.some((listObj) => {
    return listObj.some((obj) => {
        return !(obj == "[condition]")
    })
})

Like this, the loop stopped right after the condition is met

Reminder : This code run in TypeScript

Assign the values which are in comparison condition

function test(){
    for(var i=0;i<10;i++)
    {
            for(var j=0;j<10;j++)
            {
                    if(somecondition)
                    {
                            //code to Break out of both loops here
                            i=10;
                            j=10;
                    }
                    
            }
    }

    //Continue from here

}

An example with for .. of, close to the example further up which checks for the abort condition:

test()
function test() {
  var arr = [1, 2, 3,]
  var abort = false;
  for (var elem of arr) {
    console.log(1, elem)

    for (var elem2 of arr) {
      if (elem2 == 2) abort = true;  
        if (!abort) {
            console.log(2, elem2)
        }
    }
  }
}
  • Condition 1 - outer loop - will always run
  • The top voted and accepted answer also works for this kind of for loop.

Result: the inner loop will run once as expected

1 1
2 1
1 2
1 3
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