Looping through another array for conditions using forEach loop

Question

I've created a function called checkLength(input, min, max) which takes in a username and password input as shown below.

checkLength(username, 3, 15);
checkLength(password, 8, 25);

However, I'm looking to pass in an inputArray, minArray, maxArray to reduce the repetitiveness using a forEach loop. But it seems like a forEach loop can only take in one array at a time. Any suggestions on how I can improve this?

checkLength2([username,password], [3,8], [8,15])

function checkLength2 (inputArray, minArray, maxArray, i=0) {
inputArray.forEach(input => {
    if (input.value < minArray[i]) {
        //another error function
        i++;
    }
});}

I think you original approach with two function calls makes a lot more sense — Phil
– Phil, Commented Sep 30, 2020 at 5:24
[[username, 3, 8], [password, 8, 15]].forEach(args => checkLength(...args)); is an approach. — Wyck
– Wyck, Commented Sep 30, 2020 at 5:53
I'm looking to pass in an inputArray, minArray, maxArray and now you have 3 lists that you need to maintain and keep in sync — Thomas
– Thomas, Commented Sep 30, 2020 at 6:12

Sachin Singh · Accepted Answer · 2020-09-30 05:27:17Z

1

You should try index from forEach loop.

function checkLength2 (inputArray, minArray, maxArray) {
  inputArray.forEach((input, i) => {
      if (input.value < minArray[i]) {
          // do something
      } else if (input.value > maxArray[i]) {
          // do something else
      }
  });
}

answered Sep 30, 2020 at 5:27

Sachin Singh

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1 Comment

Timothy Kwok Over a year ago

Thank you! This was exactly what I was looking for.

Paul Rooney · Accepted Answer · 2020-09-30 05:31:18Z

1

if you want to show the same message you can try this out. or else you can use if/else statement for checking both the minArray and maxArray

function checkLength2(inputArray, minArray, maxArray, i = 0) {    
  for (let index = 0; index <= inputArray.length; index++) {    
    if (inputArray[index].value < minArray[index] && inputArray[index].value>maxArray[index]) {    
      //error
    }
  }
}

edited Sep 30, 2020 at 5:31

Paul Rooney

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answered Sep 30, 2020 at 5:29

Ahmad Suddle

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