Question 266321 in Javascript Question posted in
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Javascript – How to get elements from an array that do not have certain letter, just using the for loop?

I need to get the elements from an array that do not have the letter 'l' in them and push it to another array, just using the for loop. For this project no special methods are allowed. I can only use the push() and toLowerCase() methods.

The result I get is a list of all the names repeated multiple times.

var list =['sage', 'carolina', 'everly', 'carlos', 'frankie'];
var list1 =[];

for(let i in list){
  for(let j in list[i]){
//Get name without the 'l'
    if(list[i][j].toLowerCase() !== 'l'){
      list1.push(list[i]);
    }
  }
}
console.log(list1);
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2

Answers


  1. 0

    A few notes to make:

    1. First check whole word if it contains the letter
    2. Outside that loop, push result so it gets only added once
    3. Better to check it has the letter then the reverse 🙂
    var list = ['sage', 'carolina', 'everly', 'carlos', 'frankie'];
var result = [];

for(let i in list){
  var hasL = false;
  // Check every letter
  for(let j in list[i]){
    if(list[i][j].toLowerCase() === 'l') {
      hasL = true;
      break;
    }
   }
  
  // If the condition (hasL) hasn't been met, push to result
  if (!hasL) {
    result.push(list[i]);
  }

}

console.log(result);
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  2. 0
    1. In the inner loop check if at least one element has the letter
    2. If it has, add the item to the list in the outer loop
    const list = ['sage', 'carolina', 'everly', 'carlos', 'frankie'];
const list1 = [];

for (const str of list) {
  let hasLetter = false
  for (const char of str) {
    if (char.toLowerCase() === 'l') {
      hasLetter = true
      break
    }
  }
  if (!hasLetter) {
    list1.push(str)
  }
}
console.log(list1);

    Shorter version below:

    const list = ['sage', 'carolina', 'everly', 'carlos', 'frankie'];
const list1 = list.filter(item => !/l/i.test(item))
console.log(list1)
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