如果数组在javascript中具有索引的字符串值，如何获取错误

Question

function SummPositive( numbers ) {
  var negatives = [];
  var sum = 0;

  for(var i = 0; i < numbers.length; i++) {
    if(numbers[i] < 0) {
      negatives.push(numbers[i]);
    }else{
      sum += numbers[i];
    }
  }

  console.log(negatives);

  return sum;
}

var sum_result = SummPositive( [ 1, 2, 3, 4, 5, -2, 23, -1, -13, 10,-52 ] );

console.log(sum_result);

i have some trouble, i dont know how to check if array have index string, return false, thx all for help me to fix it

Answer 1

function SummPositive( numbers ) {
    let negatives = []; // hold all negative number
    let falsy = false; // boolean to hold true if string exists in the list
  let sum = 0; // to increment all the numbers in the list
  for(let num of numbers) { // loop through the list
    if(typeof num == 'string') {
        falsy = true // set boolean to true if string item exists in the list
      break // break out of the loop
    }
    if (num < 0) {
        negatives.push(num); // push all negative numbers to a negatives list
    } else {
        sum += num // update sum
    }
  }
  return falsy ? false : sum; // return false if falsy is true else return sum
}

var sum_result = SummPositive([ '44', 1, 2, 3, 4, 5, -2, 23, -1, -13, 10,-52, '3' ]);

Answer 2

The direct answer to your question of "How to check if an array element is a string" is to use the typeof operator , specifically comparing the evaluated result to string :

// const element = array[index];
const isAString = typeof element === 'string';

Here's an example showing how to use various array methods to help with the details of your problem.

 function sum (numbers) { let sum = 0; for (const n of numbers) sum += n; return sum; } function sortAndStringify (numbers) { // Create a copy so the original object is not mutated by sorting const copy = [...numbers]; return copy.sort((a, b) => a - b).join(', '); } function example (array) { const stringWasFound = array.some(element => typeof element === 'string'); console.log('string was found:', stringWasFound); // Even better (more strict): ensure that all elements are numbers const allElementsAreNumbers = array.every(element => typeof element === 'number'); console.log('all elements are numbers:', allElementsAreNumbers); // Return false if a non-number was found if (!allElementsAreNumbers) return false; // If we get this far, then we know that we are only dealing with numbers const negatives = array.filter(n => n < 0); console.log('negatives:', sortAndStringify(negatives)); const positives = array.filter(n => n >= 0); console.log('positives:', sortAndStringify(positives)); console.log('sum:', sum(positives)); } console.log('Just numbers:'); example([1, 2, 3, 4, 5, -2, 23, -1, -13, 10, -52]); console.log('With a string:'); // This one won't make it past the "if" statement example([1, 2, '3', 4, 5, -2, 23, -1, -13, 10, -52]); console.log('With a boolean:'); // Neither will this one example([1, 2, false, 4, 5, -2, 23, -1, -13, 10, -52]);

Answer 3

By using of Array.some() method you can easily find out if array contains any string value or not based on the type checking. It will return true if any of the element is string in the array else false.

Live Demo :

 const arr = [1, 2, 3, 4, '5', -2, 23, -1, -13, 10, -52]; const isArrayContainsString = arr.some(elem => typeof elem === 'string'); console.log(isArrayContainsString); // true

Then, You can add this condition in your code :

if (isArrayContainsString)
  return false
else
  // remaining logic comes here