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);

我有一些麻烦，我不知道如何检查数组是否有索引字符串，返回 false，谢谢大家帮助我修复它

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

您的“如何检查数组元素是否为字符串”问题的直接答案是使用typeof运算符，特别是将评估结果与string进行比较：

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

这是一个示例，展示了如何使用各种数组方法来帮助解决问题的细节。

 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

通过使用Array.some()方法，您可以根据type检查轻松找出数组是否包含任何字符串值。 如果任何元素是数组中的字符串，它将返回 true，否则返回 false。

现场演示：

 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

然后，您可以在代码中添加此条件：

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