根据条件查找所有索引

Question

如何根据对象数组的条件获取所有索引？我已经尝试了下面的代码，但是它只返回第一次出现的代码。

a = [
  {prop1:"abc",prop2:"yutu"},
  {prop1:"bnmb",prop2:"yutu"},
  {prop1:"zxvz",prop2:"qwrq"}];
    
index = a.findIndex(x => x.prop2 ==="yutu");

console.log(index);

Answer 1

您可以使用 normal for 循环，当 prop2 匹配时推送数组中的索引

const a = [{
    prop1: "abc",
    prop2: "yutu"
  },
  {
    prop1: "bnmb",
    prop2: "yutu"
  },
  {
    prop1: "zxvz",
    prop2: "qwrq"
  }
];

const indArr = [];
for (let i = 0; i < a.length; i++) {
  if (a[i].prop2 === 'yutu') {
    indArr.push(i)
  }

}
console.log(indArr);

Answer 2

尝试Array.reduce

js prettyprint-override">a = [
  {prop1:"abc",prop2:"yutu"},
  {prop1:"bnmb",prop2:"yutu"},
  {prop1:"zxvz",prop2:"qwrq"}];
    
index = a.reduce((acc, {prop2}, index) => prop2 ==="yutu" ? [...acc, index] : acc, []);

console.log(index);

Answer 3

< code>findIndex将只返回一个匹配的索引，您可以使用< code>filter根据属性< code>prop2检查值

a = [
  {prop1:"abc",prop2:"yutu"},
  {prop1:"bnmb",prop2:"yutu"},
  {prop1:"zxvz",prop2:"qwrq"}];
    
const allIndexes = a
  .map((e, i) => e.prop2 === 'yutu' ? i : -1)
  .filter(index => index !== -1);
  
  console.log(allIndexes);
  // This is one liner solution might not work in older IE ('flatMap')
 const  notSupportedInIE =a.flatMap((e, i) => e.prop2 === 'yutu' ? i : []);
 console.log(notSupportedInIE);