Question

JavaScript findIndex返回第一个找到的索引（如果值重复）

const arr = [{a: 10, b: 20, c: 30},{a: 15, b: 25, c: 32},{a: 10, b: 23, c: 350}]
const index = arr.findIndex(m => m.a === 10)
console.log(index);

上面的代码只会返回0索引。

我该怎么做才能获得索引2。

Answer 1

您可以filter像这样keys：

const arr = [{a: 10, b: 20, c: 30},{a: 15, b: 25, c: 32},{a: 10, b: 23, c: 350}]
const indices = [...arr.keys()].filter(i => arr[i].a === 10)
console.log(indices)

或者，只需使用for循环

const arr = [{a: 10, b: 20, c: 30},{a: 15, b: 25, c: 32},{a: 10, b: 23, c: 350}]
const output = [];

for (let i = 0; i < arr.length; i++) {
  if (arr[i].a === 10)
    output.push(i)
}

console.log(output)

Answer 2

arr.reduce((result, current, index) => {
   return result.concat(current.a == 10 ? [index]: [])
})

Answer 3

您可以使用Array.reduce循环访问arr并返回每个索引

单行：

const arr = [{
  a: 10,
  b: 20,
  c: 30,
}, {
  a: 15,
  b: 25,
  c: 32,
}, {
  a: 10,
  b: 23,
  c: 350,
}]

// Value to check for
const v = 10;

// we loop on every entry of the arr
// we start with an empty array and look if the entries
// contains the value 10, if they do, we push the index of the
// entry in the array that was empty
//
// we do that for every entry of arr, when we treat all, we return
// the list of matching entries
const idx = arr.reduce((tmp, x, xi) => (x.a === v ? [...tmp, xi] : tmp), []);

console.log(idx);

解释：

const arr = [{
  a: 10,
  b: 20,
  c: 30,
}, {
  a: 15,
  b: 25,
  c: 32,
}, {
  a: 10,
  b: 23,
  c: 350,
}]

const valToCheck = 10;

const indexes = arr.reduce((tmp, x, xi) => {
  if (x.a === valToCheck) {
    tmp.push(xi);
  }

  return tmp;
}, []);

console.log(indexes);

Answer 4

使用map和filter：

const arr = [{a: 10, b: 20, c: 30},{a: 15, b: 25, c: 32},{a: 10, b: 23, c: 350}];
const res = arr.map(({ a }, i) => a == 10 ? i : "").filter(String);
console.log(res);

Answer 5

您可以映射找到的项目的索引并仅过滤有效的索引。

const
    array = [{ a: 10, b: 20, c: 30 }, { a: 15, b: 25, c: 32 }, { a: 10, b: 23, c: 350 }],
    indices = array
        .map((m, i) => m.a === 10 ? i : -1)
        .filter(i => i != -1);

console.log(indices);

Answer 6

您可以遍历数组并将匹配索引推入新数组中。

const arr = [{a: 10, b: 20, c: 30},{a: 15, b: 25, c: 32},{a: 10, b: 23, c: 350}]
let p = []
arr.forEach((ele,index)=> ele.a===10 ? p.push(index) : '')
console.log(p)

Answer 7

您可以尝试这样：

<Button
    Panel.ZIndex="100"
    BorderThickness="0"
    BorderBrush="Transparent"
    Background="Black"
    Grid.Column="1"
    VerticalAlignment="Stretch"
    Margin="0 2 0 5"
    Command="{Binding DataContext.TestRadTreeCommand, RelativeSource={RelativeSource AncestorType=TreeView}}">
    <Image Width="20"
           Height="20"
           Source="/CCRIS;component/Resources/Images/03-Add.png"/>
</Button>

Answer 8

如果您想要直接与Array一起使用的函数，请将其添加到Array.prototype

Array.prototype.findAllIndexes = function(iteratee){
    var resultArray = [];
    this.forEach(function(element) {
        var validated = iteratee.call(this, element);
        if(validated) {
            resultArray.push(element);
        }
    });
    return resultArray;
}

然后您可以将其与数组对象一起使用

（请根据所使用的od数据类型处理极端情况）

const arr = [{a: 10, b: 20, c: 30},{a: 15, b: 25, c: 32},{a: 10, b: 23, c: 350}]
const index = arr.findAllIndexes(m => m.a === 10)
console.log(index);

您可以使用一些帮助程序库来处理诸如lodash.js之类的东西，您可以将它们包含在项目中。

Javascript findIndex用于重复值

8 个答案: