输出是字符串，我想从数组中获取数据

Question

所以我从一个数组中获取了索引，它返回了多个索引。我希望这些索引从另一个数组中获取数据。

  let withAccent = array.map(x => x.TERM);
  let withoutAccent = terms
    .map(x => x.TERM)
    .join(",")
    .normalize("NFD")
    .replace(/[\u0300-\u036f]/g, "")
    .split(",");
  let withoutAccentPosition = withoutAccent
    .map((withoutAccent, idx) =>
      withoutAccent.includes(input) ? "withAccent[" + idx + "]" : null
    )
    .filter(e => e !== null)
    .join(", ");
  console.log(withoutAccentPosition);
  console.log(withAccent[0], withAccent[1], withAccent[22]);

带口音；

[“ahoj”、“test1”、“test2”、“test3”、“test4”、“test5”、“můžete”、 “nebo”、“pak”、“postupně”、……]

没有重音：

[“ahoj”、“test1”、“test2”、“test3”、“test4”、“test5”、“muzete”、 “nebo”、“pak”、“postupne”、……] 输入是用户输入的内容

现在我从 withoutAccent 获取索引，我想获取数据取决于 with Accent 的索引

所以第一个日志返回 withAccent[] 位置字符串，第二个运行正常，但我想要来自 withoutAccentPosition 的数据。

Answer 1

如果您需要获取数组，则不应使用 join()。 join() 方法通过连接数组中的所有元素来创建并返回一个新字符串。

解决办法

var input = 'muzete' ;

var withAccent =[ "ahoj", "test1", "test2", "test3", "test4", "test5", "můžete", "nebo", "pak", "postupně"];
var withoutAccent =  [ "ahoj", "test1", "test2", "test3", "test4", "test5", "muzete", "nebo", "pak", "postupne" ];

let withoutAccentPosition = withoutAccent
    .map((withoutAccent, idx) =>
        withoutAccent.includes(input) ? "withAccent[" + idx + "]" : null
    ).filter(e => e !== null);
console.log(withoutAccentPosition);
console.log(withAccent[0], withAccent[1], withAccent[22]);

如果您需要直接值，则直接使用 withAccent[idx] 而不是将其设为字符串类型

var input = 'muzete' ;

var withAccent =[ "ahoj", "test1", "test2", "test3", "test4", "test5", "můžete", "nebo", "pak", "postupně"];
var withoutAccent =  [ "ahoj", "test1", "test2", "test3", "test4", "test5", "muzete", "nebo", "pak", "postupne" ];

let withoutAccentPosition = withoutAccent
    .map((withoutAccent, idx) =>
        withoutAccent.includes(input) ? withAccent[idx] : null
    ).filter(e => e !== null);
console.log(withoutAccentPosition);
console.log(withAccent[0], withAccent[1], withAccent[22]);