Question

上一个问题：

如何替换字符串中的间隔字符：

例如从Apple到A *** e

更新：

需要获取char位置0-4和-4（反向）

var transaction = '1234567890987651907';
console.log('1234****1907');

解决方案

var str = "01340946380001281972";
str.replace(/^(\d{0,4})(\d{4})(.*)/gi,"$1 **** $2");

Answer 1

这是使用基本字符串函数的解决方案：

var input = "Apple";
var input_masked = input.substring(0, 1) + Array(input.length - 1).join("*") +
    input.substring(input.length-1);
console.log(input_masked);

这种方法是在输入的第一个字符和最后一个字符之间夹住中间字符（掩码为*）。

Answer 2

只需替换中间字符：

const str = "Apple";
const output = `${str[0]}${"*".repeat(str.length - 2)}${[...str].pop()}`;
console.log(output);

Answer 3

var str = "01340946380001281972";
str.replace(/^(\d{0,4})(\d{4})(.*)/gi,"$1 **** $2");

Answer 4

我想你是这个意思

function maskIt(str, keep) {
  var len = str.length,
    re = new RegExp("(.{" + keep + "})(.{" + (len - keep * 2) + "})(.{" + keep + "})", "g")
  console.log(re)
  return str.replace(re, function(match, a, b, c) {
    return a + ("" + b).replace(/./g, "*") + c
  });
}
console.log(
  maskIt("1234567890", 4),
  maskIt("Apple", 1)
)

作为原型：

String.prototype.maskIt = function(keep) { // don't use arrow or lose "this"
  const re = new RegExp("(.{" + keep + "})(.{" + (this.length - keep * 2) + "})(.{" + keep + "})", "g");
  return this.replace(re, (match, a, b, c) => a + ("" + b).replace(/./g, "*") + c);
}
console.log(
  "1234567890".maskIt(4),
  "Apple".maskIt(1)
)

给定字符串范围的中间

4 个答案: