比较字符串并获得最终差异 [英] compare strings and get end difference

问题描述

我有两个字符串.
字符串 A:快速的棕色狐狸"
字符串 B:敏捷的棕色狐狸跳过懒惰的狗."

I have two strings.
String A: "The quick brown fox"
String B: "The quick brown fox jumps over the lazy dog."

字符串 B 将始终逐字包含字符串 A.永远不会有快黑狐"或快速而迅速的棕色狐狸".

String B will always contain string A verbatim. There will never be a "quick black fox" or a "quick and speedy brown fox".

如何获得字符串 C"?区别跳过懒狗".

How do I get a "String C" of the difference "jumps over the lazy dog."?

推荐答案

const A = "The quick brown fox"

const B = "The quick brown fox jumps over the lazy dog."

const diff = (diffMe, diffBy) => diffMe.split(diffBy).join('')

const C = diff(B, A)

console.log(C) // jumps over the lazy dog.

这篇关于比较字符串并获得最终差异的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！