如何消除部分字符串並保存到變量中？

Question

我有一個string，如：

/wiki/Bologna_Central_Station 

我想將它保存在一個var這樣的：

countryLinks = doSelect("Location").siblings('td').find('a').attr(href); 

但我只需要保存Bologna_Central_Station

Answer 1

let pattern = new RegExp('\/wiki\/') 

var string = '/wiki/Bologna_Central_Station' 

var newString = string.replace(pattern, '') 

Answer 2

有幾種方法做到這一點：

String.replace()會做到這一點：

var s = "/wiki/Bologna_Central_Station"; 
 
console.log(s.replace("/wiki/",""));

或者，String.lastIndexOf()和String.substring()爲能夠處理的/字符的任何量的更動態的解決方案：

var s = "/wiki/Bologna_Central_Station"; 
 

 
// Find the index position of the last "/" in the string 
 
var lastSlash = s.lastIndexOf("/"); 
 

 
// Extract a substring of the original starting at one more than 
 
// the lastSlash position and going to the end of the string 
 
var result = s.substring(lastSlash + 1); 
 

 
// Get the part you want: 
 
console.log(result);

或者，String.split()與Array.length處理斜線的任何金額：

var s = "/wiki/Bologna_Central_Station"; 
 

 
// Split on the "/" char and return an array of the parts 
 
var ary = s.split("/"); 
 
console.log(ary); 
 

 
// Get the last elmeent in the array. 
 
// This ensures that it works no matter how many slashes you have 
 
console.log(ary[ary.length-1]);

Answer 3

您可以根據/拆分它，讓你的數組，從中可以得到所需要的價值

var countryLinks = doSelect("Location").siblings('td').find('a').attr(href); 
countryLinks=countryLinks.split("/")[1]; 

Answer 4

只要做一些像'/wiki/Bologna_Central_Station'.split('/').splice(-1).join()。這（不像一些其他的解決方案）的功能與斜槓（'/foo/bar/baz/wiki/Bologna_Central_Station'.split('/').splice(-1).join()）任意數量

例子：

var last = '/wiki/Bologna_Central_Station'.split('/').splice(-1).join(); 
 
console.log(last); 
 

 
var last2 = '/foo/bar/baz/wiki/Bologna_Central_Station'.split('/').splice(-1).join(); 
 
console.log(last2);

Answer 5

var segments = "/wiki/Bologna_Central_Station".split('/'); 
console.log(segments[segments.length - 1]); 

Answer 6

你也可以做到這一點與一個簡單的RegExp並替換任意數量的/：

var href = "/wiki/Bologna_Central_Station"; 
 
var countryLinks = href.replace(/.*\//g,''); 
 
console.log(countryLinks);