如何添加千分隔符reg reg？

问题描述

我使用这个免费的，它可以查找和替换。如何搜索所有数字并添加千位分隔符？

 输入：< node num ='12345678'> 
输出：< node num = '12,345,678'> 
  

解决方案

可以这样做：

 （？< = num ='\d +）（？=（?: \d { 3}）+（？！\d））
  

但是请注意， NET正则表达式，这正是RegExp设计器使用的。大多数正则表达式的风格只允许lookbehinds匹配固定数量的字符。只要有一个明显的最大长度，Java正则表达式就允许可变长度的lookbehinds，所以你可以通过使用 {min，max}

 （？  

约翰海兰的正则表达式支持向后看。

编辑：我差点忘了;

 （num ='\d {1,3} | \G\\ （？：\ d {3}）+（？！\ d））
  

我最喜欢这个纯粹的美学原因。 ：）

编辑2：我忘了提及最后一个替换字符串是$ 1，

I'm using this free RegExp Designer which does find and replace. How do I search for all numbers and add thousand separators?

Input:      <node num='12345678'>
Output:     <node num='12,345,678'>

解决方案

To reformat numbers only in "num" attribute values you can do this:

(?<=num='\d+)(?=(?:\d{3})+(?!\d))

But note that this will only work in .NET regexes, which is what RegExp Designer uses. Most regex flavors only allow lookbehinds that match a fixed number of characters. Java regexes allow variable-length lookbehinds as long as there's an obvious maximum length, so you can fake it out by using {min,max} quantifier an arbitrary number for the maximum:

(?<=num='\d{1,20})(?=(?:\d{3})+(?!\d))

John Hyland's regex will work in any flavor that supports lookbehinds.

EDIT: I almost forgot; here's how you can do it without lookbehinds:

(num='\d{1,3}|\G\d{3})(?=(?:\d{3})+(?!\d))

I like this one best for purely aesthetic reasons. :)

EDIT2: I forgot to mention that the replacement string for the last one is "$1,"

这篇关于如何添加千分隔符reg reg？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！