在groovy中使用正则表达式进行模式匹配

发布于 2024-12-07 02:26:17 字数 718 浏览 0 评论 0原文

我有下面的常规代码

import java.util.regex.Matcher;
import java.util.regex.Pattern;

String myInput="License_All (12313)"
String myRegex="\\(\\d+\\)"


String ResultString
Pattern regex
Matcher regexMatcher

regex = Pattern.compile(myRegex, Pattern.DOTALL);

regexMatcher = regex.matcher(myInput);

List<String> matchList = new ArrayList<String>();
while (regexMatcher.find()) {
matchList.add(regexMatcher.group());
}

for(int i=0;i<matchList.size();i++)
{
myInput=myInput.replaceAll( matchList[i], '')
println matchList[i]
}
println myInput

它应该删除 (12313) 部分，但它给了我下面的输出

(12313)
License_All ()

如何删除所有括号部分 (12313)？

原文

I have the below groovy code

import java.util.regex.Matcher;
import java.util.regex.Pattern;

String myInput="License_All (12313)"
String myRegex="\\(\\d+\\)"


String ResultString
Pattern regex
Matcher regexMatcher

regex = Pattern.compile(myRegex, Pattern.DOTALL);

regexMatcher = regex.matcher(myInput);

List<String> matchList = new ArrayList<String>();
while (regexMatcher.find()) {
matchList.add(regexMatcher.group());
}

for(int i=0;i<matchList.size();i++)
{
myInput=myInput.replaceAll( matchList[i], '')
println matchList[i]
}
println myInput

It is supposed to remove (12313) part but instead it gives me the below output

(12313)
License_All ()

How do i remove all the bracket portion (12313)?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

怎会甘心 2024-12-14 02:26:17

matchList 将包含匹配的部分，即(12313)。然后，该值再次被视为 replaceAll 中的正则表达式，因此被解释为组。因此，括号不会被移除。

相反，您可以直接在 replaceAll 中使用正则表达式：

myInput.replaceAll(myRegex, "")

The matchList will contain the matched part, i.e. (12313). This one is then again taken as regular expression within replaceAll and therefore interpreted as group. Thus the brackets are not removed.

Instead you can use your regex directly in replaceAll:

myInput.replaceAll(myRegex, "")

回复收藏 0 原文

一百个冬季 2024-12-14 02:26:17

因为您正在使用 replaceAll，所以它期望 Pattern 作为第一个元素

myInput = myInput.replaceAll( Pattern.compile(myRegex, Pattern.DOTALL), '' )

应该这样做...

顺便说一句，您的原始代码可以制作成更惯用的 Groovy (而不是基本上 Java）就像这样——显然，这与你原来的问题有同样的问题，它只是一种更 Groovy 的编写代码的方式：

import java.util.regex.Pattern

String myInput = "License_All (12313)"

def regex = Pattern.compile( /\(\d+\)/, Pattern.DOTALL);

List<String> matchList = ( myInput =~ regex ) as List

matchList.each { m ->
  myInput = myInput.replaceAll( m, '')
  println m
}
println myInput

Becaue you are using replaceAll, it is expecting a Pattern as the first element

myInput = myInput.replaceAll( Pattern.compile(myRegex, Pattern.DOTALL), '' )

Should do it...

As an aside, your original code can be made into more idiomatic Groovy (rather than basically Java) like so -- obviously, this has the same issue as in your original question, it's just a more Groovy way of writing code:

import java.util.regex.Pattern

String myInput = "License_All (12313)"

def regex = Pattern.compile( /\(\d+\)/, Pattern.DOTALL);

List<String> matchList = ( myInput =~ regex ) as List

matchList.each { m ->
  myInput = myInput.replaceAll( m, '')
  println m
}
println myInput

回复收藏 0 原文

~没有更多了~