java Java正则表达式匹配方括号
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Java Regular expression matching square brackets
提问by Pruthvi Raj Nadimpalli
I'm trying to do the following using regular expression (java replaceAll):
我正在尝试使用正则表达式（java replaceAll）执行以下操作：
**Input:**
Test[Test1][Test2]Test3
**Output**
TestTest3
In short, i need to remove everything inside square brackets including square brackets.
简而言之，我需要删除方括号内的所有内容，包括方括号。
I'm trying this, but it doesn't work:
我正在尝试这个，但它不起作用：
\[(.*?)\]
Would you be able to help?
你能帮忙吗？
Thanks,
Sash
谢谢，
腰带
回答by Rizwan M.Tuman
You can try this regex:
你可以试试这个正则表达式：
\[[^\[]*\]
and replace by empty
并替换为空
Sample Java Source:
示例 Java 源代码：
final String regex = "\[[^\[]*\]";
final String string = "Test[Test1][Test2]Test3\n";
final String subst = "";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
final String result = matcher.replaceAll(subst);
System.out.println(result);
回答by Tim Biegeleisen
Your original pattern works for me:
你的原始模式对我有用：
String input = "Test[Test1][Test2]Test3";
input = input.replaceAll("\[.*?\]", "");
System.out.println(input);
Output:
输出：
TestTest3
Note that you don't need the parentheses inside the brackets. You would use that if you planned to capture the contents in between each pair of brackets, which in your case you don't need. It isn't wrong to have them in there, just not necessary.
请注意，您不需要括号内的括号。如果您计划捕获每对括号之间的内容，您将使用它，在您的情况下您不需要。把它们放在那里并没有错，只是没有必要。
Demo here:
演示在这里：
