Options: ^ and $ match at line breaks
Match the remainder of the regex with the options: dot matches newline (s); ^ and $ match at line breaks (m) «(?sm)»
Match the regular expression below and capture its match into backreference number 1 «(^(?:\s*)?((?:/\*(?:\*)?).*?(?<=\*/))|(?://).*?(?<=$))»
Match either the regular expression below (attempting the next alternative only if this one fails) «^(?:\s*)?((?:/\*(?:\*)?).*?(?<=\*/))»
Assert position at the beginning of a line (at beginning of the string or after a line break character) «^»
Match the regular expression below «(?:\s*)?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match a single character that is a “whitespace character” (spaces, tabs, line breaks, etc.) «\s*»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
Match the regular expression below and capture its match into backreference number 2 «((?:/\*(?:\*)?).*?(?<=\*/))»
Match the regular expression below «(?:/\*(?:\*)?)»
Match the character “/” literally «/»
Match the character “*” literally «\*»
Match the regular expression below «(?:\*)?»
Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
Match the character “*” literally «\*»
Match any single character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=\*/)»
Match the character “*” literally «\*»
Match the character “/” literally «/»
Or match regular expression number 2 below (the entire group fails if this one fails to match) «(?://).*?(?<=$)»
Match the regular expression below «(?://)»
Match the characters “//” literally «//»
Match any single character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=$)»
Assert position at the end of a line (at the end of the string or before a line break character) «$»
public static void main(String[] args) throws CommentRemoverException {
// root dir is: /Users/user/Projects/MyProject
// example for startInternalPath
CommentRemover commentRemover = new CommentRemover.CommentRemoverBuilder()
.removeJava(true) // Remove Java file Comments....
.removeJavaScript(true) // Remove JavaScript file Comments....
.removeJSP(true) // etc.. goes like that
.removeTodos(false) // Do Not Touch Todos (leave them alone)
.removeSingleLines(true) // Remove single line type comments
.removeMultiLines(true) // Remove multiple type comments
.startInternalPath("src.main.app") // Starts from {rootDir}/src/main/app , leave it empty string when you want to start from root dir
.setExcludePackages(new String[]{"src.main.java.app.pattern"}) // Refers to {rootDir}/src/main/java/app/pattern and skips this directory
.build();
CommentProcessor commentProcessor = new CommentProcessor(commentRemover);
commentProcessor.start();
}
# 5 楼答案
我认为eclipse支持正则表达式搜索和替换。
我想试试这样：
search: (?s)(?>\/\*(?>(?:(?>[^*]+)|\*(?!\/))*)\*\/)
replace all with no-space-character or nothing literally
# 1 楼答案
我用
Ctrl+F命令和这个正则表达式完成了这项工作这只会替换
//comments# 2 楼答案
因为没有人在*NIX中提到文本处理工具
grep、sed、awk等等，所以我在这里发布我的解决方案如果操作系统是*NIX，可以使用文本处理工具
sed删除注释# 3 楼答案
这个对我来说很好。。。我将其添加为一个IntelliJ IDEA搜索模板，用于JavaDoc和/或单行评论
定义
# 4 楼答案
我做了一个开源的library为了这个目的，你可以删除Java注释
它支持删除或不删除TODO。
它还支持JavaScript、HTML、CSS、属性、JSP和XML注释
如何使用它的小代码片段：
# 5 楼答案
我认为eclipse支持正则表达式搜索和替换。 我想试试这样：
也与主题相关： Eclipse, regular expression search and replace
我编辑了正则表达式并进行了测试： http://regex101.com/r/sU4vI2 不确定它是否适用于你的情况
# 6 楼答案
我找到了解决办法
下面是用于查找两种注释类型的正则表达式
\/\*([\S\s]+?)\*\/和(?s)/\*.*?\*/打开带有注释的
.java文件并打开搜索对话框。（Search -> File Search）并粘贴上面的一个reg ex，然后选中右侧的Regular expression勾选框。现在，您可以搜索并选择“全部替换”，将其替换为第二个框中未键入的内容使用replace-with选项，我已经清除了java文件中的所有注释