我想忽略/定位一行的非块注释段。
例如,以下字符串需要ALL得到字符串"foobar"
"foo/*comment*/bar"
"comm*/foobar/*ent"
"comment*/foobar"
"foobar/*comment"
实施此方法的最佳方法是什么?
答案 0 :(得分:1)
编辑请试试这个:
public static void main(String[] args) {
String[] input = new String[]{"foo/*comment*/bar", "comm*/foobar/*ent", "comment*/foobar", "foobar/*comment"};
String pattern = "(?:/\\*[^\\*]+(?:\\*/)?|(?:/\\*)?[^\\*]+\\*/)";
List<String> listMatches = new ArrayList<String>();
String result = "";
for (String m : input) {
result = m.replaceAll(pattern, ""); //remove matches
listMatches.add(result); // append to list
System.out.println(result);
}
}
输出:
foobar
foobar
foobar
foobar
以下是正则表达式的解释:
(?: 1st non-capturing group starts
/\\* match /* literally
[^\\*]+ 1 or more times characters except *
(?: 2nd non-capturing group starts
\\*/ match */ literally
) 2nd non-capturing group ends
? match previous non-capturing group 0 or 1 time
| Or (signals next alternative)
(?: 3rd non-capturing group starts
/\\* match /* literally
) 3rd non-capturing group ends
? match previous non-capturing group 0 or 1 time
[^\\*]+ 1 or more times characters except *
\\*/ match */ one time
) 1st non-capturing group ends
答案 1 :(得分:0)
这与this stackoverflow post中的帖子具有相同的逻辑,但是以递归的形式实现,以满足您对简单性的渴望:
public static String cleanComment(String str) {
int open = str.indexOf("/*"), close = str.indexOf("*/");
if( (open&close) < 0 ) return str;
open &= Integer.MAX_VALUE;
close &= Integer.MAX_VALUE;
if(open < close) {
if(close > str.length()) {
return str.substring(0, open);
} else {
return str.substring(0, open) + cleanComment( str.substring(close+2) );
}
} else {
return cleanComment( str.substring(close+2) );
}
}