Regex匹配字符串，如果一行在匹配字符串的末尾或中间不包含分号

Question

我想匹配一些在行尾或匹配字符串中间不包含任何分号的字符串。

如果一行:Someword; public void methodName()或仅public void methodName()

我要匹配:public void methodName()

但是如果行在我匹配的字符串的最后或中间包含分号，如:

Someword;public void; methodName();或仅public void methodName();它不匹配任何字符串。到目前为止，我在java中尝试过:

String pattern = "(public)\s+[a-zA-Z]*\s+(\bmethodName\b)\s*\(\)[^;]*$";
Matcher methodMatcher = Pattern.compile(pattern).matcher(classContent); //classContent is a java file
while (methodMatcher.find()) {
System.out.println("methodname="+methodMatcher.group());
}

Answer 1

使用

^(?:w+;)?s*publics+w+s+methodNames*()s*$

参见正则表达式证明。

--------------------------------------------------------------------------------
^                        the beginning of the string
--------------------------------------------------------------------------------
(?:                      group, but do not capture (optional
(matching the most amount possible)):
--------------------------------------------------------------------------------
w+                      word characters (a-z, A-Z, 0-9, _) (1 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
;                        ';'
--------------------------------------------------------------------------------
)?                       end of grouping
--------------------------------------------------------------------------------
s*                      whitespace (n, r, t, f, and " ") (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
public                   'public'
--------------------------------------------------------------------------------
s+                      whitespace (n, r, t, f, and " ") (1 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
w+                      word characters (a-z, A-Z, 0-9, _) (1 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
s+                      whitespace (n, r, t, f, and " ") (1 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
methodName               'methodName'
--------------------------------------------------------------------------------
s*                      whitespace (n, r, t, f, and " ") (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
(                       '('
--------------------------------------------------------------------------------
)                       ')'
--------------------------------------------------------------------------------
s*                      whitespace (n, r, t, f, and " ") (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
$                        before an optional n, and the end of the
string

样例代码:

import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
public static void main(String[] args) {
final String regex = "^(?:\w+;)?\s*public\s+\w+\s+methodName\s*\(\)\s*$";
final String string = "Someword; public void methodName()n"
+ "public void methodName()nn"
+ "Someword; public void; methodName();n"
+ "public void methodName();";

final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);

while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));

for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
}
}