我有一个字符串:
"This is AA and this is AA and this is AA and this is the END blah blah"
我想要匹配:
"AA and this is the END"
即以END结束,回到END之前的第一次AA出现。 (语言是Java)
答案 0 :(得分:4)
试试这个:
AA(?:(?!AA).)*END
演示:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
String text = "This is AA and this is AA and this is AA and this is the END blah blah";
Matcher m = Pattern.compile("AA(?:(?!AA).)*END").matcher(text);
while(m.find()) {
System.out.println("match ->"+m.group()+"<-");
}
}
}
如果AA
和END
之间可以有换行符,请在正则表达式的开头添加(?s)
(DOT-ALL标记)。
一个简短的解释:
AA # match 'AA'
(?: # open non-capturing group 1
(?!AA). # if 'AA' cannot be seen, match any char (except line breaks)
)* # close non-capturing group 1 and repeat it zero or more times
END # match 'END'
答案 1 :(得分:1)
另一个答案:
str.substring(0, str.lastIndexOf("END")).lastIndexOf("AA");
这会创建扩展为“END”的子字符串,并在该子字符串中查找搜索字符串的最后一次出现。