[英]Regex - list the characters which are not matching the pattern
我有下面的正則表達式,它說明給定的輸入是否與模式匹配
String input="The input here @$%/";
String pattern="[a-zA-Z0-9,.\\s]*";
if (!input.matches(pattern)) {
System.out.println("not matched");
}else{
System.out.println("matched");
}
我能知道如何增強它以列出輸入中與模式不匹配的字符嗎? 例如這里@$%/
正如anubhava 在評論中已經提到的那樣,只需使用input.replaceAll(pattern, "")
。
演示:
class Main {
public static void main(String[] args) {
String input = "The input here @$%/";
String pattern = "[a-zA-Z0-9,.\\s]*";
String nonMatching = input.replaceAll(pattern, "");
System.out.println(nonMatching);
}
}
Output:
@$%/
利用
[^a-zA-Z0-9,.\s]
請參閱正則表達式證明。
解釋
[^a-zA-Z0-9,.\s] any character except: 'a' to 'z', 'A' to
'Z', '0' to '9', ',', '.', whitespace (\n,
\r, \t, \f, and " ")
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
public static void main(String[] args) {
final String regex = "[^a-zA-Z0-9,.\\s]";
final String string = "The input here @$%/";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
}
}
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.