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法律研究在 Java 中解释正则表达式“s”元字符
子表达式/元字符“\s”匹配空等效项。
示例 1
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegexExample {
public static void main( String args[] ) {
String regex = "\s";
String input = "Hello how are you welcome to Tutorialspoint !";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(input);
int count = 0;
while(m.find()) {
count++;
}
System.out.println("Number of matches: "+count);
}
}输出
Number of matches: 7
示例 2
以下示例读取一个字符串，并移除其中的所有多余空格。
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
public static void main(String args[]) {
//Reading String from user
System.out.println("Enter a String");
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
//Regular expression to match spaces (one or more)
String regex = "\s+";
//Compiling the regular expression
Pattern pattern = Pattern.compile(regex);
//Retrieving the matcher object
Matcher matcher = pattern.matcher(input);
//Replacing all space characters with single space
String result = matcher.replaceAll(" ");
System.out.print("Text after removing unwanted spaces: \n"+result);
}
}输出
Enter a String hello this is a sample text with irregular spaces Text after removing unwanted spaces: hello this is a sample text with irregular spaces
更新于：2019 年 11 月 19 日
124 次浏览
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