使用正则表达式限制特殊字符的个数
package com.l.dubbo.service;
import java.util.HashSet;
import java.util.Set;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class ThreadPoolTest {
public static boolean check1(String context, String regEx) {
Pattern pattern = Pattern.compile(regEx);
Matcher matcher = pattern.matcher(context);
String ss = "";
String s2 = "";
boolean flag = true;
int count = 0;
//符合正则表达式的子段为特殊符号
//比较子段是否重复 若重复则特殊符号种类不止一种
while (matcher.find()) {
if (count == 0) {
ss = matcher.group();
// System.out.println(ss);
} else {
s2 = matcher.group();
// System.out.println(s2);
}
if (!(ss).equals(s2) && count > 0) {
flag = false;
}
count++;
}
// System.out.println(count);
return flag;
}
public static boolean check2(String context, String regEx) {
Pattern pattern = Pattern.compile(regEx);
Matcher matcher = pattern.matcher(context);
boolean flag = false;
//利用set的元素不能重复,来检查是否只有一种特殊符号
Set<String> set = new HashSet<String>();
while (matcher.find()) {
set.add(matcher.group());
}
if (set.size() == 1) {
flag = true;
}
return flag;
}
public static void main(String[] args) {
//s1中只包含'!‘一种特殊字符
String s1 = "dasd!a22!!!!";
//s2中包含'!‘和'@‘两种特殊字符
String s2 = "dasd!a22@!!!";
String regEx = "((?=[\\x21-\\x7e]+)[^A-Za-z0-9])";
System.out.println(check1(s1, regEx));
System.out.println(check1(s2, regEx));
System.out.println(check2(s1, regEx));
System.out.println(check2(s2, regEx));
}
}
输出结果
true
false
true
false
Process finished with exit code 0
java判断限定字符的正则
判断答案是否是ABCD的正则
public class Test {
public static void main(String[] args){
Pattern pattern = Pattern.compile("^[A-D]+$");
System.out.println(pattern.matcher("A").find());
System.out.println(pattern.matcher("E").find());
}
}
以上为个人经验,希望能给大家一个参考,也希望大家多多支持编程网。