模式匹配问题与java中的特殊字符

时间:2016-01-09 07:58:31

标签: java regex

我想验证字符串模式。如果字符串中没有任何特殊字符,则使用下面的代码。

例如:

    Pattern p = Pattern.compile("Dear User, .* is your One Time Password(OTP) for registration.",Pattern.CASE_INSENSITIVE);

    Matcher m = p.matcher("Dear User, 999 is your One Time Password(OTP) for registration.");

    if (m.matches()){
        System.out.println("truee");                
    }else{
        System.out.println("false");
    }  // output false 
如果我删除(和),

及以下工作正常。

Pattern p = Pattern.compile("Dear User, .* is your One Time Password OTP for registration.",Pattern.CASE_INSENSITIVE);

    Matcher m = p.matcher("Dear User, 999 is your One Time Password OTP for registration.");

    if (m.matches()){
        System.out.println("truee");                
    }else{
        System.out.println("false");
    }  // output true

3 个答案:

答案 0 :(得分:0)

试试这个:

d[!is.finite(d)] <- NA

(,)和。用于正则表达式。如果你想要正常的行为,你需要逃脱它们。

答案 1 :(得分:0)

在正则表达式中,当你需要匹配字面时,你必须始终注意特殊字符。

在这种情况下,您有3个字符:((用于打开组),)(用于关闭组)和.(匹配任何字符)。< / p>

要按字面意思匹配它们,您需要将它们转义,或放入字符类[...]

请参阅fixed demo

答案 2 :(得分:0)

package com.ramesh.test;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class PatternMatcher {

public boolean containsPattern(String input, String pattern) {
    if (input == null || input.trim().isEmpty()) {
        System.out.println("Incorrect format of string");
        return false;
    }
    // “ * ”is replaced by “ .* ” in replaceAll()
    //“ .* ” Use this pattern to match any number, any string (including the empty string) of any characters.
    String inputpattern = pattern.replaceAll("\\*", ".*");
    System.out.println("first input" + inputpattern);
    Pattern p = Pattern.compile(inputpattern);
    Matcher m = p.matcher(input);
    boolean b = m.matches();
    return b;
}

public boolean containsPatternNot(String input1, String pattern1) {
    return (containsPattern(input1, pattern1) == true ? false : true);
}

public static void main(String[] args) {

    PatternMatcher m1 = new PatternMatcher ();
    boolean a = m1.containsPattern("ma5&*%u&^()k5.r5gh^", "m*u*r*");
    System.out.println(a);// returns True

    boolean d = m1.containsPattern("mur", "m*u*r*");
    System.out.println(d);// returns True

    boolean c = m1.containsPatternNot("ma5&^%u&^()k56r5gh^", "m*u*r*");
    System.out.println(c);// returns false

    boolean e = m1.containsPatternNot("mur", "m*u*r*");
    System.out.println(e);// returns false

}

}

输出:true                 真正                 假                 假