包含Java中所有特定字母的正则表达式 [英] Regular Expression That Contains All Of The Specific Letters In Java

问题描述

我有一个正则表达式，可以选择所有包含特定字母的全部(不是！任何)的单词，在Notepad ++上也可以正常工作.

I have a regular expression, which selects all the words that contains all (not! any) of the specific letters, just works fine on Notepad++.

正则表达式模式；

^(?=.*B)(?=.*T)(?=.*L).+$

输入文本文件；

AL
BAL
BAK
LABAT
TAL
LAT
BALAT
LA
AB
LATAB
TAB

在记事本++中输出正则表达式；

And output of the regular expression in notepad++;

LABAT
BALAT
LATAB

由于对Notepad ++很有用，因此我在Java上尝试了相同的正则表达式，但失败了.

As It is useful for Notepad++, I tried the same regular expression on java but it is simply failed.

这是我的测试代码；

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import com.lev.kelimelik.resource.*;

public class Test {

    public static void main(String[] args) {
        String patternString = "^(?=.*B)(?=.*T)(?=.*L).+$";

        String dictionary = 
                "AL" + "\n"
                +"BAL" + "\n"
                +"BAK" + "\n"
                +"LABAT" + "\n"
                +"TAL" + "\n"
                +"LAT" + "\n"
                +"BALAT" + "\n"
                +"LA" + "\n"
                +"AB" + "\n"
                +"LATAB" + "\n"
                +"TAB" + "\n";

        Pattern p = Pattern.compile(patternString, Pattern.DOTALL);
        Matcher m = p.matcher(dictionary);
        while(m.find())
        {
            System.out.println("Match: " + m.group());
        }
    }

}

输出为错误，如下所示；

Match: AL
BAL
BAK
LABAT
TAL
LAT
BALAT
LA
AB
LATAB
TAB

我的问题很简单，这个正则表达式的java兼容版本是什么?

My question is simply, what is the java-compatible version of this regular expression?

推荐答案

特定于Java的答案

在现实生活中，我们很少需要验证行，而我看到实际上，您只是将输入用作测试数据的数组.最常见的情况是逐行读取输入并对其进行检查.我同意在Notepad ++中会有些不同，但是在Java中，应单独检查一行.

Java-specific answer

In real life, we rarely need to validate lines, and I see that in fact, you just use the input as an array of test data. The most common scenario is reading input line by line and perform checks on it. I agree in Notepad++ it would be a bit different solution, but in Java, a single line should be checked separately.

也就是说，您不应该在不同的平台上复制相同的方法.在Notepad ++中有什么好处，在Java中不一定是有好处的.

That said, you should not copy the same approaches on different platforms. What is good in Notepad++ does not have to be good in Java.

我建议使用这种几乎不使用正则表达式的方法( String#split()仍在使用):

I suggest this almost regex-free approach (String#split() still uses it):

String dictionary_str = 
        "AL" + "\n"
        +"BAL" + "\n"
        +"BAK" + "\n"
        +"LABAT" + "\n"
        +"TAL" + "\n"
        +"LAT" + "\n"
        +"BALAT" + "\n"
        +"LA" + "\n"
        +"AB" + "\n"
        +"LATAB" + "\n"
        +"TAB" + "\n";
String[] dictionary = dictionary_str.split("\n"); // Split into lines
for (int i=0; i<dictionary.length; i++)   // Iterate through lines
{
    if(dictionary[i].indexOf("B") > -1 && // There must be B
       dictionary[i].indexOf("T") > -1 && // There must be T
       dictionary[i].indexOf("L") > -1)   // There must be L
    {
        System.out.println("Match: " + dictionary[i]); // No need matching, print the whole line
    }
}

请参见 IDEONE演示

您永远不应依赖.* .这种构造总是导致回溯问题.在这种情况下，您可以使用 否定字符类 和所有量词:

You should not rely on .* ever. This construct causes backtracking issues all the time. In this case, you can easily optimize it with a negated character class and possessive quantifiers:

^(?=[^B]*+B)(?=[^T]*+T)(?=[^L]*+L)

正则表达式细目:

• ^ -字符串的开头
• (?= [^ B] * + B)-在字符串的开头，检查是否至少有一个 B 存在，且可能以0开头或更多 B
以外的字符
• (?= [^ T] * + T)- still 就在字符串开头，至少检查一个 T 可能以0个或多个除 T
以外的其他字符开头的状态
• (?= [^ L] * + L)-仍在字符串开头的 still ，请检查至少一个 L 可能以0个或多个除 L
以外的其他字符开头的状态

• ^ - start of string
• (?=[^B]*+B) - right at the start of the string, check for at least one B presence that may be preceded with 0 or more characters other than B
• (?=[^T]*+T) - still right at the start of the string, check for at least one T presence that may be preceded with 0 or more characters other than T
• (?=[^L]*+L)- still right at the start of the string, check for at least one L presence that may be preceded with 0 or more characters other than L

请参见 Java演示:

String patternString = "^(?=[^B]*+B)(?=[^T]*+T)(?=[^L]*+L)";
String[] dictionary = {"AL", "BAL", "BAK", "LABAT", "TAL", "LAT", "BALAT", "LA", "AB", "LATAB", "TAB"};
for (int i=0; i<dictionary.length; i++)
{
    Pattern p = Pattern.compile(patternString);
    Matcher m = p.matcher(dictionary[i]);
    if(m.find())
    {
        System.out.println("Match: " + dictionary[i]);
    }
}

输出:

Match: LABAT
Match: BALAT
Match: LATAB

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