Java正则表达式从0-9连续查找最后一位并且前6位相同

Question

我想要一个在 java 中找到的正则表达式 - 只有前 6 位数字相同和 - 最后一位数字范围从 0-9 连续。

13032501303251

1304150130415113041521304153130415413041551304156130415713041581304159

在这种情况下，表达式将匹配 130415X。

我开发了两个独立的正则表达式作为

        Pattern f6 = Pattern.compile("^......");
        Pattern last = Pattern.compile("\\d$");

Answer 1

因此，我们编写一个正则表达式来对前 6 位数字(后跟一个 0)进行分组，并检查该组后跟最多 9 的计数。我们在您的字符串中找到匹配项，然后打印出第一个分组如果找到匹配项。

代码如下:

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

public class HelloWorld{

 public static void main(String []args){
    String test = "1304150 1304151 1304152 1304153 1304154 1304155 1304156 1304157 1304158 1304159\r\n" +
                  "5304150 5304151 5304152 5304153 5304154 5304155 5304156 5304157 5304158 5304159\r\n" + 
                  "7304150 7304153 71304156";

    Pattern p = Pattern.compile("(\\d{6})0 (?:\\1)1 (?:\\1)2 (?:\\1)3 (?:\\1)4 (?:\\1)5 (?:\\1)6 (?:\\1)7 (?:\\1)8 (?:\\1)9", Pattern.MULTILINE);
    Matcher m = p.matcher(test);

    while (m.find()) {
        System.out.println(new String(m.group(1)) + "X");
    }
 }

输出:

130415X
530415X

如果找到匹配项，它会打印出相关的 6 位数字和一个“X”。 See it in action here .

Regex explanation .本质上，它将第一个匹配的 6 位数字分组，后跟一个 0。然后，它查找该组后跟一个 1，该组后跟一个 2，等等。整个字符串中的双 '\' 只是为了转义Java 字符串中的字符，正则表达式应该在没有双斜杠的情况下读取:

(\d{6})0 (?:\1)1 (?:\1)2 (?:\1)3 (?:\1)4 (?:\1)5 ( ？:\1)6 (?:\1)7 (?:\1)8 (?:\1)9

   NODE                     EXPLANATION
--------------------------------------------------------------------------------
  (                        group and capture to \1:
    \d{6}                    digits (0-9) (6 times)
  )                        end of \1
  0                        '0 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  1                        '1 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  2                        '2 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  3                        '3 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  4                        '4 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  5                        '5 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  6                        '6 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  7                        '7 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  8                        '8 '
  (?:                      group, but do not capture:
    \1                       what was matched by capture \1
  )                        end of grouping
  9                        '9'

按照评论中的要求，您可以使用相反的方法:

^(?!((\d{6})0 (?:\2)1 (?:\2)2 (?:\2)3 (?:\2)4 (?:\2)5 (?:\2)6 (?:\2)7 (?:\2)8 (?:\2)9)).*$

正如预期的那样，反向匹配有点复杂，但是 here's the explanation你也可以see it in action .