Regular expression in Java

Talking cat 2021-03-25 00:07:38
regular expression java


One 、 Introduction of common symbols

1.1 Predefined character class

Symbol

explain

.

Any character ( May or may not match the line terminator )

\d

Numbers :[0-9]

\D

The digital : [^0-9]

\s

Blank character :[ \t\n\x0B\f\r]

\S

Nonwhite space character :[^\s]

\w

Word characters :[a-zA-Z_0-9]

\W

Nonword character :[^\w]

\

Escape character , such as "\\" matching "\" ,"\{" matching "{".

1.2 quantifiers

Symbol

explain

*

Equivalent to {0,} matching 0 At most characters before it . For example, regular expressions “zo*” Can match “z” as well as “zoo”; Regular expressions “.*” It means to be able to match any string .

+

Equivalent to {1,} Match previous subexpression one or more times . For example, regular expressions 9+ matching 9、99、999 etc. .

?

Equivalent to {0,1} Match previous subexpression zero or once . for example ,"do(es)?" Can match "do" or "does" Medium "do" . This metacharacter has another use , It means non greedy pattern matching , There will be an introduction later

{n}

Matched definite n Time . for example ,“e{2}” Can't match “bed” Medium “d”, But it matches “seed” Two of them “e”.

{n,}

Match at least n Time . for example ,“e{2,}” Can't match “bed” Medium “e”, But it can match. “seeeeeeeed” All in “e”.

{n,m}

Least match n Times and at most m Time .“e{1,3}” Will match “seeeeeeeed” Top three in “e”.

1.3 Boundary matching symbols

Symbol

explain

^

The beginning of a row

$

End of line

\b

Word boundaries

\B

Non word boundary

\A

The beginning of the input

\G

The end of the last match

\Z

The end of the input , Only for the last Terminator ( If any )

\z

The end of the input

1.3 Other common symbols

[] Use -- or

explain

[]

Match any character in brackets

[abc]

a、b or c( Simple class )

[^abc]

Any character , except a、b or c( no )

[a-zA-Z]

a To z or A To Z, Both ends of the letter are included ( Range )

[a-d[m-p]]

a To d or m To p:[a-dm-p]( Combine )

[a-z&&[def]]

d、e or f( intersection )

[a-z&&[^bc]]

a To z, except b and c:[ad-z]( subtract )

[a-z&&[^m-p]]

a To z, Instead of m To p:[a-lq-z]( subtract )

() Use -- Group

()

take () The expression enclosed between is defined as “ Group ”(group), And save the characters matching the expression to a temporary area , This metacharacter is very useful for string extraction . Capture groups can be numbered by counting their open brackets from left to right .

(\d)

The first group

((A)(B(C)))

The first group ((A)(B(C))) The second group (A) The third group (B(C)) Group 4 (C)

PS: More other symbols , Referable http://www.cnblogs.com/Mustr/p/6057159.html

Two 、 Common operations

2.1 matching

String matches() Method . Match the whole string with rules , As long as there's one violation of the rules , It's the end of the match , return false.

give an example :

public static void checkQQ(){
String qq = "123a45664";
String regex = "[1-9]\\d{4,14}";
boolean flag = qq.matches(regex);
if(flag)
System.out.println(qq+"...is ok");
else
System.out.println(qq+"... illegal ");
} // illegal 
 /*
matching
The mobile phone number segment is only 13xxx 15xxx 18xxxx
*/
public static void checkTel()
{
String tel = "16900001111";
String telReg = "1[358]\\d{9}";
System.out.println(tel.matches(telReg));
}

2.2 cutting

String split() Method ; According to the given Regular expressions Split this string by matching . Returns an array .

give an example :

 public static void splitDemo()
{
String str = "avg bb geig glsd abc";
String reg = " +";// Cut by multiple spaces
String[] arr = str.split(reg);
System.out.println(arr.length);
for(String s : arr)
{
System.out.println(s);
}
} 

Group group give an example :

public static void splitDemo()
{
String str = "erkktyqqquizzzzzo";
String reg ="(.)\\1+";// Cut according to the overlapping words
// You can encapsulate rules into a group . use () complete . Groups appear with numbers .
// from 1 Start . If you want to use an existing group, you can use \n(n It's the number of the group ) In the form of .
String[] arr = str.split(reg);
System.out.println(arr.length);
for(String s : arr)
{
System.out.println(s);
}
}
// er,ty,ui,o

2.3 Replace

String replaceAll(regex,str) Method ; Using the given replacement Replace this string with all that match the given Regular expressions Substring of .

ps: If regex There are defined groups , This can be done in the second parameter by $ Symbol gets the existing group in the regular expression .

give an example :

 public static void replaceAllDemo()
{
String str = "wer1389980000ty1234564uiod234345675f";// Replace the number in the string with #.
str = str.replaceAll("\\d{5,}","#");
System.out.println(str);
}
// wer#ty#uio#f

Group group give an example :

 public static void replaceAllDemo()
{
String str1 = "erkktyqqquizzzzzo";// Replace the reduplication with $. // Replace overlapping characters with single letters .zzzz->z
str = str.replaceAll("(.)\\1+","$1");
System.out.println(str);
}
// erktyquizo

2.4 obtain

---- Take out the substring that conforms to the rule in the string .

Operation steps : 1, Encapsulating regular expressions as objects . 2, Associate the regular object with the string to be manipulated . 3, After correlation , Get regular matching engine . 4, Through the engine to operate the substring that conforms to the rules , Take out, for example .

give an example :

public static void getDemo()
{
String str = "yin yu shi wo zui cai de yu yan";
System.out.println(str);
String reg = "\\b[a-z]{3}\\b";// Match words with only three letters
// Encapsulating rules as objects .
Pattern p = Pattern.compile(reg);
// Associate the regular object with the string you want to work with . Get matcher object .
Matcher m = p.matcher(str);
//System.out.println(m.matches());// Actually String Class matches Method . It's used Pattern and Matcher Object to complete .
// Just being String After encapsulation of the method , It's easier to use . But the function is single .
// boolean b = m.find();// Apply rules to strings , And search the substring according to the rules .
// System.out.println(b);
// System.out.println(m.group());// It is used to get the result after matching .
while(m.find())
{
System.out.println(m.group());
System.out.println(m.start()+"...."+m.end());
// start() The initial subscript of the character ( contain )
//end() The ending subscript of the character ( It doesn't contain )
}
} 

Participation of this paper Tencent cloud media sharing plan , You are welcome to join us , share .

版权声明
本文为[Talking cat]所创,转载请带上原文链接,感谢
https://javamana.com/2021/03/20210309134447697o.html

  1. 【计算机网络 12(1),尚学堂马士兵Java视频教程
  2. 【程序猿历程,史上最全的Java面试题集锦在这里
  3. 【程序猿历程(1),Javaweb视频教程百度云
  4. Notes on MySQL 45 lectures (1-7)
  5. [computer network 12 (1), Shang Xuetang Ma soldier java video tutorial
  6. The most complete collection of Java interview questions in history is here
  7. [process of program ape (1), JavaWeb video tutorial, baidu cloud
  8. Notes on MySQL 45 lectures (1-7)
  9. 精进 Spring Boot 03:Spring Boot 的配置文件和配置管理,以及用三种方式读取配置文件
  10. Refined spring boot 03: spring boot configuration files and configuration management, and reading configuration files in three ways
  11. 精进 Spring Boot 03:Spring Boot 的配置文件和配置管理,以及用三种方式读取配置文件
  12. Refined spring boot 03: spring boot configuration files and configuration management, and reading configuration files in three ways
  13. 【递归,Java传智播客笔记
  14. [recursion, Java intelligence podcast notes
  15. [adhere to painting for 386 days] the beginning of spring of 24 solar terms
  16. K8S系列第八篇(Service、EndPoints以及高可用kubeadm部署)
  17. K8s Series Part 8 (service, endpoints and high availability kubeadm deployment)
  18. 【重识 HTML (3),350道Java面试真题分享
  19. 【重识 HTML (2),Java并发编程必会的多线程你竟然还不会
  20. 【重识 HTML (1),二本Java小菜鸟4面字节跳动被秒成渣渣
  21. [re recognize HTML (3) and share 350 real Java interview questions
  22. [re recognize HTML (2). Multithreading is a must for Java Concurrent Programming. How dare you not
  23. [re recognize HTML (1), two Java rookies' 4-sided bytes beat and become slag in seconds
  24. 造轮子系列之RPC 1:如何从零开始开发RPC框架
  25. RPC 1: how to develop RPC framework from scratch
  26. 造轮子系列之RPC 1:如何从零开始开发RPC框架
  27. RPC 1: how to develop RPC framework from scratch
  28. 一次性捋清楚吧,对乱糟糟的,Spring事务扩展机制
  29. 一文彻底弄懂如何选择抽象类还是接口,连续四年百度Java岗必问面试题
  30. Redis常用命令
  31. 一双拖鞋引发的血案,狂神说Java系列笔记
  32. 一、mysql基础安装
  33. 一位程序员的独白:尽管我一生坎坷,Java框架面试基础
  34. Clear it all at once. For the messy, spring transaction extension mechanism
  35. A thorough understanding of how to choose abstract classes or interfaces, baidu Java post must ask interview questions for four consecutive years
  36. Redis common commands
  37. A pair of slippers triggered the murder, crazy God said java series notes
  38. 1、 MySQL basic installation
  39. Monologue of a programmer: despite my ups and downs in my life, Java framework is the foundation of interview
  40. 【大厂面试】三面三问Spring循环依赖,请一定要把这篇看完(建议收藏)
  41. 一线互联网企业中,springboot入门项目
  42. 一篇文带你入门SSM框架Spring开发,帮你快速拿Offer
  43. 【面试资料】Java全集、微服务、大数据、数据结构与算法、机器学习知识最全总结,283页pdf
  44. 【leetcode刷题】24.数组中重复的数字——Java版
  45. 【leetcode刷题】23.对称二叉树——Java版
  46. 【leetcode刷题】22.二叉树的中序遍历——Java版
  47. 【leetcode刷题】21.三数之和——Java版
  48. 【leetcode刷题】20.最长回文子串——Java版
  49. 【leetcode刷题】19.回文链表——Java版
  50. 【leetcode刷题】18.反转链表——Java版
  51. 【leetcode刷题】17.相交链表——Java&python版
  52. 【leetcode刷题】16.环形链表——Java版
  53. 【leetcode刷题】15.汉明距离——Java版
  54. 【leetcode刷题】14.找到所有数组中消失的数字——Java版
  55. 【leetcode刷题】13.比特位计数——Java版
  56. oracle控制用户权限命令
  57. 三年Java开发,继阿里,鲁班二期Java架构师
  58. Oracle必须要启动的服务
  59. 万字长文!深入剖析HashMap,Java基础笔试题大全带答案
  60. 一问Kafka就心慌?我却凭着这份,图灵学院vip课程百度云