java regular expression tutorial

The way to use Str.matches (REG) in Java is to return a Boolean value, and we sometimes need to return the content of the matching value, the implementation code is as follows: Public Static void Main (string[] args) { = "2[0-9]{11}-{0,1}[0-9]{0,1}"; = Pattern.compile (reg); = "200000000000-2200000000000"; =P.matcher (str); while (M.find ()) System.out.println (M.group (0)); }This will

/*** The IP address in the regular advance string *@paramipstring *@return */ Public StaticListgetips (String ipstring) {string RegEx= "((2[0-4]\\d|25[0-5]| [01]?\\d\\d?] \\.) {3} (2[0-4]\\d|25[0-5]| [01]?\\d\\d?]; ListNewArraylist(); Pattern P=Pattern.compile (regEx); Matcher m=P.matcher (ipstring); while(M.find ()) {String result=M.group (); Ips.add (result); } returnIPs; } Public Static voidMain (string[] args) {Strin

=NewStringBuffer (); Try{Pattern P= Pattern.compile ("[$]" ([^$]*?) [$]"); while(Sc.hasnext ()) {System.out.println (sc.tostring ()); Matcher m=P.matcher (Sc.nextline ()); while(M.find ()) {//Find and Replace parametersSystem.out.println (M.group (1)); //get the value from the map based on keyM.appendreplacement (buf, Contentparammap.get (M.group (1))! =NULL? Contentparammap.get (M.group (1)). ToString (): ""); } m.appendtail (BUF); }//if (Stringutils.isempty (buf.tostring ())) {//b

/*** Get the address of the picture on the webpage*/ Public StaticSetgetimgstr (String htmlstr) {SetNewHashset(); String img= ""; Pattern P_image; Matcher M_image; //String regex_img = "]*?> "; //Image Link AddressString regex_img = "]*?> "; P_image=pattern.compile (regex_img, pattern.case_insensitive); M_image=P_image.matcher (HTMLSTR); while(M_image.find ()) {//get DataIMG =M_image.group (); //match src data in Matcher m = pattern.compile ("src\\s*=\\s*\"? *?) (\ "|>|\\s+)"

String str = "To goods | |?" >\\n Yangchenghu Hairy Crabs [email protected]#$%^* () _+-=?:\ ",.] \\|~.,\/?? \\\\|\\BR, great location! "1 String tradedescription = "| |? >\\n Yangchenghu Hairy Crabs [email protected]#$%^]; 2 String description = "Give the product #goodsName # likes"; 3 String result = Description.replaceall ("#goodsName #", Matcher.quotereplacement (Commonutil.tostring ( (tradedescription)));Reference:Http://www.colabug.com/thread-1131948-1-1.htmlSpecial character causes string

/*** Verify the time format and return true if correct* @param xxx* @return*/private static Boolean Checkdateformat (String xxx) { String Correctdateformat = "^ ((\\d{2} (([02468][048]) | ( [13579] [26])) [\\-\\/\\s]? (((0? [13578]) | (1[02])) [\\-\\/\\s]? ((0? [1-9]) | ([1-2][0-9]) | (3[01])) | ((0? [469]) | (11)) [\\-\\/\\s]? ((0? [1-9]) | ([1-2][0-9]) | (30))) | (0?2[\\-\\/\\s]? ((0? [1-9]) | ([1-2][0-9])))) | (\\d{2} ([02468][1235679]) | ( [13579] [01345789])) [\\-\\/\\s]? (((0? [13578])

/** * * @author Zen Johnny * @date April 29, 2018 PM 4:31:07 * */package Demo.regex;public class Regexreplacedemo {public S tatic void Replacedemo (String string, string regex,string replacement) {System.out.println (String.replaceall (Regex, replacement));} public static void Main (string args[]) {//case1: As long as the number is more than 5 consecutive, the segment string is replaced with #string string = " Bn45353453assss3444effffewtvdt4efvdfdgf455454ggt5grgfer3 "; String regex = "\\d{5,}+"

Greediness (greedy type): Maximum matchX, x*, x+, X{n,} are the maximum matches. For example you want to use "In the greediness mode, the match will be as large as possible until the entire content is matched, and when the match is found to be unsuccessful, the fallback narrowing the match until the match is successfulString test = "A= ""# # #");Output: a## #abbReluctant (laziness) (barely): Minimum matchX??、 x*, x+?, X{n,}? Is the minimum match, in fact x{n,m}? and X{n}? Some superfluous. Add a

Tips for using Java to implement MongoDB Regular Expression query: // Fuzzy searchPrivate BasicDBObject getLikeStr (String findStr ){Pattern pattern = Pattern. compile ("^. *" + findStr + ". * $", Pattern. CASE_INSENSITIVE );Return new BasicDBObject ("$ regex", pattern );} DBObject query = new BasicDBObject ();Query. put ("real_name", getLikeStr (userName ));

The judgment of the mailbox1 /**2 * Determine if email is legal3 */4 Public Static BooleanCheckloginemail (String loginstr) {5String regEx = "^ ([a-z0-9_a-z]+[-|\\.]?) +[a-z0-9_a-z]@ ([a-z0-9_a-z]+ (-[a-z0-9_a-z]+) \ \.) +[a-za-z_]{2,}$ ";6Pattern p =Pattern.compile (regEx);7Matcher m =P.matcher (LOGINSTR);8 returnM.matches ();//Boolean9}Simple format Public Static Boolean Iskachao (String kachao) { = pattern.compile ("^ (0) {5}3[0-9]{4}$"); = P.matcher (kac

/** * * A:正则表达式 * 是指一个用来描述或者匹配一系列符合某个语法规则的字符串的单个字符串。其实就是一种规则。有自己特殊的应用。 * 作用:比如注册邮箱,邮箱有用户名和密码,一般会对其限制长度,这个限制长度的事情就是正则表达式做的 * B:案例演示 * 需求：校验qq号码. * 1:要求必须是5-15位数字 * 2:0不能开头 * 3:必须都是数字 * a:非正则表达式实现 * b:正则表达式实现 */public class Demo1_regex {public static void main (string[] args) {System.out.println (CHECKQQ ("012345")); System.out.println (CHECKQQ ("a1b345")); System.out.println (CHECKQQ ("123456")); System.out.println (CHECKQQ ("1234567890987654321")); String regex = "[1-

The instance is as follows: public static string Repdomain (string source, string domain, string element, string attr ) {String img = ""; Pattern P_image; Matcher M_image; String regex_img = " The above is the Java regular expression for everyone to get the specified value of the specified attribute of the HTML tag and replace the

Occasionally found Dongdong, do not remember is that the Bolg. Very good article .... §1 Dark Years There is a string, how do I find out if there are Y and F characters? The darkest way is to: Program 1: I know if, for statements, and Charat (). Class test{public static void Main (String args[]) {String str= "For my money, the important thing" +"About the meeting is bridge-building";Char x= ' y ';Char y= ' f ';Boolean result=false;for (int i=0;iChar Z=str.charat (i); System.out.println (z);

for(inti = 0; i ) { the //ensure that a continuous string B matches a segment in a the if(A_t[i] = = B_t[j] J b_t.length) { thetemp++;117J + +;118 //The continuous string b matches a segment in a119 if(temp = =b_t.length) { -count++;121temp = 0;122j = 0;123 }124 } the //as long as there is a character mismatch, the temp count is never126 Else {127temp = 0; -j = 0;129 } the

Floating-point judgmentpublic static Boolean isdecimal (String str) {if (Str==null | | "". Equals (str))return false;Java.util.regex.Pattern Pattern = Pattern.compile ("[0-9]* (\ \.?) [0-9]* ");return Pattern.matcher (str). matches ();}Integral type judgmentpublic static boolean isinteger (String str) {if (str==null)return false;Pattern pattern = Pattern.compile ("[0-9]+");return Pattern.matcher (str). matches ();}Floating-point test cases:public void Testisdecimal () {Asserttrue ("123", Test.is

Note: 1. [] represents the character set, does not represent the inclusion () Table capture and group Eg: can only be QQ mailbox/gmail mailbox/outlook Mailbox/163 mailbox One (qq|gmail|163) {1}\.com and [QQ] | ( Gmail) | (163)] \.com; the latter is wrong. Constructs a match character X character x\\ backslash character \0n with octal value 0 characters n (0 　　Regular expression six: matching rules for

A colleague from C + + asked me to help write a regular expression: Verify that a value is a number. I simply wrote one in Java code. public class Test {public static Boolean CheckNumber (double value) {String str = string.valueof (value); String regex = "^" (-?[ 1-9]\\d*\\.? \\d*) | (-?0\\.\\d*[1-9]) | (-? [0]) | (-? [0]\\.\\d*] $ "; return str.matches (rege

Private Set getcodes (String s) { Setnew hashset(); = Codepattern.matcher (s); while (Matcher.find ()) { Resultset.add (matcher.group (1)); } return resultSet; } Private Static Pattern Codepattern = Pattern.compile ("\" code\ ": \" (\\w+) \ ",");Java Regular expression-capturing group

S:arr) * { $ Ts.add (s);Panax Notoginseng } -Regex= "0* (\\d+)"; the for(String s:ts) + { ASystem.out.println (S.replaceall (Regex, "$")); the } + } -}View CodeOperation Result:1 192.168.10.34 2 127.0.0.1 3 105.70.11.75 4 3.3.3.3 5 6 7 192.168.010.034 8 127.000.000.001 9 105.070.011.075 003.003.003.003 A 3.3.3.3 105.70.11.75 127.0.0.1 192.168.10.34 View CodeWhen the extra 0 is removed, the mode used is: "0* (\\d+)", so that even if all is 0, the