How to get patterns of groups in Java regex? - java

Assume that I am given a regex pattern, whole_pattern
Pattern p = Pattern.compile(whole_pattern);
Matcher m = p.matcher(line);
if (m.find()) {
String s1 = m.group(1);
String s2 = m.group(2);
}
Obviously, we can get matched string of each group. But can we get the pattern of each group in the whole_pattern string? For example if whole_pattern = (\\d+)(\\w+), then patterns of group 1 and group 2 are \\d+ and \\w+, respectively.


You can use regexp for regexp:
public void simpleTest() {
String whole_pattern = "(\\d+)(\\w+)";
System.out.println(patternGroups(whole_pattern));
}
private List<String> patternGroups(String patternString) {
List<String> result = new ArrayList<>();
String pattern = "\\(([^()]+)\\)";
Pattern p = Pattern.compile(pattern);
Matcher m = p.matcher(patternString);
while (m.find()) {
result.add(m.group(1));
}
return result;
}
Output of simpleTest() call will be:
[\d+, \w+]


Split your pattern string by ) and remove the first character in each elementh of the resulting array.

Related

Find all occurrences of a regex pattern in a line

My String is like this (one single line):
String input = "Details of all persons. Person=details=John Smith-age-22; Person=details=Alice Kohl-age-23; Person=details=Ram Mohan-city-Dallas; Person=details=Michael Jack-city-Boston;"
I want to find out using regex matching all the persons with its details (basically text from details upto the char prior to semicolon). I am interested in finding:
details=John Smith-age-22
details=Alice Kohl-age-23
details=Ram Mohan-city-Dallas
details=Michael Jack-city-Boston
Can someone tell me how to do this ? Sorry, I could not find any example like that over the net. Thanks.

You can try this code.
public static void main(String[] args) {
String input = "Details of all persons. Person=details=John Smith-age-22; Person=details=Alice Kohl-age-23; Person=details=Ram Mohan-city-Dallas; Person=details=Michael Jack-city-Boston;";
Pattern pattern = Pattern.compile("(?<=Person=).*?(?=;)");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
String str = matcher.group();
System.out.println(str);
}
}
No assertion
public static void main(String[] args) {
String input = "Details of all persons. Person=details=John Smith-age-22; Person=details=Alice Kohl-age-23; Person=details=Ram Mohan-city-Dallas; Person=details=Michael Jack-city-Boston;";
Pattern pattern = Pattern.compile("Person=.*?;");
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
String str = matcher.group();
System.out.println(str.substring(7, str.length()-1));
}
}

I suspect you will find it easiest if you put the fields you are looking for into groups so that you can extract the details you want.
Something like:
Pattern personPattern = Pattern.compile("Person=details=(\\w+)-(age-\\d+|city-\\w+); ");
Matcher matcher = personPattern.match(input);
while (matcher.find()) {
String name = matcher.group(1);
String field = matcher.group(2);
...
}

Java Regex compress String

I have random String for example "aaaaaaBccccCCCCd" I need make regex which searches the text for groups to get effect "a6B1c4C4d1". My regex looks like that "(\\D+)\\D*\\1" but he lost single letters, so in this sample B and d.
Maybe someone would have an idea?
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Compress {
public static void main(String[] args) {
String text = "aaaaaaBccccCCCCd";
String regex = "(\\D+)\\D*\\1"; // or (.+).*\\1
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(text);
String result = new String();
while (matcher.find()) {
String letter = matcher.group().substring(0, 1);
String numberOfLetter = String.valueOf(matcher.group().length());
result = result + letter + numberOfLetter;
}
System.out.println(result);
}
}
Thank you.

Use the following approach based on Matcher#appendReplacement:
String text = "aaaaaaBccccCCCCd"; //a6B1c4C4d1
String regex = "(.)(\\1*)";
String pattern = "test";
Pattern r = Pattern.compile(regex);
Matcher m = r.matcher(text);
StringBuffer sb = new StringBuffer();
while (m.find()) {
m.appendReplacement(sb, m.group(1) + (m.group(2).length()+1));
}
m.appendTail(sb);
System.out.println(sb);
See the Java demo
The (.)(\1*) will capture any char into Group 1 and then will capture into Group 2 zero or more repetitions of the same content. In the "callback", Group 1 is concatenated with the length of Group 2 incremented to account for the Group 1 length.

Get a particular string from a data using regular expression

I am trying to get particular string from the data below.It is too long am here with sharing sample data. From this I have to get the 'france24Id=7GHYUFGty6fdGFHyy56'
am not that much familier with regex.
How can I retreive the string 'france24Id=7GHYUFGty6fdGFHyy56' from above data?
I tried splitting the data using ',' but it is not an effective way.That's why I choose regex.
2016-07-29 12:08:46,260 s=xGuide, [xre-10-pipe#6da05f7a[,connection=WebSocketConnectionWrapper[/1.8.9]]] INFO c=c.ore., - onConnect event payload={minimumVersion='0', applicationName='shell', fetl='555', authenticationToken='6y777', sessionAuthToken='', sessionGUID='null', connectURL='http://section?ruleName=Default', partnerId='hp', nativeDimensions=null, appParams={heartbeatRequest=1, france24Id=7GHYUFGty6fdGFHyy56, service=false, networkBuffer={min=150, max=150, step=0}}, deviceCaps={platform=Mac, receiverType=Native, revisions={protocol=1, auth=1, video=1}, pixelDimensions=[1280, 720]}, forceSource=null, reconnect=false, currentCommandIndex=0, reconnectReason=7, authService=9}

You can get what you want with (france\d+Id)=([a-zA-Z0-9]+),. This will grab your string and dump the two parts of it into platform-appropriate capture group variables (for instance, in Perl, $1 and $2 respectively).
In Java, your code would look a little like this:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public String matchID(String data) {
Pattern r = new Pattern("(france\\d+Id)=([a-zA-Z0-9]+),");
Matcher m = r.matcher(data);
return m.group(2);
}

public static void main(String[] args) {
String str = "2016-07-29 12:08:46,260 s=xGuide, [xre-10-pipe#6da05f7a[,connection=WebSocketConnectionWrapper[/1.8.9]]] INFO c=c.ore., - onConnect event payload={minimumVersion='0', applicationName='shell', fetl='555', authenticationToken='6y777', sessionAuthToken='', sessionGUID='null', connectURL='http://section?ruleName=Default', partnerId='hp', nativeDimensions=null, appParams={heartbeatRequest=1, france24Id=7GHYUFGty6fdGFHyy56, service=false, networkBuffer={min=150, max=150, step=0}}, deviceCaps={platform=Mac, receiverType=Native, revisions={protocol=1, auth=1, video=1}, pixelDimensions=[1280, 720]}, forceSource=null, reconnect=false, currentCommandIndex=0, reconnectReason=7, authService=9}";
String regex = ".*(france24Id=[\\d|\\w]*),.*";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
if(matcher.matches()){
System.out.println(matcher.group(1));
}
}

You can use Pattern and Matcher classes in Java.
String data = "2016-07-29 12:08:46,260 s=xGuide, [xre-10-pipe#6da05f7a[,connection=WebSocketConnectionWrapper[/1.8.9]]] INFO c=c.ore., - onConnect event payload={minimumVersion='0', applicationName='shell', fetl='555', authenticationToken='6y777', sessionAuthToken='', sessionGUID='null', connectURL='http://section?ruleName=Default', partnerId='hp', nativeDimensions=null, appParams={heartbeatRequest=1, france24Id=7GHYUFGty6fdGFHyy56, service=false, networkBuffer={min=150, max=150, step=0}}, deviceCaps={platform=Mac, receiverType=Native, revisions={protocol=1, auth=1, video=1}, pixelDimensions=[1280, 720]}, forceSource=null, reconnect=false, currentCommandIndex=0, reconnectReason=7, authService=9}";
String regex1 = "france24Id=[a-zA-Z0-9]+"; //this matches france24Id=7GHYUFGty6fdGFHyy56
String regex2 = "(?<=france24Id=)[a-zA-Z0-9]+"; //this matches 7GHYUFGty6fdGFHyy56 or whatever after "france24Id=" and before ','
Pattern pattern1 = Pattern.compile(regex1);
Pattern pattern2 = Pattern.compile(regex2);
Matcher matcher1 = pattern1.matcher(data);
Matcher matcher2 = pattern2.matcher(data);
String result1, result2;
if(matcher1.find())
result1 = matcher1.group(); //if match is found, result1 should contain "france24Id=7GHYUFGty6fdGFHyy56"
if(matcher2.find())
result2 = matcher2.group(); //if match is found, result1 should contain "7GHYUFGty6fdGFHyy56"

You can also try this one:
String str = "france24Id=7GHYUFGty6fdGFHyy56";
Pattern pattern = Pattern.compile("(?<=france24Id=)([a-zA-Z0-9]+)");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
System.out.println("ID = " + matcher.group());
}
And the result is:
ID = 7GHYUFGty6fdGFHyy56

Got Exception in thread "main" java.lang.IndexOutOfBoundsException: No group 1

I have a string like this
"abc:def"
I need to check if it matches. If yes, I want to apply a regex and retrieve first part of it ("abc").
Here is my try:
Pattern pattern = Pattern.compile(".*:.*");
String name = "abc:def"
Matcher matcher = pattern.matcher(name);
if (matcher.find()) {
String group = matcher.group(1);
System.out.println(group);
}
It gives me
Exception in thread "main" java.lang.IndexOutOfBoundsException: No group 1

You need to add a capturing group inside your regular expression. This is done by putting what you want to capture inside parentheses:
Pattern pattern = Pattern.compile("(.*):.*"); // <-- parentheses here
String name = "abc:def";
Matcher matcher = pattern.matcher(name);
if (matcher.find()) {
String group = matcher.group(1);
System.out.println(group); // prints "abc"
}

You only have a group (0) (full match) because you don't define any capturing groups (enclosed in ( and )) in your regex.
If you change your code to include a capturing group like this
Pattern pattern = Pattern.compile("([^:]*):.*");
String name = "abc:def";
Matcher matcher = pattern.matcher(name);
if (matcher.find()) {
String fullMatch = matcher.group(0);
String group = matcher.group(1);
System.out.println(group);
}
you'll have full match "abc:def" and first group (until first colon) "abc"

There is no capturing group in your Pattern:
Pattern pattern = Pattern.compile(".*:.*");
See the documentation of Pattern for more details.
I assume you need this instead:
Pattern pattern = Pattern.compile("(.*):.*");

Try this:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class HelloWorld
{
public static void main(String[] args)
{
String patternString = ".*:.*";
Pattern pattern = Pattern.compile(patternString);
String name = "abc:def";
Matcher matcher = pattern.matcher(name);
while(matcher.find()) {
System.out.println("found: " + matcher.group(0));
}
}
}

How split a string using regex pattern

How split a [0] like words from string using regex pattern.0 can replace any integer number.
I used regex pattern,
private static final String REGEX = "[\\d]";
But it returns string with [.
Spliting Code
Pattern p=Pattern.compile(REGEX);
String items[] = p.split(lure_value_save[0]);

You have to escape the brackets:
String REGEX = "\\[\\d+\\]";

Java doesn't offer an elegant solution to extract the numbers. This is the way to go:
Pattern p = Pattern.compile(REGEX);
String test = "[0],[1],[2]";
Matcher m = p.matcher(test);
List<String> matches = new ArrayList<String>();
while (m.find()) {
matches.add(m.group());
}

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