String TextValue = "hello{MyVar} Discover {MyVar2} {MyVar3}";
String[] splitString = TextValue.split("\\{*\\}");
What I'm getting output is [{MyVar, {MyVar2, {MyVar3] in splitString
But my requirement is to preserve those delimiters {} i.e. [{MyVar}, {MyVar2}, {MyVar3}].
Required a way to match above output.
Use something like so:
Pattern p = Pattern.compile("(\\{\\w+\\})");
String str = ...
Matcher m = p.matcher(str);
while(m.find())
System.out.println(m.group(1));
Note, the code above is untested but that will look for words within curly brackets and place them in a group. It will then go over the string and output any string which matches the expression above.
An example of the regular expression is available here.
Thanks kelvin & npinti.
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class CreateMatcherExample {
public static void main(String[] args) {
String TextValue = "hello{MyVar} Discover {My_Var2} {My_Var3}";
String patternString = "\\{\\w+\\}";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(TextValue);
while(matcher.find()) {
System.out.println(matcher.group());
}
}
}
Related
I'm pretty rusty with regex, but I have the requirement to extract the first token of the following string:
Input: /token1/token2/token3
Required output: /token1
I have tried:
List<String> connectorPath = Splitter.on("^[/\\w+]+")
.trimResults()
.splitToList(actionPath);
Doesn't work for me, any ideas?
Instead of split, you can match
^/\\w+
Or if the string has 3 parts, use a capture group for the first part.
^(/\\w+)/\\w+/\\w+$
Java example
Pattern pattern = Pattern.compile("^/\\w+");
Matcher matcher = pattern.matcher("/token1/token2/token3");
if (matcher.find()) {
System.out.println(matcher.group(0));
}
Output
/token1
You can split on the / that is not at the string start using the (?!^)/ regex:
String[] res = "/token1/token2/token3".split("(?!^)/");
System.out.println(res[0]); // => /token1
See the Java code demo and the regex demo.
(?!^) - a negative lookahead that matches a location not at the start of string
/ - a / char.
Using Guava:
Splitter splitter = Splitter.onPattern("(?!^)/").trimResults();
Iterable<String> iterable = splitter.split(actionPath);
String first = Iterables.getFirst(iterable, "");
You are over-complicating it.
Try the following regular expression: ^(\/\w+)(.+)$
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class PathSplitter {
public static void main(String args[]) {
String input = "/token1/token2/token3";
Pattern pattern = Pattern.compile("^(\\/\\w+)(.+)$");
Matcher matcher = pattern.matcher(input);
if (matcher.find()) {
System.out.println(matcher.group(1)); // /token1
System.out.println(matcher.group(2)); // /token2/token3
} else {
System.out.println("NO MATCH");
}
}
}
from this -> contractor:"Hi, this is \"Paul\", how are you?" client:"Hi ...." <-
I want to get just -> Hi, this is \"Paul\", how are you? <-
I need a regular expression in java to do that I try it but I m struggle with the inner quotation (\") is driving me mad.
Thanks for any hint.
Java supports lookbehinds, so vanilla regex:
"(.*?(?<!\\))"
Inside a Java string (see https://stackoverflow.com/a/37329801/1225328):
\"(.*?(?<!\\\\))\"
The actual text will be contained inside the first group of each match.
Demo: https://regex101.com/r/8OXujX/2
For example, in Java:
String regex = "\"(.*?(?<!\\\\))\"";
String input = "contractor:\"Hi, this is \\\"Paul\\\", how are you?\" client:\"Hi ....\"";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
if (matcher.find()) { // or while (matcher.find()) to iterate through all the matches
System.out.println(matcher.group(1));
} else {
System.out.println("No matches");
}
Prints:
Hi, this is \"Paul\", how are you?
The regexp should be like this: "(?:\\.|[^"\\])*"
Online demo
It uses non-capturing group ?:, matching any character . or a single character NOT in the list of double quote and backslash.
var text1 = "contractor:\"Hi, this is \\\"Paul\\\", how are you?\" client:\"Hi ....\" <-";
var regExWithQuotation = "contractor:(.+\".+\".+) client:";
Pattern p = Pattern.compile(regExWithQuotation);
var m = p.matcher(text1);
;
if (m.find()) {
var res = m.group(1);
System.out.println(res);
}
var regExWithoutQuotation = "contractor:\"(.+\".+\".+)?\" client:";
p = Pattern.compile(regExWithoutQuotation);
m = p.matcher(text1);
if (m.find()) {
var res = m.group(1);
System.out.println(res);
}
Output is:
"Hi, this is "Paul", how are you?"
Hi, this is "Paul", how are you?
You can use the regex, (?<=contractor:\").*(?=\" client:)
Description of the regex:
(?<=contractor:\") specifies positive lookbehind for contractor:\"
.* specifies any character
(?=\" client:) specifies positive lookahead for \" client:
In short, anything preceded by contractor:\" and followed by \" client:
Demo:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
String str = "contractor:\"Hi, this is \\\"Paul\\\", how are you?\" client:\"Hi ....\"";
String regex = "(?<=contractor:\").*(?=\" client:)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group());
}
}
}
Output:
Hi, this is \"Paul\", how are you?
i have a string like "Hi ${CUSTOMER_NAME} your address is ${CUSTOMER_ADDRESS}
".
I need to remove every text which starts with $and ends with }. so the output of above should be like Hi Your address is.
In this case you should use regex, here is an example you can use:
import java.util.regex.*;
public class HelloWorld{
public static void main(String []args){
String regex = "(\\$\\{\\w+\\})";
String str = "Hi ${CUSTOMER_NAME} your address is ${CUSTOMER_ADDRESS}";
Pattern p = Pattern.compile(regex);
Matcher matcher = p.matcher(str);
String replaceAll = matcher.replaceAll("");
System.out.println(str);
System.out.println(replaceAll);
}
}
You are welcome to read more about regex and metcher.
Hope this helps
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);
...
}
I am using regex in java to get a specific output from a list of rooms at my University.
A outtake from the list looks like this:
(A55:G260) Laboratorium 260
(A55:G292) Grupperom 292
(A55:G316) Grupperom 316
(A55:G366) Grupperom 366
(HDS:FLØYEN) Fløyen (appendix)
(ODO:PC-STUE) Pulpakammeret (PC-stue)
(SALEM:KONF) Konferanserom
I want to get the value that comes between the colon and the parenthesis.
The regex I am using at the moment is:
pattern = Pattern.compile("[:]([A-Za-z0-9ÆØÅæøå-]+)");
matcher = pattern.matcher(room.text());
I've included ÆØÅ, because some of the rooms have Norwegian letters in them.
Unfortunately the regex includes the building code also (e.g. "A55") in the output... Comes out like this:
A55
A55
A55
:G260
:G292
:G316
Any ideas on how to solve this?
The problem is not your regular expression. You need to reference group(1) for the match result.
while (matcher.find()) {
System.out.println(matcher.group(1));
}
However, you may consider using a negated character class instead.
pattern = Pattern.compile(":([^)]+)");
You can try a regex like this :
public static void main(String[] args) {
String s = "(HDS:FLØYEN) Fløyen (appendix)";
// select everything after ":" upto the first ")" and replace the entire regex with the selcted data
System.out.println(s.replaceAll(".*?:(.*?)\\).*", "$1"));
String s1 = "ODO:PC-STUE) Pulpakammeret (PC-stue)";
System.out.println(s1.replaceAll(".*?:(.*?)\\).*", "$1"));
}
O/P :
FLØYEN
PC-STUE
Can try with String Opreations as follows,
String val = "(HDS:FLØYEN) Fløyen (appendix)";
if(val.contains(":")){
String valSub = val.split("\\s")[0];
System.out.println(valSub);
valSub = valSub.substring(1, valSub.length()-1);
String valA = valSub.split(":")[0];
String valB = valSub.split(":")[1];
System.out.println(valA);
System.out.println(valB);
}
Output :
(HDS:FLØYEN)
HDS
FLØYEN
import java.util.regex.Matcher;
import java.util.regex.Pattern;
class test
{
public static void main( String args[] ){
// String to be scanned to find the pattern.
String line = "(HDS:FLØYEN) Fløyen (appendix)";
String pattern = ":([^)]+)";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher(line);
while (m.find()) {
System.out.println(m.group(1));
}
}
}