Unable to Parse reqular expression in JAVA - java

This is my code,
String xyz = "{\"status\":\"ok\",\"data\":[{\"RatingCount\":4}], [{\"RatingCount\":1}], [{\"RatingCount\":1}]\"code\":1}";
String pattern = ".*],\\s*\\[.*";//"(.*)(],\\s.*\\[)(.*)";
Pattern p1 = Pattern.compile(pattern);
Matcher m1 = p1.matcher(xyz);
boolean b = m1.matches();
System.out.println(b);
I would like to replace pattern '], [' with "".
I used replaceAll but no luck


Check it in working Fiddle
Use the regex as below
String pattern = "\],[ ]*\[";


xyz =xyz.replaceAll("]\s*,\s*\[", "],[");
This worked like charm and also this link helped me alot http://regexr.com/
Thank you for all your inputs!


The output of this block is: {"status":"ok","data":[{"RatingCount":4}{"RatingCount":1}{"RatingCount":1}]"code":1}
public static void main(String[] args) {
String str = "{\"status\":\"ok\",\"data\":[{\"RatingCount\":4}], [{\"RatingCount\":1}], [{\"RatingCount\":1}]\"code\":1}";
System.out.println(replace(str, "\\],\\s+\\["));
}
public static String replace(String str, String regex) {
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
return matcher.replaceAll("");
}

Related

How to extract a substring from a sentence until a delimeter

Hi I have some string like this:
location/city/home-a-berlin?/someNewAdress
I want to extract word berlin which placed between "-a-" and "?". How can i do that with regex in java?
I can do it by using string API but kinda stuck with regex.
String cityName = url.substring(url.lastIndexOf("-a-")+3, url.indexOf('?')) //berlin

You can use a capture group with a negated character class.
-a-([^\?]+)\?
Regex demo | Java demo
In Java:
String regex = "-a-([^\\?]+)\\?";
String string = "location/city/home-a-berlin?/someNewAdress\n";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
if (matcher.find()) {
System.out.println(matcher.group(1));
}
Output
berlin

Or
s = s.replaceAll(".*-(.*?)\\?.*", "$1");

Alternative regex:
"-a-(.+?)\\?"
Regex in testbench and context:
public static void main(String[] args) {
String input1 = "location/city/home-a-berlin?/someNewAdress";
List<String> inputs = Arrays.asList(input1);
Pattern pattern = Pattern.compile("-a-(.+?)\\?");
List<String> results = inputs.stream().map(s -> pattern.matcher(s))
.filter(Matcher::find).map(m -> m.group(1)).collect(Collectors.toList());
//Output:
results.forEach(System.out::println);
}
Output:
berlin
Summary of regular-expression constructs:
https://docs.oracle.com/javase/10/docs/api/java/util/regex/Pattern.html

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);
...
}

How to preserve delimeters while using String.split() in Java?

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());
}
}
}

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

Regex after a special character in Java

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));
}
}
}

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