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Java Regular expression for exacted matched case

I have been struggling to find the matched string(s) with Java Regular expression for the syntax {//<some string>/<some String>}
My regular expression should return with these matched cases: {//data/process_id}
Below is the String which i want to find matched syntax:
#process_id={//data/process_id}##history_id={//data/history_id}##Pdataxml={//data/dataxml}##Prules =_UNESCAPEXMLVALUE({//data/rules})##submitted_by={//data/submitted_by}##table_definition={//data/table_definition}
I have tried with below regx pattern but it did not work:
[a-zA-Z_/\\[\\]\\(\\)0-9|]+
Can someone please help me to solve this issue?

You can use the following regex:
\{\/\/[^\/{}\s]*\/[^\/{}\s]*\}
Demo on regex101
code:
String input = "#process_id={//data/process_id}##history_id={//data/history_id}##Pdataxml={//data/dataxml}##Prules =_UNESCAPEXMLVALUE({//data/rules})##submitted_by={//data/submitted_by}##table_definition={//data/table_definition}";
List<String> allMatches = new ArrayList<String>();
Matcher m = Pattern.compile("\\{\\/\\/[^\\/{}\\s]*\\/[^\\/{}\\s]*\\}").matcher(input);
while (m.find()) {
allMatches.add(m.group());
}
System.out.println(allMatches);
output:
[{//data/process_id}, {//data/history_id}, {//data/dataxml}, {//data/rules}, {//data/submitted_by}, {//data/table_definition}]

Try this regex with a Matcher:
"\\{//([^/]+)/([^/}]+)}"
The parts are captured in groups 1 and 2.
Like this:
Matcher m = Pattern.compile("\\{//([^/]+)/([^/}]+)}").matcher(str);
while (m.find()) {
String part1 = m.group(1);
String part2 = m.group(2);
// do something with the parts
}
To just grab the whole thing, which would be got from m.group(), use this regex:
"(?<=\\{)//[^/]+/[^/}]+(?=})"

Java: Appended string need to remove certain parts

I would like to know i wish to remove appended strings previously how can i do that.
Example getValue() value is
-------------------------- ------------------------
information1.getValue() Test A </n>abc = 1234
information2.getValue() Test A </n>def = Test B
information3.getValue() 123 </n>jkl = Test B
information4.getValue() 123 </n>abc = Test A
Expected output i should get is
Output
-------
abc
def
jkl
abc
whereby i should ignore characters before < /n> and characters after =

System.out.println(information1.getValue().substring(
information1.getValue().indexOf(">")+1,
information1.getValue().indexOf("=")).trim());
Use Below link to see example
https://repl.it/repls/AmazingSmoothAnalyst
Using StringBuffer
StringBuffer sb = new StringBuffer(information1.getValue());
sb.delete(0,sb.indexOf(">")+1).delete(sb.indexOf("=")-1,sb.length());
System.out.println(sb);

I'd use Regex.
Pattern p = Pattern.compile(".*<\\n>([^ ]*). =. *");
String replaced = p.matcher(s.getValue()).group(1));

You should use regex to capture words, lookahead (?<=) and lookbehind (?=) tokens can capture words
String regx="(?<=</n>)(.*)(?==)";
Pattern datePatt = Pattern.compile(regx);
Matcher m = datePatt.matcher("Test A </n>abc = 1234");
System.out.println("checking");
if(m.find()){
System.out.println(m.group(0));
}
Output
abc

Regex: how to extract a JSESSIONID cookie value from cookie string?

I might receive the following cookie string.
hello=world;JSESSIONID=sdsfsf;Path=/ei
I need to extract the value of JSESSIONID
I use the following pattern but it doesn't seem to work. However https://regex101.com shows it's correct.
Pattern PATTERN_JSESSIONID = Pattern.compile(".*JSESSIONID=(?<target>[^;\\n]*)");

You can reach your goal with a simpler approach using regex (^|;)JSESSIONID=(.*);. Here is the demo on Regex101 (you have forgotten to link the regular expression using the save button). Take a look on the following code. You have to extract the matched values using the class Matcher:
String cookie = "hello=world;JSESSIONID=sdsfsf;Path=/ei";
Pattern PATTERN_JSESSIONID = Pattern.compile("(^|;)JSESSIONID=(.*);");
Matcher m = PATTERN_JSESSIONID.matcher(cookie);
if (m.find()) {
System.out.println(m.group(0));
}
Output value:
sdsfsf
Of course the result depends on the all of possible variations of the input text. The snippet above will work in every case the value is between JSESSIONID and ; characters.

You can try below regex:
JSESSIONID=([^;]+)
regex explanation
String cookies = "hello=world;JSESSIONID=sdsfsf;Path=/ei;submit=true";
Pattern pat = Pattern.compile("\\bJSESSIONID=([^;]+)");
Matcher matcher = pat.matcher(cookies);
boolean found = matcher.find();
System.out.println("Sesssion ID: " + (found ? matcher.group(1): "not found"));
DEMO

You can even get what you aiming for with Splitting and Replacing the string aswell, below I am sharing which is working for me.
String s = "hello=world;JSESSIONID=sdsfsf;Path=/ei";
List<String> sarray = Arrays.asList(s.split(";"));
String filterStr = sarray.get(sarray.indexOf("JSESSIONID=sdsfsf"));
System.out.println(filterStr.replace("JSESSIONID=", ""));

First and second tocen regex

How could I get the first and the second text in "" from the string?
I could do it with indexOf but this is really boring ((
For example I have a String for parse like: "aaa":"bbbbb"perhapsSomeOtherText
And I d like to get aaa and bbbbb with the help of Regex pattern - this will help me to use it in switch statement and will greatly simplify my app/

If all that you have is colon delimited string just split it:
String str = ...; // colon delimited
String[] parts = str.split(":");
Note, that split() receives regex and compilies it every time. To improve performance of your code you can use Pattern as following:
private static Pattern pColonSplitter = Pattern.compile(":");
// now somewhere in your code:
String[] parts = pColonSplitter.split(str);
If however you want to use pattern for matching and extraction of string fragments in more complicated cases, do it like following:
Pattert p = Patter.compile("(\\w+):(\\w+):");
Matcher m = p.matcher(str);
if (m.find()) {
String a = m.group(1);
String b = m.group(2);
}
Pay attention on brackets that define captured group.

Something like this?
Pattern pattern = Pattern.compile("\"([^\"]*)\"");
Matcher matcher = pattern.matcher("\"aaa\":\"bbbbb\"perhapsSomeOtherText");
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Output
aaa
bbbbb

String str = "\"aaa\":\"bbbbb\"perhapsSomeOtherText";
Pattern p = Pattern.compile("\"\\w+\""); // word between ""
Matcher m = p.matcher(str);
while(m.find()){
System.out.println(m.group().replace("\"", ""));
}
output:
aaa
bbbbb

there are several ways to do this
Use StringTokenizer or Scanner with UseDelimiter method

Extracting part of URL using java regular expression

I'm trying to extract part of the URL in the text files.
for example:
/p/gnomecatalog/bugs/search/?q=status%3Aclosed-accepted+or+status%3Awont-fix+or+status%3Aclosed" class="search_bin"><span>Closed Tickets</span></a>
I would like to extract only
/p/gnomecatalog/bugs/search/?q=status%3Aclosed-accepted+or+status%3Awont-fix+or+status%3Aclosed
HOW I COULD DO THAT BY USING REGULAR Expression. I tried with regex
"/p/*./bugs/*."
but it didn't work.

Try this:
"\/p.*\/bugs[^"]*"
it means: "/p"
then: all chars,
then: "/bugs",
then: all chars except "

You can use :
(\/p\/.*\/bugs\/.*?(?="))
Java Code :
String REGEX = "(\\/p\\/.*\\/bugs\\/.*?(?=\"))";
Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(line);
while (m.find()) {
String matched = m.group();
System.out.println("Mached : "+ matched);
}
OUTPUT
Mached : /p/gnomecatalog/bugs/search/?q=status%3Aclosed-accepted+or+status%3Awont-fix+or+status%3Aclosed
DEMO
Explanation:

Here's another way:
(?i)/p/[a-z/]+bugs/[^ "]+
The (?i) in the beginning makes the regex case insensitive so you don't have to worry about that. Then after bugs/ it will continue until it reaches either a space or a ".