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Regex replacing everything before a predefined range of chars - Java

I have string values where I want to remove or replace everything that comes before "TV|TH". My problem is that despite using the correct syntax, the string seems to stay the same.
String test = "10TH";
String replaceBeforeSide = test.replaceAll("^\\(TH|TV)+", "");
System.out.println(replaceBeforeSide);
//Desired result = "TH";

Converting my comment to answer so that solution is easy to find for future visitors.
You could use a simple regex with a capture group:
replaceBeforeSide = test.replaceAll(".+?(TH|TV)", "$1");
or even shorter:
replaceBeforeSide = test.replaceAll(".+?(T[HV])", "$1");
Using .+?, we are matching 1+ of any character (non-greedy) before matching (TH|TV) that we capture in group #1.
In replacement we just put $1 back so that only string before (TH|TV) is removed.
We could also use a lookahead and avoid capture group:
replaceBeforeSide = test.replaceAll(".+?(?=T[HV])", "");
If you want to match ignore case then use inline modifier (?i):
replaceBeforeSide = test.replaceAll("(?i).+?(?=T[HV])", "");

Matching groups with lookahead expression

I have problem with matching groups that contain lookahead expression. I don't know why this expressions doesn't work:
"""((?<=^)(.*)(?=\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s%))((?<=[\w:]\s)(\w+)(?=\s[cr]))"""
When I compile them separately, for example:
"""(?<=^)(.*)(?=\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s%)"""
I get the correct result
My sample text:
May 5 23:00:01 10.14.3.10 %ASA-6-302015: Built inbound UDP connection
Expressions have been checked with this tool: http://regex-testdrive.com/en/dotest
My Scala code:
import scala.util.matching.Regex
val text = "May 5 23:00:01 10.14.3.10 %ASA-6-302015: Built inbound UDP connection"
val regex = new Regex("""((?<=^)(.*)(?=\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s%))((?<=[\w:]\s)(\w+)(?=\s[cr]))""")
val result = regex.findAllIn(text)
Does anyone know solution of this problem?

Multiple matching
You may fix the pattern as
^.*?(?=\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s%)|(?<=[\w:]\s)\w+(?=\s[cr])
See the regex demo. The main point is to introduce the | alternation operator to match either of the 2 subpatterns. Note you do not need to put the ^ start of string anchor into a lookbehind, as ^ is already a zero-width assertion. Also, there are too many groupings that you do not seem to use any way. Also, to match a literal dot you need to escape it (. -> \.).
To obtain the multiple matches, you may use the following code snippet:
val text = "May 5 23:00:01 10.14.3.10 %ASA-6-302015: Built inbound UDP connection"
val regex = """^.*?(?=\s\d{1,3}.\d{1,3}.\d{1,3}.\d{1,3}\s%)|(?<=[\w:]\s)\w+(?=\s[cr])""".r
val result = regex.findAllIn(text)
result.foreach { x => println(x) }
// => May 5 23:00:01
// UDP
See the Scala online demo.
Note that once a pattern is used with .FindAllIn, it is not anchored by default, so you will get all the matches there are in the input string.
Capturing groups
Another approach you may use is matching the whole line while capturing the necessary bits with capturing groups:
val text = "May 5 23:00:01 10.14.3.10 %ASA-6-302015: Built inbound UDP connection"
val regex = """^(.*?)\s+\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s%.*[\w:]\s+(\w+)\s+[cr].*""".r
val results = text match {
case regex(date, protocol) => Array(date, protocol)
case _ => Array[String]()
}
// Demo printing
results.foreach { m =>
println(m)
}
See another Scala demo. Since match requires a full string match, .* is added at the end of the pattern, and only relevant pairs of unescaped (...) are kept in the pattern. See the regex demo here.

your matches are not next to each other,
try this:
"""((?<=^)(.*)(?=\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s%)).*((?<=[\w:]\s)(\w+)(?=\s[cr]))"""
I just added the .* between them, it works on the link you sent :)

how to get all names and date of births from a specific file using java

Hi below is my text file
welcome to java training
program
Name rtrti*&*
John
address india say^%$7
Date of Birth
11/12/1989
I have 100 files like above.The above text is the extracted text from the image files so it is not in order, from this i need to get the names and date of births can you please suggest me how to do this, I am new to this task.
Required output
John
11/12/1989
I have tried
Pattern p = Pattern.compile("Name");
Matcher matcher = p.matcher(content);
matcher.find();
But I have know idea how to get the next line of matched pattern, I cant not read this file line by line because my need is to store entire text in a single string.

I'll give a few hints that will get you on track. Without more details regarding the expected input, it will be difficult to give you a solid solution. First, I trust that you are already familiar with the Pattern and Matcher javadocs. You will need to understand the Groups and capturing section. Finally, you can utilize DOTALL mode which will allow the . character to match newlines.
To get you started, the following should work to find the name:
Pattern p = Pattern.compile(
"(?s)" + // DOTALL
".*" + // Match anything (to consume everything before 'Name')
"Name" + // Match the literal 'Name'
".*?" + // Reluctantly grab everything until...
"\n" + // Newline is reached
"\\s*" + // Consume leading whitespace
"(\\S+)" // Capture at least one non-whitespace character
);
Matcher m = p.matcher(content);
if(m.find()) {
String name = m.group(1); // The first capturing group contains "John"
}

java regexp for reluctant matching

need to find an expression for the following problem:
String given = "{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"answer 4\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"answer 5\"}";
What I want to get: "{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"*******\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"******\"}";
What I am trying:
String regex = "(.*answer\"\\s:\"){1}(.*)(\"[\\s}]?)";
String rep = "$1*****$3";
System.out.println(test.replaceAll(regex, rep));
What I am getting:
"{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"answer 4\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"******\"}";
Because of the greedy behaviour, the first group catches both "answer" parts, whereas I want it to stop after finding enough, perform replacement, and then keep looking further.

The pattern
("answer"\s*:\s*")(.*?)(")
Seems to do what you want. Here's the escaped version for Java:
(\"answer\"\\s*:\\s*\")(.*?)(\")
The key here is to use (.*?) to match the answer and not (.*). The latter matches as many characters as possible, the former will stop as soon as possible.
The above pattern won't work if there are double quotes in the answer. Here's a more complex version that will allow them:
("answer"\s*:\s*")((.*?)[^\\])?(")
You'll have to use $4 instead of $3 in the replacement pattern.

The following regex works for me :
regex = "(?<=answer\"\\s:\")(answer.*?)(?=\"})";
rep = "*****";
replaceALL(regex,rep);
The \ and " might be incorrectly escaped since I tested without java.
http://regexr.com?303mm

How to extract CSS color using regex?

I have a CSS style that I need to extract the color from using a Java regex.
eg
color:#000;
I need to extract the thing after : to ;. Can anyone give an example?

I'm not sure how to apply it to Java, but one regex to do this would be:
^color:\s*(#[0-9a-f]+);?$

To just extract from : up to ; do something like:
Pattern pattern = Pattern.compile("[^:]*:(.*);");
Matcher matcher = pattern.matcher(text);
if (matcher.matches()) {
String value = matcher.group(1);
System.out.println("'" + value+ "'"); // do something with value
}
[^:]* - any number of chars that are not ':'
: - one ':'
(...) - a capturing group
.*- any number of any character
;- the terminating ';'
use color:(.*); for only accepting values for 'color'.

/(?<=:).+(?=;)/
That will do it for you
Not sure how you implement regex in Java though.
www.regexr.com to help you text out your regex in real time.

The expression
":(#.+);"
should do it