Regex Pattern - Repeat group [duplicate] - java

This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?

Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:

Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.

Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;

I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm

(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*

Related

Java regex to filter strings starting with a pattern and end after first ] appearance [duplicate]

This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?
Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:
Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.
Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;
I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm
(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*

Extracting multiple labels from a string with java regex [duplicate]

This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?
Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:
Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.
Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;
I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm
(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*

Java Regexp for matching all the content between "<" and ">" in a paragraph [duplicate]

This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 2 years ago.
I have this gigantic ugly string:
J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM
J0000010: Project name: E:\foo.pf
J0000011: Job name: MBiek Direct Mail Test
J0000020: Document 1 - Completed successfully
I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time).
Here's the regex I've been playing with:
Project name:\s+(.*)\s+J[0-9]{7}:
The problem is that it doesn't stop until it hits the J0000020: at the end.
How do I make the regex stop at the first occurrence of J[0-9]{7}?
Make .* non-greedy by adding '?' after it:
Project name:\s+(.*?)\s+J[0-9]{7}:
Using non-greedy quantifiers here is probably the best solution, also because it is more efficient than the greedy alternative: Greedy matches generally go as far as they can (here, until the end of the text!) and then trace back character after character to try and match the part coming afterwards.
However, consider using a negative character class instead:
Project name:\s+(\S*)\s+J[0-9]{7}:
\S means “everything except a whitespace and this is exactly what you want.
Well, ".*" is a greedy selector. You make it non-greedy by using ".*?" When using the latter construct, the regex engine will, at every step it matches text into the "." attempt to match whatever make come after the ".*?". This means that if for instance nothing comes after the ".*?", then it matches nothing.
Here's what I used. s contains your original string. This code is .NET specific, but most flavors of regex will have something similar.
string m = Regex.Match(s, #"Project name: (?<name>.*?) J\d+").Groups["name"].Value;
I would also recommend you experiment with regular expressions using "Expresso" - it's a utility a great (and free) utility for regex editing and testing.
One of its upsides is that its UI exposes a lot of regex functionality that people unexprienced with regex might not be familiar with, in a way that it would be easy for them to learn these new concepts.
For example, when building your regex using the UI, and choosing "*", you have the ability to check the checkbox "As few as possible" and see the resulting regex, as well as test its behavior, even if you were unfamiliar with non-greedy expressions before.
Available for download at their site:
http://www.ultrapico.com/Expresso.htm
Express download:
http://www.ultrapico.com/ExpressoDownload.htm
(Project name:\s+[A-Z]:(?:\\w+)+.[a-zA-Z]+\s+J[0-9]{7})(?=:)
This will work for you.
Adding (?:\\w+)+.[a-zA-Z]+ will be more restrictive instead of .*

Regex to find variables and ignore methods

I'm trying to write a regex that finds all variables (and only variables, ignoring methods completely) in a given piece of JavaScript code. The actual code (the one which executes regex) is written in Java.
For now, I've got something like this:
Matcher matcher=Pattern.compile(".*?([a-z]+\\w*?).*?").matcher(string);
while(matcher.find()) {
System.out.println(matcher.group(1));
}
So, when value of "string" is variable*func()*20
printout is:
variable
func
Which is not what I want. The simple negation of ( won't do, because it makes regex catch unnecessary characters or cuts them off, but still functions are captured. For now, I have the following code:
Matcher matcher=Pattern.compile(".*?(([a-z]+\\w*)(\\(?)).*?").matcher(formula);
while(matcher.find()) {
if(matcher.group(3).isEmpty()) {
System.out.println(matcher.group(2));
}
}
It works, the printout is correct, but I don't like the additional check. Any ideas? Please?
EDIT (2011-04-12):
Thank you for all answers. There were questions, why would I need something like that. And you are right, in case of bigger, more complicated scripts, the only sane solution would be parsing them. In my case, however, this would be excessive. The scraps of JS I'm working on are intented to be simple formulas, something like (a+b)/2. No comments, string literals, arrays, etc. Only variables and (probably) some built-in functions. I need variables list to check if they can be initalized and this point (and initialized at all). I realize that all of it can be done manually with RPN as well (which would be safer), but these formulas are going to be wrapped with bigger script and evaluated in web browser, so it's more convenient this way.
This may be a bit dirty, but it's assumed that whoever is writing these formulas (probably me, for most of the time), knows what is doing and is able to check if they are working correctly.
If anyone finds this question, wanting to do something similar, should now the risks/difficulties. I do, at least I hope so ;)
Taking all the sound advice about how regex is not the best tool for the job into consideration is important. But you might get away with a quick and dirty regex if your rule is simple enough (and you are aware of the limitations of that rule):
Pattern regex = Pattern.compile(
"\\b # word boundary\n" +
"[A-Za-z]# 1 ASCII letter\n" +
"\\w* # 0+ alnums\n" +
"\\b # word boundary\n" +
"(?! # Lookahead assertion: Make sure there is no...\n" +
" \\s* # optional whitespace\n" +
" \\( # opening parenthesis\n" +
") # ...at this position in the string",
Pattern.COMMENTS);
This matches an identifier as long as it's not followed by a parenthesis. Of course, now you need group(0) instead of group(1). And of course this matches lots of other stuff (inside strings, comments, etc.)...
If you are rethinking using regex and wondering what else you could do, you could consider using an AST instead to access your source programatically. This answer shows you could use the Eclipse Java AST to build a syntax tree for Java source. I guess you could do similar for Javascript.
A regex won't cut in this case because Java isn't regular. Your best best is to get a parser that understands Java syntax and build onto that. Luckily, ANTLR has a Java 1.6 grammar (and 1.5 grammar).
For your rather limited use case you could probably easily extend the variable assignment rules and get the info you need. It's a bit of a learning curve but this will probably be your best best for a quick and accurate solution.
It's pretty well established that regex cannot be reliably used to parse structured input. See here for the famous response: RegEx match open tags except XHTML self-contained tags
As any given sequence of characters may or may not change meaning depending on previous or subsequent sequences of characters, you cannot reliably identify a syntactic element without both lexing and parsing the input text. Regex can be used for the former (breaking an input stream into tokens), but cannot be used reliably for the latter (assigning meaning to tokens depending on their position in the stream).

Java Regex, capturing groups with comma separated values

InputString: A soldier may have bruises , wounds , marks , dislocations or other Injuries that hurt him .
ExpectedOutput:
bruises
wounds
marks
dislocations
Injuries
Generalized Pattern Tried:
".[\s]?(\w+?)"+ // bruises.
"(?:(\s)?,(\s)?(\w+?))*"+ // wounds marks dislocations
"[\s]?(?:or|and) other (\w+)."; // Injuries
The pattern should be able to match other input strings like: A soldier may have bruiser or other injuries that hurt him.
On trying the generalized pattern above, the output is:
bruises
dislocations
Injuries
There is something wrong with the capturing group for "(?:(\s)?,(\s)?(\w+?))*". The capturing group has one more occurences.. but it returns only "dislocations". "marks" and "dislocation: are devoured.
Could you please suggest what should be the right pattern, and where is the mistake?
This question comes closest to this question, but that solution didn't help.
Thanks.
When the capture group is annotated with a quantifier [ie: (foo)*] then you will only get the last match. If you wanted to get all of them then you need to quantifier inside the capture and then you will have to manually parse out the values. As big a fan as I am of regex, I don't think it's appropriate here for any number of reasons... even if you weren't ultimately doing NLP.
How to fix: (?:(\s)?,(\s)?(\w+?))*
Well, the quantifier basically covers the whole regex in that case and you might as well use Matcher.find() to step through each match. Also, I'm curious why you have capture groups for the whitespace. If all you are trying to do is find a comma-separated set of words then that's something like: \w+(?:\s*,\s*\w+)* Then don't bother with capture groups and just split the whole match.
And for anything more complicated re: NLP, GATE is a pretty powerful tool. The learning curve is steep at times but you have a whole industry of science-guys to draw from: http://gate.ac.uk/
Regex in not suited for (natural) language processing. With regex, you can only match well defined patterns. You should really, really abandon the idea of doing this with regex.
You may want to start a new question where you specify what programming language you're using to perform this task and ask for pointers there.
EDIT
PSpeed posted a promising link to a 3rd party library, Gate, that's able to do many language processing tasks. And it's written in Java. I have not used it myself, but looking at the people/institutions working on it, it seems pretty solid.
The pattern that works is: \w+(?:\s*,\s*\w+)* and then manually separate CSV
There is no other method to do this with Java Regex.
Ideally, Java regex is not suitable for NLP. A useful tool for text mining is: gate.ac.uk
Thanks to Bart K. , and PSpeed.

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