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Different result between Javascript and Java regular expression matches

Now I am trying to match some patterns from a String containing elasticsearch's structured bulk requests. Here is an example:
index {[event_20191209][event][null], source[{"haha":"haha","jaja":"jaja"}]}, update {[event_20191209][event][xxx], doc_as_upsert[false], doc[index {[null][_doc][null], source[{"haha":"haha","jaja":"jaja"}]}], scripted_upsert[false], detect_noop[true]}, delete {[event_20191208][_doc][sjdos]}, update {[event_20191209][event][yyy], doc_as_upsert[false], upsert[index {[null][_doc][null], source[{"haha":"haha","jaja":"jaja"}]}], scripted_upsert[false], detect_noop[true]}
My goal is to match every separate request out of the bulk requests string, i.e to get strings like:
index {[event_20191209][event][null], source[{"haha":"haha","jaja":"jaja"}]},
update {[event_20191209][event][xxx], doc_as_upsert[false], doc[index {[null][_doc][null], source[{"haha":"haha","jaja":"jaja"}]}], scripted_upsert[false], detect_noop[true]},
delete {[event_20191208][_doc][sjdos]},
update {[event_20191209][event][yyy], doc_as_upsert[false], upsert[index {[null][_doc][null], source[{"haha":"haha","jaja":"jaja"}]}], scripted_upsert[false], detect_noop[true]}
And my pattern expression is [a-z]+\s\{.+?\}[,\w\t\r\n]+? which works fine on a Javascript based regular expression online tester like below:
However, when I copied this pattern expression to my Java code, the output was not what I expected. It was like this:
So I realized there exists some differences between Javascript and Java regular expression engine, but I cannot figure out how to update my expression so that it could work well in Java after so much coding and googling.
I would be so grateful if someone could give me some favor or hint for this.

After a short nap, I found epiphany. I was a fool in the morning....
The workaround is so easy to implement. Elasticsearch has well overridden toString() for us.

At first glance, I wouldn't suggest using regex right away. It looks like those lines follow some kind of pattern that you could parse and split up first.
After that, if you're talking about regex, I'd try:
Taking a look at the java regex format: https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html
How about using an online java regex tool instead?

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 .*

Are regex engines of Java and Groovy the same?

Now I am doing some code basing on regex in Groovy. But for creation and testing my regexes I use books referencing to Java regex engine and Java-oriented http://www.regexplanet.com/advanced/java/index.html.
And I am a bit afraid - Is Groovy regex engine really the same as Java one? I know that they are very close. But have they some differences nevertheless? If you know the answer - can you kindly give me some reference on the subject?

From the language documentation:
The pattern operator (~) provides a simple way to create a java.util.regex.Pattern instance.
I can't find phrasing where the documentation guarantees this is the regular expression engine used for pattern matching throughout Groovy; I do however find it very, very, very, very unlikely Groovy would use two RE engines in its implementation now or switch the RE engine in the future.

"Because Groovy is based on Java, you can use Java's regular expression package with Groovy. Simply put import java.util.regex.* at the top of your Groovy source code. Any Java code using regular expressions will then automatically work in your Groovy code too."
Source: regular-expressions.info

Here is a groovy example to match regex along with find :
assert ['abc'] == ['def', 'abc', '123'].findAll { it =~ /abc/ }
You can find more examples from here (including above sample), thanks to Mr. Haki.

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 .*

understanding regex if then statements

So I'm not sure if I understand how this works and would like
a simple explanation to how they work is all. I probably have it way off. A pure regex solution is required, and I don't know if this is possible. If it is, a solution would be awesome too, but a shove in the right direction would be good for my learning process ^_^
This is how I thought the if/then/else option built into my regex engines was formatted:
?(condition)if regex|else regex
I want it to capture a string from a very specific location only when this string exists within a certain section of javascript. Because this is how I thought it worked after a decent amount of research I tried out a few variations of this code but they all ended up something like this.
((?^view_large$)Tables-137(.*?)search.htm)
Also of relevance: I'm using an java based app that has regex searches which pull the data I need so I cannot write an if statement in java which would be my preferred method. It's a pain to have to do it this way, but at the moment I have no other choice. I'm trying really hard for them to allow java code functionality instead of pure regex for more versatile options.
So to summarize, is there even a if/then option in regex and if so how is it formatted for what I'm trying to accomplish?
EDIT: The string that I want to be the "if condition" is like this: if view_large string exists and is not null then capture the exact string 500/ which is captured within the catch all group I used: (.*?)

There is no conditionals in Java regexp, but you can simulate them by writing two expressions that include mutually exclusive look-behind constructs, like this:
((?<=if )then)|((?<!if )end)
This expression will match "then" when it is preceded by an "if "; it will match "end" when it is not preceded by an "if "

The Javadoc for java.util.regex.Pattern mentions, in its list of "Perl constructs not supported by this class":
The conditional constructs (?(condition)X) and (?(condition)X|Y).
So, no dice. But you should look through the Javadoc to see if you can achieve what you need by using regex features that it does support. (Or, if you post some more detailed examples, we can try to help.)

Try lookaround assertions.
For example, say you want to capture FOOBAR only if there is a 4+ digit number somewhere:
(?=.*\d{4}).*(FOOBAR)