Develop Reference

How do I get a regex expression to contain only uppercase letters or numbers?

Regex expression: [A-Z]([^0-9]|[^A-Z])+[A-Z]
The requirements are that the string should start and end with a capital letter A-Z, and contain at least one number in between. It should not have anything else besides capital letters on the inside. However, it's accepting spaces and punctuation too.
My expression fails the following test case A65AJ3L 3F,D due to the comma and whitespace.
Why does this happen when I explicitly said only numbers and uppercase letters can be in the string?

Starting the character class with [^ makes is a negated character class.
Using ([^0-9]|[^A-Z])+ matches any char except a digit (but does match A-Z), or any char except A-Z (but does match a digit).
This way it can match any character.
If you would turn it into [A-Z]([0-9]|[A-Z])+[A-Z] it still does not make it mandatory to match at least a single digit on the inside due to the alternation | and it can still match AAA for example.
You might use:
^[A-Z]+[0-9][A-Z0-9]*[A-Z]$
The pattern matches:
^ Start of string
[A-Z]+ Match 1+ times A-Z
[0-9] Match a single digit
[A-Z0-9]* Optionally match either A-Z or 0-9
[A-Z] Match a single char A-Z
$ End of string
Regex demo

Use
^(?=\D*\d\D*$)[A-Z][A-Z\d]*[A-Z]$
See regex proof.
(?=\D*\d\D*$) requires only one digit in the string, no more no less.
EXPLANATION
--------------------------------------------------------------------------------
^ the beginning of the string
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
\D* non-digits (all but 0-9) (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
\D* non-digits (all but 0-9) (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of
the string
--------------------------------------------------------------------------------
) end of look-ahead
--------------------------------------------------------------------------------
[A-Z] any character of: 'A' to 'Z'
--------------------------------------------------------------------------------
[A-Z\d]* any character of: 'A' to 'Z', digits (0-9)
(0 or more times (matching the most amount
possible))
--------------------------------------------------------------------------------
[A-Z] any character of: 'A' to 'Z'
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string

Java String match using the regex for optional String

I have the following Java String which I need to compare with the regex but these string can consist of the optional values which may be present or not present I need to perform different things based on their availability:
String 1: With serial
uri = https://myid.com/123/1234567890128/456/1111
String 2: Without Serial
uri = https://myid.com/123/1234567890128
As we can see the incoming string can have the /456/1111 or it may not have. How can I write a single regex function which checks whether it is present or not? I have written a regex but it would work only if the /456/1111 is present:
uri.matches("(http|https)://.*/123/[0-9]{13}.*)")
I tried adding the optional values after looking at some of the answers here something like this:
uri.matches("(http|https)://.*/123/[0-9]{13}+([/456/[0-9]{1,20}]?)")
But for some reason, it does not work. Can someone please help me how can I verify whether there are any strings present in uri /456/1111 or not. I feel like I am missing some small thing.

Use
^https?:\/\/.*\/123\/[0-9]{1,13}(?:/456/[0-9]{1,20})?$
See proof.
Explanation
--------------------------------------------------------------------------------
^ the beginning of the string
--------------------------------------------------------------------------------
http 'http'
--------------------------------------------------------------------------------
s? 's' (optional (matching the most amount
possible))
--------------------------------------------------------------------------------
: ':'
--------------------------------------------------------------------------------
\/ '/'
--------------------------------------------------------------------------------
\/ '/'
--------------------------------------------------------------------------------
.* any character except \n (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
\/ '/'
--------------------------------------------------------------------------------
123 '123'
--------------------------------------------------------------------------------
\/ '/'
--------------------------------------------------------------------------------
[0-9]{1,13} any character of: '0' to '9' (between 1
and 13 times (matching the most amount
possible))
--------------------------------------------------------------------------------
(?: group, but do not capture (optional
(matching the most amount possible)):
--------------------------------------------------------------------------------
/456/ '/456/'
--------------------------------------------------------------------------------
[0-9]{1,20} any character of: '0' to '9' (between 1
and 20 times (matching the most amount
possible))
--------------------------------------------------------------------------------
)? end of grouping
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string

regex101.com is your friend in this regard.
When looking at your regex on that site, you can see some errors like:
you have a lone ] at the end which seems off
and at last your ? at the end targets the wrong group, move it out of the parenthesis.
Something like https?://[^/]+/123/\d{13}(?:/456/\d{1,20})? should work for you.
The good thing about regex101 is that on the right side you see a detailed explanation about your regex, and it highlights exactly which character does what.

The reason the pattern that you tried does not work, is because in the last part of your pattern you have ([/456/[0-9]{1,20}]?) which means:
( Capture group
[/456/[0-9]{1,20} Match 1-20 repetitions of either / or a digit 0-9 (as 0-9 also matches 456)
]? Match optional ]
) Close group
What you could do instead, is making the last group as a whole optional without a character class use https? making the s optional.
^https?://.*/123/[0-9]{13}(?:/456/[0-9]{1,20})?$
Regex demo | Java demo
As you use matches() it should match the whole string and you can omit the anchors ^ and $
String uri1 = "https://myid.com/123/1234567890128/456/1111";
String uri2 = "https://myid.com/123/1234567890128";
String uri3 = "https://myid.com/123/1234567890128/456/111122222222222222222";
String pattern = "https?://.*/123/[0-9]{13}(?:/456/[0-9]{1,20})?";
System.out.println(uri1.matches(pattern));
System.out.println(uri2.matches(pattern));
System.out.println(uri3.matches(pattern));
Output
true
true
false

How to match a block from start to end using regex

I want to pick up the entire block from the starting title to the end title, but not include the end title. Example is :
<section1>
Base_Currency=EUR
Description=Revaluation
Grouping_File
<section2>
the match result should be:
<section1>
Base_Currency=EUR
Description=Revaluation
Grouping_File
Problem is that how can I formulate the Pattern for this match using Regex in java?

If your input is something like below
<section1>
Base_Currency=EUR
Description=Revaluation
Grouping_File
<section2>
Base_Currency=EUR
Description=Revaluation
Grouping_File
<section3>
Base_Currency=EUR
Description=Revaluation
Grouping_File
Then you can use the following regex
(?s)(<section\d+>.*?)(?=<section\d+>|$)
Explanation for the regex is
NODE EXPLANATION
--------------------------------------------------------------------------------
(?s) set flags for this block (with . matching
\n) (case-sensitive) (with ^ and $
matching normally) (matching whitespace
and # normally)
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
<section '<section'
--------------------------------------------------------------------------------
\d+ digits (0-9) (1 or more times (matching
the most amount possible))
--------------------------------------------------------------------------------
> '>'
--------------------------------------------------------------------------------
.*? any character (0 or more times (matching
the least amount possible))
--------------------------------------------------------------------------------
) end of \1
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
<section '<section'
--------------------------------------------------------------------------------
\d+ digits (0-9) (1 or more times (matching
the most amount possible))
--------------------------------------------------------------------------------
> '>'
--------------------------------------------------------------------------------
| OR
--------------------------------------------------------------------------------
$ before an optional \n, and the end of
the string
--------------------------------------------------------------------------------
) end of look-ahead
If you want to match only for one tag then you can use
(?s)(<section\d+>[^<]*)
Explanation for this regex is
NODE EXPLANATION
--------------------------------------------------------------------------------
(?s) set flags for this block (with . matching
\n) (case-sensitive) (with ^ and $
matching normally) (matching whitespace
and # normally)
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
<section '<section'
--------------------------------------------------------------------------------
\d+ digits (0-9) (1 or more times (matching
the most amount possible))
--------------------------------------------------------------------------------
> '>'
--------------------------------------------------------------------------------
[^<]* any character except: '<' (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
) end of \1

If your entire input is of this format, you can simply split:
String[] sections = input.split("\\R(?=<)");
\R is "any newline sequence" and (?=<) means "the next char is a '<'".
However if that's not the case, from the regex toolbox you're going to need:
the DOTALL flag so dot matches newlines too
the MULTILINE flag so ^ matches start of line too
a negative look ahead so you stop consuming at the start of the next section
Assuming "sections" start with a "<" at the start of a line:
"(?sm)^<\\w+>(.(?!^<))*"
Here's how you could use it:
String input = "<section1>\nBase_Currency=EUR\nDescription=Revaluation\nGrouping_File\n<section2>\nfoo";
Matcher matcher = Pattern.compile("(?sm)^<\\w+>(.(?!^<))*").matcher(input);
while (matcher.find()) {
String section = matcher.group();
}

Java - regular expression for get number format

I have this:
110121 NATURAL 95 1570,40
110121 NATURAL 95 1570,40*
41,110 1 x 38,20 CZK)[A] *
' 31,831 261,791 1308,61)
>01572 PRAVO SO 17,00
1,000 ks x 17,00
1570,40
Every line of this output is saved in List and I want to get number 1570,40
My regular expressions looks like this for this type of format
"([1-9][0-9]*[\\.|,][0-9]{2})[^\\.\\d](.*)"
"^([1-9][0-9]*[\\.|,][0-9]{2})$"
I have a problem that 1570,40 at the last line if founded (by second regular expression), also 1570,40 (from line with 1570,40* at the end) but the first line is not founded.. do you know where is the problem?

Not sure I well understand your needs, but I think you could use word boundaries like:
\b([1-9]\d*[.,]\d{2})\b
In order to not match dates, you can use:
(?:^|[^.,\d])(\d+[,.]\d\d)(?:[^.,\d]|$)
explanation:
The regular expression:
(?-imsx:(?:^|[^.,\d])(\d+[,.]\d\d)(?:[^.,\d]|$))
matches as follows:
NODE EXPLANATION
----------------------------------------------------------------------
(?-imsx: group, but do not capture (case-sensitive)
(with ^ and $ matching normally) (with . not
matching \n) (matching whitespace and #
normally):
----------------------------------------------------------------------
(?: group, but do not capture:
----------------------------------------------------------------------
^ the beginning of the string
----------------------------------------------------------------------
| OR
----------------------------------------------------------------------
[^.,\d] any character except: '.', ',', digits
(0-9)
----------------------------------------------------------------------
) end of grouping
----------------------------------------------------------------------
( group and capture to \1:
----------------------------------------------------------------------
\d+ digits (0-9) (1 or more times (matching
the most amount possible))
----------------------------------------------------------------------
[,.] any character of: ',', '.'
----------------------------------------------------------------------
\d digits (0-9)
----------------------------------------------------------------------
\d digits (0-9)
----------------------------------------------------------------------
) end of \1
----------------------------------------------------------------------
(?: group, but do not capture:
----------------------------------------------------------------------
[^.,\d] any character except: '.', ',', digits
(0-9)
----------------------------------------------------------------------
| OR
----------------------------------------------------------------------
$ before an optional \n, and the end of
the string
----------------------------------------------------------------------
) end of grouping
----------------------------------------------------------------------
) end of grouping
----------------------------------------------------------------------

The "([1-9][0-9]*[\\.|,][0-9]{2})[^\\.\\d](.*)" has [^\\.\\d], it means it expects one non-digit, non-dot symbol right after the number. The second line has * which matches it. First line has the number at the end of line, so nothing matches. I think you need just one regexp which will catch all numbers: [^.\\d]*([1-9][0-9]*[.,][0-9]{2})[^.\\d]*. Also, you should use find instead of match to find any substring in a string instead of matching the whole string. Also, maybe it has a sense to find all matches in case if a line has two such numbers in it, not sure if it is a case for you or not.
Also, use either [0-9] or \d. At the moment it is confusing - it means the same, but looks differently.

Try this:
String s = "41,110 1 x 38,20 CZK)[A] * ";
Matcher m = Pattern.compile("\\d+,\\d+").matcher(s);
while(m.find()) {
System.out.println(m.group());
}

Regex expression in plain english

I'm working on a new Java project and therefore im reading the already existing code. On a very important part of the code if found the following regex expression and i can't really tell what they are doing. Anybody can explain in plain english what they do??
1)
[^,]*|.+(,).+
2)
(\()?\d+(?(1)\))

Next time you need a regex explained, you can use the following explain.pl service from Rick Measham:
Regex: [^,]*|.+(,).+
NODE EXPLANATION
--------------------------------------------------------------------------------
[^,]* any character except: ',' (0 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
| OR
--------------------------------------------------------------------------------
.+ any character except \n (1 or more times
(matching the most amount possible))
--------------------------------------------------------------------------------
( group and capture to \1:
--------------------------------------------------------------------------------
, ','
--------------------------------------------------------------------------------
) end of \1
--------------------------------------------------------------------------------
.+ any character except \n (1 or more times
(matching the most amount possible))
Regex: (\()?\d+(?(1)\))
NODE EXPLANATION
--------------------------------------------------------------------------------
( group and capture to \1 (optional
(matching the most amount possible)):
--------------------------------------------------------------------------------
\( '('
--------------------------------------------------------------------------------
)? end of \1 (NOTE: because you're using a
quantifier on this capture, only the LAST
repetition of the captured pattern will be
stored in \1)
--------------------------------------------------------------------------------
\d+ digits (0-9) (1 or more times (matching
the most amount possible))
--------------------------------------------------------------------------------
(?(1) if back-reference \1 matched, then:
--------------------------------------------------------------------------------
\) ')'
--------------------------------------------------------------------------------
| else:
--------------------------------------------------------------------------------
succeed
--------------------------------------------------------------------------------
) end of conditional on \1
Links
http://rick.measham.id.au/paste/explain.pl
Note on conditionals
JAVA DOES NOT SUPPORT CONDITIONALS! An unconditionalized regex for the second pattern would be something like:
\d+|\(\d+\)
i.e. a non-zero repetition of digits, with or without surrounding parentheses.
Links
regular-expressions.info/If-then-else conditionals
Conditionals are supported by the JGsoft engine, Perl, PCRE and the .NET framework.
The patterns in depth
Here's a test harness for the first pattern
import java.util.regex.*;
//...
Pattern p = Pattern.compile("[^,]*|.+(,).+");
String[] tests = {
"", // [] is a match with no commas
"abc", // [abc] is a match with no commas
",abc", // [,abc] is not a match
"abc,", // [abc,] is not a match
"ab,c", // [ab,c] is a match with separating comma
"ab,c,", // [ab,c,] is a match with separating comma
",", // [,] is not a match
",,", // [,,] is not a match
",,,", // [,,,] is a match with separating comma
};
for (String test : tests) {
Matcher m = p.matcher(test);
System.out.format("[%s] is %s %n", test,
!m.matches() ? "not a match"
: m.group(1) != null
? "a match with separating comma"
: "a match with no commas"
);
}
Conclusion
To match, the string must fall into one of these two cases:
Contains no comma (potentially an empty string)
Contains a comma that separates two non-empty strings
On a match, \1 can be used to distinguish between the two cases
And here's a similar test harness for the second pattern, rewritten without using conditionals (which isn't supported by Java):
Pattern p = Pattern.compile("\\d+|(\\()\\d+\\)");
String[] tests = {
"", // [] is not a match
"0", // [0] is a match without parenthesis
"(0)", // [(0)] is a match with surrounding parenthesis
"007", // [007] is a match without parenthesis
"(007)", // [(007)] is a match with surrounding parenthesis
"(007", // [(007] is not a match
"007)", // [007)] is not a match
"-1", // [-1] is not a match
};
for (String test : tests) {
Matcher m = p.matcher(test);
System.out.format("[%s] is %s %n", test,
!m.matches() ? "not a match"
: m.group(1) != null
? "a match with surrounding parenthesis"
: "a match without parenthesis"
);
}
As previously said, this matches a non-zero number of digits, possibly surrounded by parenthesis (and \1 distinguishes between the two).

1)
[^,]* means any number of characters that are not a comma
.+(,).+ means 1 or more characters followed by a comma followed by 1 or more characters
| means either the first one or the second one
2)
(\()? means zero or one '(' note* backslash is to escape '('
\d+ means 1 or more digits
(?(1)\)) means if back-reference \1 matched, then ')' note* no else is given
Also note that parenthesis are used to capture certain parts of the regular expression, except, of course, if they are escaped with a backslash

1) Anything that doesn't starts with a comma, or anything that contains a comma in between.
2) Any number that ends with a 1, and is between parenthesis, possible closed before and opened again after the number.