I have the following string
someaddres.com/?f=[B]a-test,a test,Test[C]test a,test2
I'm trying to pull two groups from it:
[B]a-test,a test,Test
and
[C]test a,test2
How would I repeat the capture group until a character not present in the group is found?
My current regex is: f=(\[[A-Z]\][a-zA-Z0-9,-\s]+)
You may use this regex with a captured group that will match twice:
(\[\w+\][^[]+)
RegEx Demo
If you want 2 capture groups in single match then use:
(\[\w+\][^[]+)(\[\w+\][^[]+)
RegEx Demo 2
RegEx Details:
(: Start capture group #1
\[: Match a [
\w+: Match 1+ word characters
\]: Match a ]
[^[]+: Match 1+ of any characters that is not [`
): End capture group #1
Related
For example I want to match three values, required text, optional times and id, and the format of id is [id=100000], how can I match data correctly when text contains spaces.
my reg: (?<text>[\s\S]+) (?<times>\d+)? (\[id=(?<id>\d+)])?
example source text: hello world 1 [id=10000]
In this example, all of source text are matched in text
The problem with your pattern is that matches any whitespace and non whitespace one and unlimited times, which captures everything without getting the other desired capture groups. Also, with a little help with the positive lookahead and alternate (|) , we can make the last 2 capture groups desired optional.
The final pattern (?<text>[a-zA-Z ]+)(?=$|(?<times>\d+)? \[id=(?<id>\d+)])
Group text will match any letter and spaces.
The lookahead avoid consuming characters and we should match either the string ended, or have a number and [id=number]
Said that, regex101 with further explanation and some examples
You could use:
:\s*(?<text>[^][:]+?)\s*(?<times>\d+)? \[id=(?<id>\d+)]
Explanation
: Match literally
\s* Match optional whitespace chars
(?<text> Group text
[^][:]+? match 1+ occurrences of any char except [ ] :
) Close group text
\s* Match optional whitespace chars
(?<times>\d+)? Group times, match 1+ digits
\[id= Match [id=
(?<id>\d+) Group id, match 1+ digirs
] Match literally
Regex demo
I want to capture all the strings from multi lines data. Supposed here the result and here’s my code which does not work.
Pattern: ^XYZ/[0-9|ALL|P] I’m lost with this part anyone can help?
Result
XYZ/1
XYZ/1,2-5
XYZ/5,7,8-9
XYZ/2-4,6-8,9
XYZ/ALL
XYZ/P1
XYZ/P2,3
XYZ/P4,5-7
XYZ/P1-4,5-7,8-9
Changed to
XYZ/1
XYZ/1,2-5
XYZ/5,7,8-9
XYZ/2-4,6-8,9
XYZ/A12345 after the slash limited to 6 alphanumeric chars
XYZ/LH-1234567890 after the /LH- limited to 10 numeric chars
The pattern could be:
^XYZ\/(?:ALL|P?[0-9]+(?:-[0-9]+)?(?:,[0-9]+(?:-[0-9]+)?)*)$
The pattern in parts matches:
^ Start of string
XYZ\/ Match XYX/ (You don't have to escape the / depending on the pattern delimiters)
(?: Outer on capture group for the alternatives
ALL Match literally
| Or
P? Match an optional P
[0-9]+(?:-[0-9]+)? Match 1+ digits with an optional - and 1+ digits
(?: Non capture group to match as a whole
,[0-9]+(?:-[0-9]+)? Match ,and 1+ digits and optional - and 1+ digits
)* Close the non capture group and optionally repeat it
) Close the outer non capture group
$ End of string
Regex demo
You can use this regex pattern to match those lines
^XYZ\/(?:P|ALL|[0-9])[0-9,-]*$
Use the global g and multiline m flags.
Btw, [P|ALL] doesn't match the word "ALL".
It only matches a single character that's a P or A or L or |.
For example, let's say I have a list of words:
words.txt
accountable
accountant
accountants
accounted
I want to match "accountant\naccountants"
I've tried /(\n\w+){2}s/, but \w+ seems to be perfectly matching different things.
My RegEx also matches the following undesirable texts:
action
actionables
actionable
actions
Am I reaching out too far in what regex can do?
You could for example use a capture group, and match a newline followed by a backreference to the same captured text and an s char.
If the first word can also be at the start of the string, instead of being preceded by a newline, you can use an anchor ^ instead.
^(\w+)\n\1s$
^ Start of string
(\w+) Capture group 1, match 1+ word chars
\n\1s Match a newline, backreference \1 to match the same text as group 1 and an s char
$ End of string
Regex demo
I have the following example of numbers, and I need to add a zero after the second period (.).
1.01.1
1.01.2
1.01.3
1.02.1
I would like them to be:
1.01.01
1.01.02
1.01.03
1.02.01
I have the following so far:
Search:
^([^.])(?:[^.]*\.){2}([^.].*)
Substitution:
0\1
but this returns:
01 only.
I need the 1.01. to be captured in a group as well, but now I'm getting confuddled.
Does anyone know what I am missing?
Thanks!!
You may try this regex replacement with 2 capture groups:
Search:
^(\d+\.\d+)\.([1-9])
Replacement:
\1.0\2
RegEx Demo
RegEx Details:
^: Start
(\d+\.\d+): Match 1+ digits + dot followed by 1+ digits in capture group #1
\.: Match a dot
([1-9]): Match digits 1-9 in capture group #2 (this is to avoid putting 0 before already existing 0)
Replacement: \1.0\2 inserts 0 just before capture group #2
You could try:
^([^.]*\.){2}\K
Replace with 0. See an online demo
^ - Start line anchor.
([^.]*\.){2} - Negated character 0+ times (greedy) followed by a literal dot, matched twice.
\K - Reset starting point of reported match.
EDIT:
Or/And if \K meta escape isn't supported, than see if the following does work:
^((?:[^.]*\.){2})
Replace with ${1}0. See the online demo
^ - Start line anchor.
( - Open 1st capture group;
(?: - Open non-capture group;
`Negated character 0+ times (greedy) followed by a literal dot.
){2} - Close non-capture group and match twice.
) - Close capture group.
Using your pattern, you can use 2 capture groups and prepend the second group with a dot in the replacement like for example \g<1>0\g<2> or ${1}0${2} or $10$2 depending on the language.
^((?:[^.]*\.){2})([^.])
^ Start of string
((?:[^.]*\.){2}) Capture group 1, match 2 times any char except a dot, then match the dot
([^.].*) Capture group 2, match any char except a dot
Regex demo
A more specific pattern could be matching the digits
^(\d+\.\d+\.)(\d)
^ Start of string
(\d+\.\d+\.) Capture group 1, match 2 times 1+ digits and a dot
(\d) Capture group 2, match a digit
Regex demo
For example in JavaScript
const regex = /^(\d+\.\d+\.)(\d)/;
[
"1.01.1",
"1.01.2",
"1.01.3",
"1.02.1",
].forEach(s => console.log(s.replace(regex, "$10$2")));
Obviously, there will be tons of solutions for this, but if this pattern holds (i.e. always the trailing group that is a single digit)... \.(\d)$ => \.0\1 would suffice - to merely insert a 0, you don't need to match the whole thing, only just enough context to uniquely identify the places targeted. In this case, finding all lines ending in a . followed by a single digit is enough.
I have to collect two informantion from a text using regex. The name and the database and relate then in one table. But a can only collect then individually.
This is an example, i have many blocks of these, and two of then don't have a database value, these i need to ingnore
[SCD] {I need the name between []}
Driver=/opt/pcenter/pc961/ODBC7.1/lib/DWmsss27.so
Description=
Database=scd {I need the value after Defaut|Database}
Address=#######
LogonID=######
Password=######
QuoteId=No
AnsiNPW=No
ApplicationsUsingThreads=1
The regex to find the name is:
(?<=\[)(.*)(?=\])
The regex to find the value after database is
(?<=Defaut|Database=)(.*)
How can i combine both of then into onde regex ?
To match both values you could use 2 capturing groups instead and use a repeating pattern and a negative lookahead to check if a line do not start with Default of Database until the line does.
\[([^]]+)\](?:\r?\n(?!Default|Database).*)*\r?\n(?:Default|Database)=(\S+)
About the pattern
\[ Match [
( Capture group 1
[^]]+ match 1+ times not ]
) Close group 1
\] Match ]
(?: Non capturing group
\r?\n Match newline,
(?! Negative lookahead, assert what is directly on the right is not
Default|Database Match one of the options
).* Close negative lookahead and match any char except a newline 0+ times
)* Close non capturing group and repeat 0+ times
\r?\n(?:Default|Database)= Match newline, any of the options and =
(\S+) Capturing group 2, match 1+ times a non whitespace char (or use (.+) to match any char 1+ times)
regexstorm demo