I want to build a RegEx that will match with a string of IPs separated by comma (,) OR string will have only *. String should not have both IP address & *
Validate IP i.e 1.1.1.1 (Numbers and . dot char). Also, * alone is allowed
* is present, no other IPs should be present.
This is the regex
(((25[0-5]|2[0-4]\d|[01]?\d\d?)\.(25[0-5]|2[0-4]\d|[01]?\d\d?)\.(25[0-5]|2[0-4]\d|[01]?\d\d?)\.(25[0-5]|2[0-4]\d|[01]?\d\d?)(,\n|,?))|(,*))
Testing string:
192.168.1.1,192.56.3.23,189.35.2.2,198.23.45.56,198.168.1.255
How do I check for *?
You may use
^(?:(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.(?:25[0-5]|2[0-4]\d|[01]?\d\d?)|\*)(?:,\s*(?:(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\.(?:25[0-5]|2[0-4]\d|[01]?\d\d?)|\*))*$
See the regex demo
Expanded//verbose/free-spacing version:
^ # start of string
(?: # start of a grouping
(?: # start of another grouping
(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\. # First octet and .
(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\. # Second octet and .
(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\. # Third octet and .
(?:25[0-5]|2[0-4]\d|[01]?\d\d?) # Fourth octet
|\* # or just a * char instead of an IP
) # end of another grouping
) # end of grouping
(?:,\s* # a group that will repeat 0+ times, matches , then 0+ whitespaces
(?: # an IP matching grouping
(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\. # First octet and .
(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\. # Second octet and .
(?:25[0-5]|2[0-4]\d|[01]?\d\d?)\. # Third octet and .
(?:25[0-5]|2[0-4]\d|[01]?\d\d?) # Fourth octet
|\*) # Or a *
)* # ... zero or more times
$ # end of string
Related
I want to change one of the _ to another character, for example to -, the reason is there are problems reading in these filenames. I want a to become like b. So I want to change the second last underscore(_), how to specify this in an efficient way?
gsub("_", "-"), it must also be specified to a certain location.
a <- c("2018-01-09_B2_HILIC_POS_123_-14b_090.mzML", "2018-01-09_B2_HILIC_POS_243_-12a_026.mzML", "2020-01-09_B2_HILIC_POS_415_893a_059.mzML", "2020-01-18_B3_HILIC_POS_LV7001248356_040.mzML")
b <- c("2018-01-09_B2_HILIC_POS_123--14b_090.mzML", "2018-01-09_B2_HILIC_POS_243--12a_026.mzML", "2020-01-09_B2_HILIC_POS_415-893a_059.mzML", "2020-01-18_B3_HILIC_POS_LV4004365711_040.mzML")
Here is one base R option using sub :
sub('(.*)(_)(.*_.*)$', '\\1-\\3', a)
#[1] "2018-01-09_B2_HILIC_POS_123--14b_090.mzML"
#[2] "2018-01-09_B2_HILIC_POS_243--12a_026.mzML"
#[3] "2020-01-09_B2_HILIC_POS_415-893a_059.mzML"
#[4] "2020-01-18_B3_HILIC_POS-LV7001248356_040.mzML"
Here we divide data into 3 groups -
The 1st group is everything until second last underscore which is captured using (.*) and used as a backreference (\\1).
The 2nd group is second last underscore which us replaced with -.
The 3rd one is everything after second last underscore which is captured using (.*_.*) and used as a backreference (\\3).
Use
sub("_(?=[^_]*_[^_]*$)", "-", a, perl=TRUE)
See regex proof.
Explanation
--------------------------------------------------------------------------------
_ '_'
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
[^_]* any character except: '_' (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
_ '_'
--------------------------------------------------------------------------------
[^_]* any character except: '_' (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
$ before an optional \n, and the end of
the string
--------------------------------------------------------------------------------
) end of look-ahead
See R proof:
a <- c("2018-01-09_B2_HILIC_POS_123_-14b_090.mzML", "2018-01-09_B2_HILIC_POS_243_-12a_026.mzML", "2020-01-09_B2_HILIC_POS_415_893a_059.mzML", "2020-01-18_B3_HILIC_POS_LV7001248356_040.mzML")
sub("_(?=[^_]*_[^_]*$)", "-", a, perl=TRUE)
Results:
[1] "2018-01-09_B2_HILIC_POS_123--14b_090.mzML"
[2] "2018-01-09_B2_HILIC_POS_243--12a_026.mzML"
[3] "2020-01-09_B2_HILIC_POS_415-893a_059.mzML"
[4] "2020-01-18_B3_HILIC_POS-LV7001248356_040.mzML"
How do I match the year such that it is general for the following examples.
a <- '"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}'
b <- 'Þegar það gerist (1998/I) (TV)'
I have tried the following, but did not have the biggest success.
gsub('.+\\(([0-9]+.+\\)).?$', '\\1', a)
What I thought it did was to go until it finds a (, then it would make a group of numbers, then any character until it meets a ). And if there are several matches, I want to extract the first group.
Any suggestions to where I go wrong? I have been doing this in R.
You could use
library(stringr)
strings <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)')
years <- str_match(strings, "\\((\\d+(?: B\\.C\\.)?)")[,2]
years
# [1] "1953" "1998"
The expression here is
\( # (
(\d+ # capture 1+ digits
(?: B\.C\.)? # B.C. eventually
)
Note that backslashes need to be escaped in R.
Your pattern contains .+ parts that match 1 or more chars as many as possible, and at best your pattern could grab last 4 digit chunks from the incoming strings.
You may use
^.*?\((\d{4})(?:/[^)]*)?\).*
Replace with \1 to only keep the 4 digit number. See the regex demo.
Details
^ - start of string
.*? - any 0+ chars as few as possible
\( - a (
(\d{4}) - Group 1: four digits
(?: - start of an optional non-capturing group
/ - a /
[^)]* - any 0+ chars other than )
)? - end of the group
\) - a ) (OPTIONAL, MAY BE OMITTED)
.* - the rest of the string.
See the R demo:
a <- c('"You Are There" (1953) {The Death of Socrates (399 B.C.) (#1.14)}', 'Þegar það gerist (1998/I) (TV)', 'Johannes Passion, BWV. 245 (1725 Version) (1996) (V)')
sub("^.*?\\((\\d{4})(?:/[^)]*)?\\).*", "\\1", a)
# => [1] "1953" "1998" "1996"
Another base R solution is to match the 4 digits after (:
regmatches(a, regexpr("\\(\\K\\d{4}(?=(?:/[^)]*)?\\))", a, perl=TRUE))
# => [1] "1953" "1998" "1996"
The \(\K\d{4} pattern matches ( and then drops it due to \K match reset operator and then a (?=(?:/[^)]*)?\\)) lookahead ensures there is an optional / + 0+ chars other than ) and then a ). Note that regexpr extracts the first match only.
I thought I had a great regex for limiting the number of words entered into a TextBox, however I discovered, it fails when there is punctuation in the text.
How can I modify this regex (or use a different one) that correctly counts words that may be made up of several sentences or contain other symbols?
^(?:\b\w+\b[\s\r\n]*){1,10}$
This limits the words to 10.
I think this is the ONLY WAY it can be done.
I would like to see a better way if it exists.
(requires an atomic group)
For Unicode:
^\s*(?>[^\pL\pN]*[\pL\pN](?:[\pL\pN_-]|\pP(?=[\pL\pN\pP_-])|[?.!])*\s*){1,10}$
Explained
^
\s*
(?>
[^\pL\pN]* [\pL\pN] # Not letters/numbers, followed by letter/number
(?:
[\pL\pN_-] # Letter/number or '-'
|
\pP # Or, punctuation if followed by punctuation/letter/number or '-'
(?= [\pL\pN\pP_-] )
|
[?.!] # Or, (Add) Special word ending punctuation
)*
\s*
){1,10}
$
For Ascii:
^\s*(?>[\W_]*[^\W_](?:\w|[[:punct:]_-](?=[\w[:punct:]-])|[?.!])*\s*){1,10}$
Expanded
^
\s*
(?>
[\W_]* [^\W_]
(?:
\w
|
[[:punct:]_-]
(?= [\w[:punct:]-] )
|
[?.!]
)*
\s*
){1,10}
$
Regular Expression To exclude sub-string name(job corps)
Includes at least 1 upper case letter, 1 lower case letter, 1 number and 1 symbol except "#"
I have written something like below :
^((?!job corps).)(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[!#$%^&*]).*$
I tested with the above regular expression, not working for special character.
can anyone guide on this..
If I understand well your requirements, you can use this pattern:
^(?![^a-z]*$|[^A-Z]*$|[^0-9]*$|[^!#$%^&*]*$|.*?job corps)[^#]*$
If you only want to allow characters from [a-zA-Z0-9^#$%&*] changes the pattern to:
^(?![^a-z]*$|[^A-Z]*$|[^0-9]*$|[^!#$%^&*]*$|.*?job corps)[a-zA-Z0-9^#$%&*]*$
details:
^ # start of the string
(?! # not followed by any of these cases
[^a-z]*$ # non lowercase letters until the end
|
[^A-Z]*$ # non uppercase letters until the end
|
[^0-9]*$
|
[^!#$%^&*]*$
|
.*?job corps # any characters and "job corps"
)
[^#]* # characters that are not a #
$ # end of the string
demo
Note: you can write the range #$%& like #-& to win a character.
stribizhev, your answer is correct
^(?!.job corps)(?=.[0-9])(?=.[a-z])(?=.[A-Z])(?=.[!#$%^&])(?!.#).$
can verify the expression in following url:
http://www.freeformatter.com/regex-tester.html
The currency required format looks like
1,100,258
100,258
23,258
3,258
Or all integers like
123456 or 2421323 and so on.
I type below in ValidationExpression
(^[0-9]{1,3}(,\d{3})*) | (^[0-9][0-9]*)
But it doesn't work.
Do you have ignore pattern whitespace on? If not, remove the two spaces on each side of the pipe.
Since you're trying to match either, you should stick a marker $ at the end of the string, like so
Also what is the point of ^[0-9][0-9]*, when you can use ^[0-9]+?
^([0-9]{1,3}(?:,\d{3})*|[0-9]+)$
or
^(\d{1,3}(?:,\d{3})*|\d+)$
Explanation:
^ # Anchors to the beginning to the string.
( # Opens CG1
\d{1,3} # Token: \d (digit)
(?: # Opens NCG
, # Literal ,
\d{3} # Token: \d (digit)
# Repeats 3 times.
)* # Closes NCG
# * repeats zero or more times
| # Alternation (CG1)
\d+ # Token: \d (digit)
# + repeats one or more times
) # Closes CG1
$ # Anchors to the end to the string.