What should be the regex pattern if my texts contain the characters like "\ / > <" etc and I want to find them. That's because regex treats "/" like it's part of the search pattern and not an individual character.
For example, I want to find Super Kings from the string <span>Super Kings</span>, using VB 2010.
Thanks!
Just try this:
\bYour_Keyword_to_find\b
\b is used in RegEx for matching word boundary.
[EDIT]
You might be looking for this:
(?<=<span>)([^<>]+?)(?=</span>)
Explanation:
<!--
(?<=<span>)([^<>]+?)(?=</span>)
Options: case insensitive; ^ and $ match at line breaks
Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=<span>)»
Match the characters “<span>” literally «<span>»
Match the regular expression below and capture its match into backreference number 1 «([^<>]+?)»
Match a single character NOT present in the list “<>” «[^<>]+?»
Between one and unlimited times, as few times as possible, expanding as needed (lazy) «+?»
Assert that the regex below can be matched, starting at this position (positive lookahead) «(?=</span>)»
Match the characters “</span>” literally «</span>»
-->
[/EDIT]
In regex you must escape the / with \.
For instance, try: <span>(.*)<\/span> <span>([^<]*)<\/span> or <span>(.*?)<\/span>
Read more from:
http://www.regular-expressions.info/characters.html
Related
I want to replace anything other than character, spaces and number only in end with empty string or in other words: we replace any number or spaces comes in-starting or in-middle of the string replace with empty string.
Example
**Input** **Output**
Ndd12 Ndd12
12Ndd12 Ndd12
Ndd 12 Ndd 12
Nav G45up Nav Gup
Attempted Code
regexp_replace(df1[col_name]), "(^[A-Za-z]+[0-9 ])", ""))
You may use:
\d+(?!\d*$)|[^\w\n]+(?!([A-Z]|$))
RegEx Demo
Explanation:
\d+(?!\d*$): Match 1+ digits that are not followed by 0+ digits and end of line
|: OR
[^\w\n]+(?!([A-Z]|$)): Match 1+ non-word characters that are not followed by an uppercase letter or and end of line
if you use python, you can use regular expressions.
You can use the re module.
import re
new_string = re.sub(r"[^a-zA-Z0-9]","",s)
Where ^ means exclusion.
Regular expressions exist in other languages. So it would be helpful to find a regular expression.
I came up with this regex to capture all characters that you want to remove from the string.
^\d+|(?<=\w)\d+(?![\d\s])|(?<=\s)\s+
Do
regexp_replace(df1[col_name]), "^\d+|(?<=\w)\d+(?![\d\s])|(?<=\s)\s+", ""))
Regex Demo
Explanation:
^\d+ - captures all digits in a sequence from the start.
(?<=\w)\d+(?![\d\s]) - Positive look behind for a word character with a negative look ahead for a number followed by space and capturing a sequence of digits in the middle. (Captures digits in G45up)
(?<=\s)\s+ - positive look behind for a space followed by one or more spaces, capturing all additional spaces.
Note : This regex could be inefficient when matching large strings as it uses expensive look-arounds.
^\d+|(?<=\w)\d+(?![\d\s])|(?<=\s)\s+|(?<=\w)\W|\W(?=\w)|(?<!\w)\W|\W(?!\w)
I have a string and would like to match a part of it.
The string is Accept: multipart/mixedPrivacy: nonePAI: <sip:4168755400#1.1.1.238>From: <sip:4168755400#1.1.1.238>;tag=5430960946837208_c1b08.2.3.1602135087396.0_1237422_3895152To: <sip:4168755400#1.1.1.238>
I want to match PAI: <sip:4168755400#
the whitespace can be a word so i would like to use .* but if i used that it matches most of the string
The example on that link is showing what i'm matching if i use the whitespace instead of .*
(PAI: <sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
The example on that link is showing what i'm trying to achieve with .* but it should only match PAI: <sip:4168755400#
(PAI:.*<sip:)((?:\([2-9]\d{2}\)\ ?|[2-9]\d{2}(?:\-?|\ ?))[2-9]\d{2}[- ]?\d{4})#
I tried lookaround but failing.
Any idea?
thanks
Matching the single space can be updated by using a character class matching either a space or a word character and repeat that 1 or more times to match at least a single occurrence.
Note that you don't have to escape the spaces, and in both occasions you can use an optional character class matching either a space or hyphen [ -]?
If you want the match only, you can omit the 2 capturing groups if you want to.
(PAI:[ \w]+<sip:)((?:\([2-9]\d{2}\) ?|[2-9]\d{2}[ -]?)[2-9]\d{2}[- ]?\d{4})#
Regex demo
The regex should be like
PAI:.*?(<sip:.*?#)
Explanation:
PAI:.*? find the word PAI: and after the word it can be anything (.*) but ? is used to indicate that it should match as few as possible before it found the next expression.
(<sip:.*?#) capturing group that we want the result.
<sip:.*?# find <sip: and after the word it can be anything .*? before it found #.
Example
I'm trying to create a simple Grammar correction tool.
I want to create a regular expression that finds fullstops (" . ") that are not followed by a space so I can replace that with a fullstop and space.
For e.g. This is a sentence.This is another sentence.
Only the first fullstop in the above example should be matched in the expression.
I've tried /\.[^\s]/g but it returns an additional character after the matched fullstop. I would like to match only the fullstop.
How can I do this?
The negated character class [^\s] in the pattern expects a match (any character except a whitespace character), that is why you have the additional character.
If you want to match the dot only, you could use a negative lookahead to assert what is on the right is not a whitspace char or the end of the string:
\.(?!\s|$)
Regex demo
To not match a dot that is not followed by a whitespace char excluding a newline:
\.(?![^\S\r\n])
Regex demo
You can look for all dots using:
(\.)
This will match all dots on below examples:
This is a sentence.This is another sentence.
i am looking. for dots. . ...
You can add a |$ to seek for end of line, and with a little tweak, you get a regex that match all dots not followed by whitespace nor being on the end of a line:
(\.(?!\ |$))
Note that there's a whitespace as literal here. The "must-work-everywhere" example will be like:
(\.(?![[:space:]]|$))
If not, search on the regex reference on the language you use.
I am looking for a regular expression to catch a whole word or expression within a sentence that contains dots:
this is an example test.abc.123 for what I am looking for
In this case i want to catch "test.abc.123"
I tried with this regex:
(.*)(\b.+\..++\b)(.*)
(.*) some signs or not
(\b.+\..++\b) a word containing some signs followed by at least on dot that is followed by some signs and this at least once
(.*) some more signs nor not#
but it gets me: "abc.123 for what I am looking for"
I see that I got something completely wrong, can anyone enlighten me?
If you need to match part of a string you don't need to match entire string (unless you are restricted by a functionality).
Your regex is so greedy. It also has dots every where (.+ is not a good choice most of the time). It doesn't have a precise point to start and finish either. You only need:
\w+(?:\.+\w+)+
It looks for strings that begin and end with word characters and contain at least a period. See live demo here
This regex pattern matches strings with two or more dots:
.*\..*\..*
"." matches any character except line-breaks
"*" repeats previous tokens 0 or more times
"." matches a single dot, slash is used for escape
.* Match any character and continue matching until next token
test.abc.123
(.) Match a single dot
test. abc.123
.* Again, any character and continue matching until next token
test.example.com
. Matches a single dot
test.example. com
.* Matches any character and continue matching until next token
test.example.com
Try this pattern: (?=\w+\.{1,})[^ ]+.
Details: (?=\w+\.{1,}) - positive lookahead to locate starting of a word with at least one dot (.). Then, start matching from that position, until space with this pattern [^ ]+.
Demo
A perforce depot path is of the following format:
//depot/solution/project/file.cs#232
How can I extract just the "file.cs". I have tried the following.
[^//]*$
Not sure how to eliminate "#1" part. Could anyone help?
This will find file names even if they don't have a # after them.
(\w+\.\w+)[^/]*$
Explanation:
(\w+\.\w+)
This matches the file name itself, \w is a word character (same as [a-zA-Z0-9_]). So its 1+ word character, a full stop (. on its own matches any character, you need \. to match an actual .), then 1+ more word characters.
[^/]*
Matches 0+ characters that are not /. But all the word characters will get put into the \w+ match before (because it is evaluated first and + will try to match as much as it can) so in your example this matches the #200
$
matches the end of the line. Which is needed so a.directory wouldn't get matched in /a.directory/file.txt
You can use this regex:
/\/([^\/#]*)#/
And use matched group #1 for your value file.cs
Assuming you're using PCRE, you can use the pattern:
'[^/]*(?=#)'