I built an expression to find any instances of '[###]' in Notepad++ for a given file:
[1]0x0a[22]0xbb[33]0xcc
and replace them with whitespace. Seems easy enough...
(\[[0-9]\]|\[[0-9][0-9]\]|\[[0-9][0-9][0-9]\])
and permutations of
/(\[[0-9]\]|\[[0-9][0-9]\]|\[[0-9][0-9][0-9]\])g
They did not yield any matches.
I tested this out on http://extendsclass.com/regex-tester.html and it worked as expected.
Am I missing some nuance for Regex in Notepad++? And, yes, I did have Search mode set for "Regular Expression".
My question is not a duplicate of this one: Delete brackets and numbers via Notepad++
My Regex works, just not in the version of Notepad++ I am using.
Ctrl+H
Find what: \[\d+\]
Replace with: A SPACE
CHECK Wrap around
CHECK Regular expression
Replace all
Explanation:
\[ # opening bracket
\d+ # 1 or more digits
\] # closing bracket
Screenshot (before):
Screenshot (after):
Related
Using Notepad++, how do I replace a word with a number within ()?
For instance, replace the word "serial" with "(1234)"? When I try to do this, the word "serial" is replaced with "1234", but without the parens.
In Notepad++ the parenthesis have to be escaped in the replacement part.
Ctrl+H
Find what: serial
Replace with: \(1234\)
CHECK Wrap around
CHECK Regular expression
Replace all
Screen capture (before):
Screen capture (after):
Choosing Extended (\n,\r,\t,\0, \x...) as the Search Mode and just specifying (1234) as the Replace with value worked fine.
The answer by Toto solves my problem.
In Notepad++ the parenthesis have to be escaped in the replacement part.
Ctrl+H
Find what: serial
Replace with: (1234)
CHECK Wrap around
CHECK Regular expression
Replace all
How would you use the regex in Notepad++ to format replacing a single character that it finds in every line excepts for the duplicate ones in the certain line further?
test1:_|TEST:-TEST.|
test2:_|TEST:-TEST.|
test3:_|TEST:-TEST.|
As shown in the test code, there are two colons; I'm trying to replace the first colon with each line to a ; and NOT the second one found; the result of me doing the regex should equal to this:
test1;_|TEST:-TEST.|
test2;_|TEST:-TEST.|
test3;_|TEST:-TEST.|
Ctrl+H
Find what: ^.+?\K:
Replace with: ;
CHECK Wrap around
CHECK Regular expression
UNCHECK . matches newline
Replace all
Explanation:
^ # beginning of line
.+? # 1 or more any character but newline, not greedy
\K # forget all we have seen until this position
: # colon
Screen capture (before):
Screen capture (after):
I'm guessing that maybe this expression,
(\w+)\s*(?::)(\s*_\s*\|\s*\w+\s*:\s*-\w+\.\|)
with a replacement of $1;$2 might work.
DEMO 1
Or with less boundaries, this expression:
([^:]+):(.*)
with the same replace.
DEMO 2
It's done like this
Find (?m)^[^:\r\n]*\K:
Replace ;
https://regex101.com/r/rT1vG9/1
I have a N++ file with the following lines:
asm-java-2.0.0-lib
cib-slides-3.1.0
lib-hibernate-common-4.0.0-beta
I want to remove everything from the '-' before the numbers begin so the results look like:
asm-java
cib-slides
lib-hibernate-common
So far I've come up with [0-9]+ but that ignores the '.' and the trailing alphabets. Does anyone know a correct command for find and replace?
Ctrl+H
Find what: -\d.*$
Replace with: LEAVE EMPTY
check Wrap around
check Regular expression
UNCHECK . matches newline
Replace all
Explanation:
- # a dash
\d # a digit
.* # 0 or more any character but newline
$ # end of line
Result for given example:
asm-java
cib-slides
lib-hibernate-common
Use regex to find and replace
Find: ^(.+)-\d.*$
Replace: $1
Here's regex I used in VSCode to find and replace to get your task done:
(.*)?-\d.*
And replace with $1
Not sure about notepad++ but should get it done for you as well.
Good Morning in my timezone
I want to replace a character that is in the beginning of each line
So i had used the following regular expression to find the text
^\d
And it works fine in finding all the characters
The problem is in the replace with
I want to replace with single quote followed by the same character found above
How can i do it ?
Thanks in advance
You may try this option:
Find:
^(?=\d)
Replace:
' <-- just a single quote
The find pattern uses a positive lookahead which asserts that the first character is a digit, but nothing is ever matched. Then, the replacement is a single quote.
You may use
Find What: ^\d
Replace With: '$0
where $0 is the backreference to the match value.
Another one would be:
Find:
^(\d)
Replace:
'\1
In this example \1 would be 1st captured group.
I have the following:
itemid=44'>Red Flower</a>
I need it to be this:
_ITEMID_START_44_ITEMID_END_
Can this be done with regular expressions? I need to keep the id (44 in the example), and replace everything on the left with _ITEMID_START_and everything on the right with _ITEMID_END_.
Note: The itemid is one digit or two but never no more than two.
I found something about tagged regular expressions and backreferences which seems like it would work but the syntax is killing me.
I tried this (and other attempts):
Find What: ^(\bitemid=\b)^([0-9][0-9]^)\b'>\b[a-z]+\b</a>\b)
Replace With: ^(\b_ITEMID_START_\b^2^(\b_ITEMID_END_\b
I am using UltraEdit to do the find and replace in over 20,000 *.html files. Any help would be very much appreciated.
The solution of Casimir et Hippolyte and also first solution of Avinash Raj work both in UltraEdit with selecting Perl as regular expression engine. The second search string of Avinash Raj requires removing backslash left of character ' in search string to work in UltraEdit.
UltraEdit has 3 regular expression engines: UltraEdit, Unix and Perl.
The search string in the question is a mixture of UltraEdit and Perl regular expression syntax and therefore does not work.
With UltraEdit reguar expression engine:
Find what: itemid=^([0-9]+^)*</a>
Replace with: _ITEMID_START_^1_ITEMID_END_
With Unix or Perl regular expression engine:
Find what: itemid=([0-9]+).*</a>
Replace with: _ITEMID_START_\1_ITEMID_END_
More secure because non greedy, but only with Perl regex engine:
Find what: itemid=(\d+).*?</a>
Replace with: _ITEMID_START_\1_ITEMID_END_
IDM published the power tips tagged expressions for UltraEdit regex engine and Perl regular expressions: Backreferences for Perl regex engine.
You can try this:
Find What: \bitemid=([0-9][0-9]?)'>[^<]*</a>
Replace With: _ITEMID_START_\1_ITEMID_END_
A replacement string is a normal string, and all the regex special characters (except for the backreference) loose their special meaning.
\b the word boundary is the limit between a character that come from the \w character class (a shortcut for [A-Za-z0-9_]) and an other character.
Note: I can't try it with ultraedit, if you obtain a literal \1, replace it with $1
The below regex would match everything and capture only the digits which was just after to the itemid=. And in the replacement part, the whole line is replaced with _ITEMID_START_\1_ITEMID_END_ (\1 represents the first captured group. It may vary for different languages)
.*(?<=\bitemid=)([0-9]{1,2}).*
And the substitution would be,
_ITEMID_START_\1_ITEMID_END_
DEMO
If you just want to replace only,
itemid=44'>Red Flower</a>
with
_ITEMID_START_44_ITEMID_END_
Then your regex would be,
\bitemid=([0-9]{1,2})\'>[^<]*<\/a>
And the substitution would be,
_ITEMID_START_\1_ITEMID_END_