I have here a text file
with
245. asdasd
45. asdasd
42. gfhfgh
5353. sdfsdf
want to remove all numbers in front of it.
try it already with find and replace [0-9].
Ctrl+F - Replace - Search mode: regular expression
Find What: [0-9]+
Replace With:
// Replace with is empty
Ctrl-H - shows replace.
Ensure that 'regular expression' is on.
Use ^[0-9]+
Note begging ^ used to mark begin of line. Remove it if need remove all numbers
use RegEx search (should be ^\d+ for numbers with more than 1 digit):
On the replace tab switch to regular expressions; Search for ^\d+\.\s and leave Replace With empty.
(Matches start of line, any number, a dot, trailing space)
You should try to use this regular expression to remove what you want in the replace dialog:
^[0-9]*\.\ (.*)
And replace with:
\1
To search for digits:
Find what: [[:d:]]
Select: o Regular Expression
To remove only "line numbers with dots" use
[0-9]+.
To remove digits only from start of string
Input : 123. asdasd 3434
Output: . asdasd 3434
Replace All using below regex with o or more spaces in start of string
^\s*\d+
For Numbers
Ctrl H
Find what: [0-9]
Replace With:
(leave replace with blank)
For Period
Ctrl H
Find what: .
Replace With:
(leave replace with blank)
Related
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.
How do I replace ) when it comes after 1, 2, or 3 digits (not chars, and without removing the digit(s) themselves)?
Find what: ((?<!\d)\d{1,3})\)
Replace with: $1
This ensures that the ) comes after 1 to 3 digits (no more, no less).
Just append your replacement text to the end of $1. For example, if you want to replace it with the word TEST, your replacement would be $1TEST
As simple as:
Find: ((^|[^\d])\d{1,3})\)
Replace \1
And don't forget to enable te regular expressions in the panel.
Visit this link to try a working demo.
I have a plain text file with content like this:
prežrať/RN
prežrieť/Z
prežrúc/zZ
prežuť/c
...
Q: How can I remove all strings after / symbol in every row in Notepad++?
Desired output:
prežrať
prežrieť
prežrúc
prežuť
...
I am doing this with Find-and-Replace for every different string after /, but there are too many combinations.
Search for: /.*, replace with nothing.
The character / matches just /. ., however, matches any character except newlines, so .* will match a sequence of characters up until the first newline. You can find a demonstration here: http://regex101.com/r/kT0uE3.
If you want to remove characters only after the last on the line /, you should use the regex /[^/]*$. You can find an explanation and demonstration here: https://regex101.com/r/sZ6kP7/74.
In regular expression mode
Find:
/.*
Replace:
(empty)
Set find and replace to regular expression mode.
Find string: /.*
Replace String: (empty string)
Notepad++ find and replace is by default line ended (it won't go over multiple lines)
Using find and replace:
Hit CTRL-H to open the Replace dialogue box
enter /.* into "Find what"
leave "Replace with" empty
Select "Regular expression" (and .matches newline if it is single line)
Click on Replace
Here we go... You are done.