I am trying to update specific rows starting with 0500 and update positions 31 to 35, 41 to 45 with 00000 in notepad ++ using regex. Is it something possible ? positive feedback will be really appreciated.
Code tried and got no luck
Find: ∆(0500){1}(.{5,30}).{5}(.{36,40}).{5}
Replace: \1\200000\300000
Mode: Regular Expression
You can use
Find: ^(0500.{26}).{5}(.{5}).{5}
Replace: \100000\200000
Details:
^ - start of a line
(0500.{26}) - Group 1 (\1): 0500 and any 26 chars other than line break chars
.{5} - any 5 chars other than line break chars
(.{5}) - Group 2 (\2): any 5 chars other than line break chars
.{5} - any 5 chars other than line break chars
Related
I need a pattern that would accept
minimun two characters and only numbers
OR
only numbers with dash in third position and first two numbers should be either 56 or 78. So if there is dash then there should be 56 or 78 next to it
Valid matches:
12
123456789
423432423423423423423432
56-1
56-23456789
78-12
78-34234234234234234234234
Invalid matches:
1
1-
11-
1-112
44-2342424
64-4345334
55-sdrfewrwe
56-
5678-234324123423154
Here is my pattern so far:
[0-9]{2}-[0-9]{1,}|^[0-9]+$
You can use
^(?:(?:56|78)-[0-9]+|[0-9]{2,})$
See the regex demo.
Details:
^ - start of string
(?: - start of a non-capturing group with two alternatives:
(?:56|78)-[0-9]+-56or78, -`, one or more digits
| - or
[0-9]{2,} - two or more digits
) - end of the group
$ - end of string.
How about that idea...
^(?:56-|78-|\d)\d+$
See this demo at regex101
I'm using regex re.findall(r"[0-9]+(.*?)\.\s(.*?)[0-9]+", text) to get below text
8 EXT./INT. MONORAIL - MORNING 8
9 EXT. CITY SCAPE/MONORAIL - CONTINUOUS 9
But my current output doesn't have the prefix and suffix numbers. I'm trying to have the prefix digits also in the output as follows.
9 EXT. CITY SCAPE/MONORAIL - CONTINUOUS
Any help greatly appreciated! Thanks in advance.
(The current output is given below)
You can use
(?m)^([0-9]+)\s*(.*?)\.\s(.*?)(?:\s*([0-9]+))?$
See the regex demo. *Details:
(?m) - a multiline modifier
^ - start of string
([0-9]+) - Group 1: one or more digits
\s* - zero or more whitespaces
(.*?) - Group 2: zero or more chars other than line break chars as few as possible
\.\s - a dot and a whitespace
(.*?) - Group 3: zero or more chars other than line break chars as few as possible
(?:\s*([0-9]+))? - an optional occurrence of zero or more whitespaces and then Group 4 capturing one or more digits
$ - end of line.
I have strings like is below,
nn"h11p3ppppvxq3b288N1 m 227"]
{vanxtageendganmesbhorgtgt(1702)}' d3zd6xf8dz8xd6dz8f6zd8`
[nn"5rvh11p3ppppvxq3b288N1 n 227"]
{vanxtageendganmesbhorgtgt(1802)}
d3zd6xf8dz8xd6dz8f6zd8
I start my 1st capturing group from m 227 till end of third line,
And my 2nd group from n 227 till end of third line .....
Now I want to add some digits to end of first captured group , say it -22
And some digits to end of second captured group, say it -11
My first regex can match and works separately so 2nd as well .... but to make them combine with | it doesn't .....
Search: (m\s.*\n.*\n.*)
Replace: $1 -22
My combined regex is as below
(m\s.*\n.*\n.*|n\s.*\n.*\n.*)
Replace: $1-22 $2-11
But this will add (-22 -11) to both intendeds ...
I want the output to be as below
nn"h11p3ppppvxq3b288N1 m 227"]
{vanxtageendganmesbhorgtgt(1702)}
d3zd6xf8dz8xd6dz8f6zd8 -22
[nn"5rvh11p3ppppvxq3b288N1 n 227"]
{vanxtageendganmesbhorgtgt(1802)}
d3zd6xf8dz8xd6dz8f6zd8 -11
I have used | or for to combine both regexes to works as one for the purpose of time Savage ....
Any help will be appreciated
You can use
Find What: ([mn])\s.*\R.*\R.*
Replace With: $& -$1
Details:
([mn]) - Group 1 ($1): m ior n
\s - a whitespace
.*\R.*\R.* - a line, a line break, then again a line and a line break and then a line.
The $& in the replacement is the backreference to the whole match.
I am simply trying to take the following text:
Password length (length ordered)
5 = 1 (0.37%)
6 = 1 (0.37%)
7 = 1 (0.37%)
8 = 157 (58.58%)
9 = 55 (20.52%)
10 = 33 (12.31%)
11 = 12 (4.48%)
12 = 6 (2.24%)
13 = 2 (0.75%)
And find every new line that exists between Password Length and \n\n. Here's what I was currently doing
data[/(?<=Password length)(.*?)(?=\n\n)/m]
but that captures (length ordered) in the first line.
I have tried to do something like this:
44] pry(main)> data[/(?<=Password length.*?\n)(.*?)(?=\n\n)/m]
(eval):2: invalid pattern in look-behind: /(?<=Password length.*?\n)(.*?)(?=\n\n)/m
To basically capture everything after Password length up to the new line, but as you can see above I get an error about the invalid pattern in look-behind.
What should I be doing instead of this to fix this?
You can use
data[/Password length.*\R(?m:(.*?))\R{2}/, 1]
See the Rubular demo. Details:
Password length - a literal string
.* - the rest of the line
\R - a line break sequence
(?m:(.*?)) - An inline modifier group where . matches any char including line break chars, capturing group 1 matching any zero or more chars but as few as possible
\R{2} - double line break sequence.
The 1 argument returns the value inside the first capturing group only (see the str[regexp, capture] → new_str or nil reference).
An alternative:
data[/Password length.*\R\K.*(?:\R(?!\R).*)*/]
See this Rubular demo. Details:
Password length.*\R - Password length, the rest of the line and a line break sequence
\K - match reset operator, it removes all text matched so far from the match memory buffer
.* - a line, any zero or more chars other than line break chars as many as possible
(?:\R(?!\R).*)* - zero or more lines that do not end with double line break sequence where \R(?!\R).* matches a line break not immediately followed with another line break sequence, and .* matches the rest of the line.
So I got a text file with over 100000 lines to find and replace. Here is what I need to find:
>>DiskNum: 118
adfad
adfadf
adfadf
adfad
adafdd
>>FilePath: \\xxx\xxx\abc.tif
Replace with:
>>DiskNum: 118
adfad
adfadf
adfadf
adfad
adafdd
>>FullPath: C:\abc\abc.tif
The goal is to find >>DiskNum: 118 and the replace will happen on 6 lines after >>DiskNum: 118
Is there anyway to do it in notepadd++??
Use the following regex:
Find What: (^>>DiskNum: 118(?:\R.*){5}\R>>FilePath:\h*).*
Replace With: \1C:\\abc\\abc.tif
Regular expression: CHECKED
Details:
(^>>DiskNum: 118(?:\R.*){5}\R>>FilePath:\h*) - Group 1 (referenced with \1 from the replacement pattern) capturing a sequence of:
^ - start of a line
>>startoDiskNum: 118 - a literal char sequence
(?:\R.*){5} - 5 lines (\R is a line break, and .* matches any 0+ chars other than line break chars)
\R - line break
>>FilePath: - a literal char sequence
\h* - 0+ horizontal whitespaces
.* - the rest of the line