In Notepad++, is there a special char I could use in search to only give me letters which are lower cased?
Notepad++ supports searching regular expressions. You can search for [a-z] in your file. This will find any single lower case character.
You expand the regular expression to make it more specific - such as a certain # of lower case characters.
Related
I would like to mark in Notepad++ the sql scripts in a text log. The sql files have this format in the text:
AAAAAAAA.BBBBBBBBBBB.sql
So what I execute is this sentence in search menu:
\w*.sql
As I should get BBBBBBBBBBB.sql. The point is that in some script names there are dashes (-), and when that happens I dont get the whole name, but just the end after the last dash.
For example, in:
AAAAAAAA.BBBBB-CCCCCCC.sql
I would like to get BBBBB-CCCCCCC.sql, but I just get CCCCCCC.sql
Is there any possible formula to get them?
If the match can not start and end with a hyphen:
\w+(?:-\w+)*\.sql
\w+ Match 1+ word characters
(?:-\w+)* Optionally match - and 1+ word characters
\.sql Match .sql
See a regex demo.
Note that in your pattern the \w* can also match 0 occurrences and that the . can match any character if it is not escaped.
Another option could be using a character class to match either - or a word character, but this would also allow to mix and match like --a--.sql
[\w-]+\.sql
See another regex demo.
I´m working with VBA and trying to split a string into three columns, almost all strings are like Company Name 3567782 Agent Name.pdf
With this pattern I want to match all the text before a space and digits (1st group), the digits (2nd group) and all the text after the space and before the .pdf (3rd group).
strPattern = "^(.+)\n(\d{4,10})\n(.+).pdf"
I recall spaces in python are \s but saw in VBA are \n.
Can you help me find the right pattern for what I´m looking for?
As I put in my comment, I use the https://regex101.com site. There are others but I find this one the most helpful to me.
When I put in your regex
^(.+)\n(\d{4,10})\n(.+).pdf
and test string
Company Name 3567782 Agent Name.pdf
the first thing I notice is that the regex does not match the test string (see right side under MATCH INFORMATION).
Here are a couple things that I saw:
\n is newline, not space. In regex, space is " ".
Your last "." in ".pdf" is not registering as a literal period, it's a token that matches any character. To match a literal period, you need \.
If we change those two things it returns three groups that seem to match what you are looking for.
^(.+) (\d{4,10}) (.+)\.pdf
It looks like for the digits, you are looking for between 4 and 10 digits. If that's correct, it looks like your regex is good. You could put in a handful of example strings into the TEST STRING area and make sure that it works in all cases.
I'd use either of these:
(?:(?:([a-zA-Z]+\.?)|(\d+)))
capture a-Z greedy with a possible . to allow for the .pdf or capture digits
this version excludes the space [ ] or \s
or keep the search structured so you can control what goes in and out of each column
^(\w+\s\w+)|(\d+)|(\w+\s\w+\.\w+$)
\b or ^ - word boundary or start of string
(\w+\s\w+) - 1st capture \w+ - any alpha numeric char greedily, followed by 1 x space (use \s* or \s+ for more), followed again by alpha numeric greedily
|(\d+) - alteration - \d+ - capture just digits
`|(\w+\s\w+.\w+$) - similar to 1st group but allows for the '.' of pdf and bounds to the end of string (\G or $).
you could optionally build the '.' into the 1st group like my top answer, but for neatness and better control I prefer the 2nd.
I need to replace a character with a pattern but the characters before the selected character also need to be replaced, using the search and replace function.
For instance: . has to be replace with 1/2, so hugo.ignatz becomes hug1/2gnatz.
How should I go about this?
If I understood you right, do you want this?
:%s#.[.].#1/2#g
this will replace x.y by 1/2
I have a very large number (a couple hundred digits long), and I'd like to use vim to add commas to the number in the appropriate manner, i.e. after each group of three digits, moving from right to left. How can I do this efficiently?
Taken from here
Substitue command that adds commas in the right spot.
:%s/\(\d\)\(\(\d\d\d\)\+\d\#!\)\#=/\1,/g
This uses a zero width lookahead to match any number that isn't followed by groups of three numbers followed by one number. (or 3n+1 numbers)
So the numbers that match in are marked with ^. These are then replaced with a comma after it the match.
31415926
^ ^
Which replaces to
31,415,926
A friend of mine suggests using the printf program: ciw<C-r>=system("printf \"%'d\" ".shellescape(#"))<CR>.
This is one way of doing it:
s/\d\{-1,}\ze\(\d\{3}\)\+\s/&,/g
Notes:
\{-1,} is saying match at least 1 but in a non-greedy way (Vim doesn't seem to support the usual \+\? syntax; also, for quantifiers, you just need to escape the opening curly brace)
\ze is saying match the pattern behind this but don't store the match in & (equivalent to positive look-ahead)
\(\d\{3}\)\+\> matches groups of 3 digits that ends with word-nonword boundary (word in this sense means alphanumerical + underscore).
Alternatively, you can use \s for space/tab, or \D for non-digit instead of \>, whichever fits your needs better
The way that I used is to create a macro that adds one single comma, and then invoke the macro a whole bunch of times, like qahhi,<ESC>hq#a#a#a#a…
In a Notepad++ or an editor with similar features, is there an easy to search for a word which contains a number? For instance, suppose numbers were denoted by "~", then if I searched for "abc~" in the following text:
abc4
abc not a number
I would simply get the first word.
Use the "regular expression" search mode. If you just want to find groups of numbers, use \d+. If you want it to select the whole word containing one or more digits, use \w*\d+\w*.
Type this regular expression in the Find dialog box (select the regex option):
\w*\d\w*