I have application where few labels are written like
ui-label-Display Not Masked
Now I want to replace it by
ui-label-Display_Not_Masked
so i have written search regex by
ui-label-(\w+ )*
This searches all expression but I am not able to create a expression to replace this text as required.
I have written one regex
$1_
which replaces
ui-label-Display Not Masked
by
ui-label-Display Not_Masked
This cannot be done with a single regex in a single iteration.
You have two choices:
Replace (ui-label-\w+) (note the space at the end) with $1_ until it no longer matches anything.
Make a looong regex with as many capture groups as necessary, i.e. (ui-label-\w+) (?:(\w+)(?: (\w+))?)? and replace with $1_$2_$3.
Related
Trying to do this in the Atom editor (1.39.1 x64, uBuntu 18.04), though assume this applies to other text editors using regular expressions.
Say we have this text:
This text has some double-spaces. Lets try to remove them.
But not after a full-stop or if three or more spaces.
Which we would like to change to:
This text has some double-spaces. Lets try to remove them.
But not after a full-stop or if three or more spaces.
Using Find with Regex enabled (.*), all occurrences are correctly found using: [a-zA-Z] [a-zA-Z]. But what goes in the Replace row to enforce the logic:
1st letter, single space, 2nd letter?
You can use this
([a-z])\s{2}([a-z])
and replace by $1 $2
Regex Demo
If your editor supports lookarounds you can use
(?<=[a-z])\s{2}(?=[a-z])
Replace by single space character
Regex demo
Note:- don't forget to use i flag for case insensitivity or just change the character class to [a-zA-Z]
I want to use a regular expression in VBA to do a search, capture part of it, and then use that part in the replacement. For example, I want to run the search and replace on these lines:
(a4a)
(aHa)
And get a result of:
(b4b)
(bHb)
How do I capture the 2nd character and use it again in the replacement?
In VBA, capture parts of the search with parenthesis () and use them in replacement with $ and the number of the capture occurrence. Note, normal parenthesis need to be escaped, which is the opposite of vim.
So in this case:
searchPattern = "\(a(.)a\)"
replacement = "(b$1b)"
I have something in a text file that looks like '%r'%XXXX, where the XXXX represents some name at the end. Examples include '%r'%addR or '%r'%removeA. I can match these patterns using regex '%r'%\w+. What I would like to replace this with is '{!r}'.format(XXXX). Note that the name has to stay the same in the replace so I'd get '{!r}.format(addR) or '{!r}.format(removeA). Is there a way to replace parts of the matched string in this way while retaining the unknown variable name pulled out with \w+ in the regex search?
I'm specifically looking for a solution using the find and replace features in Notepad++.
You can use
'%r'%(\w+)
and replace with '{!r}.format\(\1\)
The '%r'%(\w+) pattern contains a pair of unescaped parentheses that create a capturing group. Inside the replacement pattern, we use a \1 backreference to restore that value.
NOTE: The ( and ) in the replacement must be escaped because otherwise they are treated as Boost conditional replacement pattern functional characters.
See more on capturing groups and backreferences.
Search on:
'%r'%(XXXX)
Replace with:
Whatever You like \1
\1 will match the first set of grouping parentheses.
I have a document which has been copy/pasted from MS Word. All the quotations are copied as ''something'' which basically is creating a mess in my LaTeX document, hence they have to be ``something''.
Is it possible to make a regular expression that finds all these ''something'' where something can be anything (including symbols, numbers etc.), and a regular expression that replaces it with the correct quotation? I am using Sublime Text which is able to use RegEX directly in the editor.
The below regex would match all the double single quoted strings and capture all the characters except the first two single quotes(only in the matched string). Replacing the matched characters with double backticks plus the characters inside group index 1 will give you the desired result.
Regex:
''(.*?'')
Replacemnet string:
``$1
DEMO
I've a huge text file with lines like this:
080012;Bovalino;RC;CAL;0964;89034;B098;9021;http://www.website-most.en/000/000/
And i would like extract only:
080012;***Bovalino***;***RC***;CAL;***0964***;***89034***;B098;9021;http://www.website-most.en/000/000/
And delete all other text.
Can this be done with regular expressions?
You can capture the stuff you want to keep and use a backreference in the replacement string:
Find what: ^\d*;(\w*;\w*);\w*;(\d*;\d*).*
Replace with: \1;\2
And make sure you do not tick the . matches newline option.
With Notepad++ 6 you can also use $1;$2 for the replacement (with the same meaning).
If the different fields may contain all sorts of characters and not just digits and letters, this is probably your best bet:
Find what: ^[^;]*;([^;]*;[^;]*);[^;]*;([^;]*;[^;]*).*