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]
Related
I have a couple of sentences that need processing using regular expressions. They're in a text file and I'm opening it in notepad++.
<tag>There are two tags here</tag>
<tag>How am i supposed to
feel when this is happening?</tag>
<tag>I'm not sure.
But oh well<tag>
Is it possible to use notepad++'s regular expressions and replace functionality to produce an output like so:
<tag>There are two tags here</tag>
<tag>How am i supposed to feel when this is happening?</tag>
<tag>I'm not sure. But oh well<tag>
So that sentences that span over two or more lines are joined based on the fact that there is a > at the end of the sentence. Thanks.
Replace this:
[\r\n]+(?!<)
with a space
Click for Demo
Explanation:
[\r\n]+ - matches 1+ occurrences of a \r or \n
(?!<) - negative lookahead to validate that the above match is not followed by an opening tag <
Before Replacement with space:
After replacing the matches with space:
I want to replace quotation marks in a latex document. It's written in German, which means that all quotation marks should be of the form "´text"' but some editors of the document have used these: "text", ´´text''.
The complication here is, that the document contains highlighted code using the lstlisting enviroment. In there the quotation marks should not be replaced.
I have a regex, that matches text inside the unwanted quotes, even if there are multiple words:
((``((\w+\s*)+)'')|("((\w+\s*)+)"))
I also have a regex, that matches a string ("asdf" in this case), only if it is not inside the lstlisting environment:
"asdf"(?=((?!\\end\{lstlisting\}).)*\\begin\{lstlisting\}?)
They work fine on their own, but when I combine them like this:
((``((\w+\s*)+)'')|("((\w+\s*)+)"))(?=((?!\\end\{lstlisting\}).)*\\begin\{lstlisting\}?)
some of the quoted strings, that should be matched are not and additionally the whole document is matched.
PS: I am currently using notepad++ for matching, because it allows . to match \n
[EDIT]: It works fine, as long as I limit the first part to single words:
((``((\w)+)'')|("((\w)+)"))(?=((?!\\end\{lstlisting\}).)*\\begin\{lstlisting\}?)
To match words with whitespaces, you can use
(``[\w\s]+''|"[\w\s]+")(?=(?:(?!\\end\{lstlisting\}).)*\\begin\{lstlisting\}?)
See regex demo
If you have spaces only between `` and '', or between "s, you will need to unroll the [\w\s]+ part as \w+(?:\s+\w+)*.
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.
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: ^[^;]*;([^;]*;[^;]*);[^;]*;([^;]*;[^;]*).*
I'd like to use Notepad++ to replace all leading spaces on a line with a like number of given characters. So for instance, I want to change:
zero
one
two
three
into:
zero
#one
##two
###three
I haven't been successful at getting this working. I did find Regex to replace html whitespace and leading whitespace in notepad++, but wasn't able to get the result I wanted.
Is this possible with Notepad++? I'd rather not have to write code to do this...
As Tim's answer indicates, this can't be done in a single search/replace, however here is how you can accomplish the same task fairly quickly using multiple replacements:
Find: ^( *)[ ]
Replace with: \1#
Now just spam the "Replace All" button until it indicates that there were no matches to replace. This will replace a single space at the beginning of each line on each click, so it will require the same number of clicks as your most-indented line.
Make sure "Regular expression" is selected as the search mode.
You would need variable-length lookbehind assertions to do this in a single regex, and Notepad++ doesn't support these.
For the record, in EditPadPro you can search for (?<=^ *)\s and replace with #.