Replacing char in a String with Regular Expression - regex

I got a string like this:
PREFIX-('STRING WITH SPACES TO REPLACE')
and i need this:
PREFIX-('STRING_WITH_SPACES_TO_REPLACE')
I'm using Notepad++ for the Regex Search and Replace, but i'm shure every other Editor capable of regex replacements can do it to.
I'm using:
PREFIX-\('(.*)(\s)(.*)'\)
for search and
PREFIX-('\1_\3')
for replace
but that replaces only one space from the string.

The regex search feature in Notepad++ is very, very weak. The only way I can see to do this in NPP is to manually select the part of the text you want to work on, then do a standard find/replace with the In selection box checked.
Alternatively, you can run the document through an external script, or you can get a better editor. EditPad Pro has the best regex support I've ever seen in an editor. It's not free, but it's worth paying for. In EPP all I had to do was this:
search: ((?:PREFIX-\('|\G)[^\s']+)\s+
replace: $1_
EDIT: \G matches the position where the previous match ended, or the beginning of the input if there was no previous match. In other words, the first time you apply the regex, \G acts like \A. You can prevent that by adding a negative lookahead, like so:
((?:PREFIX-\('|(?!\A)\G)[^\s']+)\s+
If you want to prevent a match at the very beginning of the text no matter what it starts with, you can move the lookahead outside the group:
(?!\A)((?:PREFIX-\('|\G)[^\s']+)\s+
And, just in case you were wondering, a lookbehind will work just as well as a lookahead:
((?:PREFIX-\('|(?<!\A)\G)[^\s']+)\s+

You have to keep matching from the beggining of the string untill you can match no more.
find /(PREFIX-\('[^\s']*)\s([^']*'\))/
replace $1_$2
like: while (/(PREFIX-\('[^\s']*)\s([^']*'\))/$1_$2/) {}

How about using Replace all for about 20 times? Or until you're sure no string contains more spaces

Due to nature of regex, it's not possible to do this in one step by normal regular expression.
But if I be in your place, I do such replaces in several steps:
find such patterns and mark them with special character
(Like replacing STRING WITH SPACES TO REPLACE with #STRING WITH SPACES TO REPLACE#
Replace #([^#\s]*)\s to #\1_ server times.
Remove markers!

I studied a little the regex tool in Notepad++ because I didn't know their possibilities.
I conclude that they aren't powerful enough to do what you want.
Your are obliged to learn and use a programming language having a real regex capability. There are a number of them. Personnaly, I use Python. It would take 1 mn to do what you want with it

You'd have to run the replace several times for each space but this regex will work
/(?<=PREFIX-\(')([^\s]+)\s+/g
Replace with
\1_ or $1_
See it working at http://refiddle.com/10z

Related

Regex in search & replace: avoid fixed length of lookaround

In a long corpus of text, I want to make some corrections in certain
environments. However, I am encountering problems when using regex with text
editors. I switched to gedit to have an editor which supports regex in
search & replace.
Crucially, I only want to make changes if the line starts with a certain
pattern (\nm or \mb). The problem is that the element that I want to
replace (o' -> o'o) is not at a fixed length from the beginning of the line
and I can't include the regex in the lookbehind (the lookbehind fails).
Is there any way to include what I am looking for in a simple text editor
regex? Or is this already a step where I have to learn how to script in, for
example, Python?
This is what the regex looks like so far.
(?<=\\(nm|mb)).*o'(?=(q|w|r|t|z|p|s|d|f|g|h|j|k|l|x|c|v|b|n|m|a|i|u|e))
Of course, I can't apply .* in the replace without losing its content.
Put a capture group around .* and a back-reference in the replacement.
Find: (?<=\\(nm|mb))(.*)o'(?=(q|w|r|t|z|p|s|d|f|g|h|j|k|l|x|c|v|b|n|m|a|i|u|e))
Replace: \1o'o

Regular expression to remove lines containing word with exceptions

I am using this regex in PowerGrep, (this regex search for strings LAB RAD TRAN)
.*((LAB)|(RAD)|(TRAN)).*\r\n
to search and remove lines in plain text that contains strings or part of a string and it works great.
Now I need something more. I want to keep the word LABER, but remove every other string containing LAB, such as LABOR, LAB1, ALAB, ALABA, etc.
Is there a way to "protect" a string LABER and remove every other string containing LAB?
Tried to alter the above regex using * but it always includes the word LABER that I need to keep. Any solution?
I think PowerGrep supports lookaround assertions; if so, this should work:
.*((LAB(?!ER\b))|(RAD)|(TRAN)).*\r\n
Although that will keep anything ending with LABER, not just the whole word.
You can add exclusions to regex in the form by means of a look-ahead:
(?m)^.*(LAB(?!(?:ER|OV)\b)|RAD|TRAN).*$
The (?!(?:ER|OV)\b) lookahead will check if the sequence LAB is not followed by ER or OV and a word boundary.
I am adding the alternation into look-ahead because your ask to "protect" LABER and LABOV.
Also, since you are looking for whole lines, you can make use of the multiline mode (?m) and ^/$ anchors.

How to search (using regex) for a regex literal in text?

I just stumbled on a case where I had to remove quotes surrounding a specific regex pattern in a file, and the immediate conclusion I came to was to use vim's search and replace util and just escape each special character in the original and replacement patterns.
This worked (after a little tinkering), but it left me wondering if there is a better way to do these sorts of things.
The original regex (quoted): '/^\//' to be replaced with /^\//
And the search/replace pattern I used:
s/'\/\^\\\/\/'/\/\^\\\/\//g
Thanks!
You can use almost any character as the regex delimiter. This will save you from having to escape forward slashes. You can also use groups to extract the regex and avoid re-typing it. For example, try this:
:s#'\(\\^\\//\)'#\1#
I do not know if this will work for your case, because the example you listed and the regex you gave do not match up. (The regex you listed will match '/^\//', not '\^\//'. Mine will match the latter. Adjust as necessary.)
Could you avoid using regex entirely by using a nice simple string search and replace?
Please check whether this works for you - define the line number before this substitute-expression or place the cursor onto it:
:s:'\(.*\)':\1:
I used vim 7.1 for this. Of course, you can visually mark an area before (onto which this expression shall be executed (use "v" or "V" and move the cursor accordingly)).

Regular expression question

I have some text like this:
dagGeneralCodes$_ctl1$_ctl0
Some text
dagGeneralCodes$_ctl2$_ctl0
Some text
dagGeneralCodes$_ctl3$_ctl0
Some text
dagGeneralCodes$_ctl4$_ctl0
Some text
I want to create a regular expression that extracts the last occurrence of dagGeneralCodes$_ctl[number]$_ctl0 from the text above.
the result should be: dagGeneralCodes$_ctl4$_ctl0
Thanks in advance
Wael
This should do it:
.*(dagGeneralCodes\$_ctl\d\$_ctl0)
The .* at the front is greedy so initially it will grab the entire input string. It will then backtrack until it finds the last occurrence of the text you want.
Alternatively you can just find all the matches and keep the last one, which is what I'd suggest.
Also, specific advice will probably need to be given depending on what language you're doing this in. In Java, for example, you will need to use DOTALL mode to . matches newlines because ordinarily it doesn't. Other languages call this multiline mode. Javascript has a slightly different workaround for this and so on.
You can use:
[\d\D]*(dagGeneralCodes\$_ctl\d+\$_ctl0)
I'm using [\d\D] instead of . to make it match new-line as well. The * is used in a greedy way so that it will consume all but the last occurrence of dagGeneralCodes$_ctl[number]$_ctl0.
I really like using this Regular Expression Cheatsheet; it's free, a single page, and printed, fits on my cube wall.

Notepad++ regexp to search and replace with exceptions

I'm a regexp newbie and I would like to know how to do a search and replace for the following case:
A file contains many occurrences of the following:
L1234_XL3.ext
and also many occurrences of:
L1234_XL3
I only want to find and replace L1234_XL3 occurrences with XL3 without affecting instances that have an extension.
I am using notepad++ to do the regular expression.
If Notepad++ supports lookaheads, you can simply use L1234_XL3(?!\.ext) for the search and "XL3" for the replacement.
EDIT: Looks like it doesn't support lookaheads after all. A pity; you'll have to do it the hard way without regexes (regexen?):
Replace L1234_XL3.ext with QQQ (or any other string that doesn't appear in the file)
Replace L1234_XL3 with XL3.
Replace QQQ with L1234_XL3.ext.
Step 1.
Change all occurences of L1234_XL3.ext to L-1-2-3-4_XL3.ext (for example)
Step 2.
Change all occurences of L1234_XL3 to XL3
Step 3.
Change all occurences of L-1-2-3-4_XL3.ext back to L1234_XL3.ext
As far as I understand Notepad++ 5.4.5 doesn't support positive lookahead