Regular Expressions are a complete void for me.
I'm dealing with one right now in TextMate that does what I want it to do...but I don't know WHY it does what I want it to do.
/[[:alpha:]]+|( )/(?1::$0)/g
This is used in a TextMate snippet and what it does is takes a Label and outputs it as an id name. So if I type "First Name" in the first spot, this outputs "FirstName".
Previously it looked like this:
/[[:alpha:]]+|( )/(?1:_:/L$0)/g (it might have been \L instead)
This would turn "First Name" into "first_name".
So I get that the underscore adds an underscore for a space, and that the /L lowercases everything...but I can't figure out what the rest of it does or why.
Someone care to explain it piece by piece?
EDIT
Here is the actual snippet in question:
<column header="$1"><xmod:field name="${2:${1/[[:alpha:]]+|( )/(?1::$0)/g}}"/></column>
This regular expression (regex) format is basically:
/matchthis/replacewiththis/settings
The "g" setting at the end means do a global replace, rather than just restricting the regex to a particular line or selection.
Breaking it down further...
[[:alpha:]]+|( )
That matches an alpha numeric character (held in parameter $0), or optionally a space (held in matching parameter $1).
(?1::$0)
As Roger says, the ? indicates this part is a conditional. If a match was found in parameter $1 then it is replaced with the stuff between the colons :: - in this case nothing. If nothing is in $1 then the match is replaced with the contents of $0, i.e. any alphanumeric character that is not a space is output unchanged.
This explains why the spaces are removed in the first example, and the spaces get replaced with underscores in your second example.
In the second expression the \L is used to lowercase the text.
The extra question in the comment was how to run this expression outside of TextMate. Using vi as an example, I would break it into multiple steps:
:0,$s/ //g
:0,$s/\u/\L\0/g
The first part of the above commands tells vi to run a substitution starting on line 0 and ending at the end of the file (that's what $ means).
The rest of the expression uses the same sorts of rules as explained above, although some of the notation in vi is a bit custom - see this reference webpage.
I find RegexBuddy a good tool for me in dealing with regexs. I pasted your 1st regex in to Buddy and I got the explanation shown in the bottom frame:
I use it for helping to understand existing regexs, building my own, testing regexs against strings, etc. I've become better # regexs because of it. FYI I'm running under Wine on Ubuntu.
it's searching for any alpha character that appears at least once in a row [[:alpha:]]+ or space ( ).
/[[:alpha:]]+|( )/(?1::$0)/g
The (?1 is a conditional and used to strip the match if group 1 (a single space) was matched, or replace the match with $0 if group 1 wasn't matched. As $0 is the entire match, it gets replaced with itself in that case. This regex is the same as:
/ //g
I.e. remove all spaces.
/[[:alpha:]]+|( )/(?1:_:/\L$0)/g
This regex is still using the same condition, except now if group 1 was matched, it's replaced with an underscore, and otherwise the full match ($0) is used, modified by \L. \L changes the case of all text that comes after it, so \LABC would result in abc; think of it as a special control code.
Related
I managed to do most of my conversion in VBA Macro (Word > txt) but some changes were made also that I could not forego or get around. Unfortunately, I had not been in the habit of using styles and precise formatting in my docs... (Which is why a PanDoc conversion did not "pan" out well, if you'll excuse the pun.)
In my docs, I was using bold text/lines for in-text titles (not Heading 2 alas) but as I was converting mid-sentence one or two-word bold phrases into phrases to go between double square brackets, the makeshift titles/headings were also changed to [[some title]] format in the process.
With Find and Replace (a batch script that goes through all files in a folder would also do), I would like to search for each and any number of instances of CRLF [[some title CRLF]]CRLF and replace the brackets with ** (to make the title bold), or perhaps ## to make the headings I was missing back in MS Word (I would of course need the line breaks as well).
For better understanding, please see attached picture here:
I am fairly sure that all instances are similarly syntaxed. If not, I may be able to tailor your regex code to differing instances later on.
As you can see, I was trying to do it in two steps but that's not good, because the second step (which I couldn't even get right) would propably have altered other texts I need intact (there must be sentences that start with double brackets after CRLF).
I would need the two steps in one so that only the targeted double bracketed text would be changed to bold or Heading 2.
Basically what I could not do is: find the proper regex solution for matching double CRLF-ed and square-bracketed text for any number of words than may occupy more than one line and starts with a capital letter. I would need an empty line above and below the title as indicated in the image (the VBA macro somehow made two instances of CRLF and carried the brackets to a new line, which I do not like, either).
EDIT.
In the meantime I managed to cook something up but now I couldn't insert the CRLF in front of the match string. At this point this is not enough as other instances are also changed, even lowercase in-line items, for some reason...
Regex:
\[\[([A-Z][\S\s]+?)\]\]
Substitution:
## $1\r\n
https://regex101.com/r/mH6B9N/1
Since then, I made improvements towards what I wanted (I had to test in NotePad++ and not Regex101, for different results), but now in multiple documents I have found match across spill-over lines, as described in here:
Single line regex search in Notepad++
Is it possible that I cannot do what I want? The problem is having non-title text strings having line-break, double brackets and capitalized letters.
What it looks like in other documents:
See here.
I circled around with red in image for clarification. See also:
https://regex101.com/r/8XsIGx/1
Is it possible to match a certain word like "címnél" and not execute on that match if that word is present in a line?
Thanks very much in advance,
F.
You can use
(?s)\R\K\[\[((?:(?!\[\[|]]).)*)\R*]](?=\R)
Replace with ## $1. See the regex demo.
Details:
(?s) - equivalent of the . matches newline option
\R - a line break sequence
\K - omit the text matched so far (the newlines)
\[\[ - a [[ text
((?:(?!\[\[|]]).)*) - Group 1: any char, as many as possible occurrences, that does not start a [[ or ]] char sequence
\R* - zero or more line breaks
]] - a ]] text
(?=\R) - immediately to the right, there must be a line break.
I'm normally ok with regex but I'm struggling with this.
I have a simple file with two words that start and end a set of data. The data between the words changes but - start and status are always in the same place.
Example :
start
Everything in between
status
I'm trying to work out how to delete (replace) everything between and including start and status
I'm sure I had it working with this at one time
(?i)^start.+?status
set(#replaceAll,$replace regular expression(#textTest,"(?i)^start.+?status"," "),"Global")
but its just not working anymore.
You could use the regular expression
\bstart\b.+?\bstatus\b
which does not require "status" to be on the same line as "start". Two flags should be set:
case indifference (/i)
single-line mode, which allows . to match a newline (/s)
Demo
The regex reads, "match 'start' with a word break fore and aft (to avoid matching 'starting' or 'jumpstart', for example), then match one or more characters lazily, then match 'status' with wordbreaks". The middle match must be lazy so that the regex engine will stop at the next (rather than last) instance of 'status'.
If the regex engine being used does not support single-line mode, or something comparable, one can replace .+ with [\s\S]+.
So my original expression works and so dose Cary's
The files have changed since I last used the expression. They contain some white-space in the form of newlines that needed to be removed first
set(#cleanup,$replace(#text2,$new line," "),"Global")
set(#text2,$replace regular expression(#cleanup,"\\bstart\\b.*?\\bstatus\\b",""),"Global")
set(#cleanup,$replace regular expression(#cleanup,"(?i)^start.+?status:",""),"Global")
Sorry about that but thanks to all who looked and helped :)
This is test
There are two tabs (\t) in this line. I want to get rid of the part from the beginning to the first tab key, which is "This ", and I used the following pattern:
:s/.\{-}\t//g
It says it can't find the pattern. If I use the following, both tabs are replaced, which isn't what I want. Why doesn't the first pattern work?
:s/.*\t//g
Your first attempt does not work because you are matching the fewest number of any character followed by a tab. The fewest number of any character is zero (0). So both of your tabs match without any other characters.
Based on the comments, the above explanation was incorrect.
Here is one possible solution.
:s/^[^\t]*\t//
This goes from the beginning ^, capturing any number of non-tab characters [^\t]* until it reaches a tab \t.
Your pattern /.\{-}\t didn't work because of the g flag in the :s command. This flag enables global matching so it matches twice. Just remove the flag and it will work. In addition, when deleting something you can omit the replacement part in :s:
:s/.\{-}\t
The full :s/.\{-}\t// is fine as well. Note that in either case it should not say "pattern not found" as you described. If you see that message, there is something else different between your example and your actual text.
I've got a practical application for a vim regex where I'd like to remove numbers from the end of file location links. For example, if the developer is sloppy and just adds files and doesn't reuse file locations, you'll end up with something awful like this:
PATH_TO_MY_FILES>
PATH_TO_MY_FILES1>
...
PATH_TO_MY_FILES22>
PATH_TO_MY_FILES_ELSEWHERE>
PATH_TO_MY_FILES_ELSEWHERE1>
...
So all I want to do is to S&R and replace PATH_TO_MY_FILES*\d+ with PATH_TO_MY_FILES* using regex. Obviously I am not doing it quite right, so I was hoping someone here could not spoon feed the answer necessarily, but throw a regex buzzword my way to get me on track.
Here's what I have tried:
:%s\(PATH_TO_MY_FILES\w*\)\(\d+\)>:gc
But this doesn't work, i.e. if I just do a vim search on that, it doesn't find anything. However, if I use this:
:%s\(PATH_TO_MY_FILES\w*\)\(\d\)>:gc
It will match the string, but the grouping is off, as expected. For example, the string PATH_TO_MY_FILES22 will be grouped as (PATH_TO_MY_FILES2)(2), presumably because the \d only matches the 2, and the \w match includes the first 2.
Question 1: Why doesn't \d+ work?
If I go ahead and use the second string (which is wrong), Vim appears to find a match (even though the grouping is wrong), but then does the replacement incorrectly.
For example, given that we know the \d will only match the last number in the string, I would expect PATH_TO_MY_FILES22> to get replaced with PATH_TO_MY_FILES2>. However, instead it replaces it with this:
PATH_TO_MY_FILES2PATH_TO_MY_FILES22>gt
So basically, it looks like it finds PATH_TO_MY_FILES22>, but then replaces only the & with group 1, which is PATH_TO_MY_FILES2.
I tried another regex at Regexr.com to see how it would interpret my grouping, and it looked correct, but maybe a hack around my lack of regex understanding:
(PATH_TO_\D*)(\d*)>
This correctly broke my target string into the PATH part and the entire number, so I was happy. But then when I used this in Vim, it found the match, but still replaced only the &.
Question 2: Why is Vim only replacing the &?
Answer 1:
You need to escape the + or it will be taken literally. For example \d\+ works correctly.
Answer 2:
An unescaped & in the replacement portion of a substitution means "the entire matched text". You need to escape it if you want a literal ampersand.
I have a set of files that i need to loop through and find all the files that does not have a specific string between 2 other specific strings. How can i do that?
I tried this but it didnt work:
grep -lri "\(stringA\).*\(?<!stringB\).*\(stringC\)" ./*.sql
EDIT:
the file could have structure as following:
StringA
StringB
StringA
StringC
all i want i s to know if there is any occurences where string A and stringC has no stringC in between.
You can use the -L option of grep to print all files which don't match and look for the specific combination of strings:
grep -Lri "\(stringA\).*\(stringB\).*\(stringC\)" ./*.sql
The short answer is along the lines of:
grep "abc[^(?:def)]*ghi" ./testregex
That's based on a testregex file like so:
abcghiabc
abcdefghi
abcghi
The output will be:
$ grep "abc[^(?:def)]*ghi" ./testregex
abcghiabc
abcghi
Mapped to your use-case, I'd wager this translates roughly to:
grep -lri "stringA[^(?:stringB)]*stringC" ./*.sql
Note that I've removed the ".*" between each string, since that will match the very string that you're attempting to exclude.
Update: The original question now calls out line breaks, so use grep's -z flag:
-z
suppress newline at the end of line, subtituting it for null character. That is, grep knows where end of line is, but sees the input as one big line.
Thus:
grep -lriz "stringA[^(?:stringB)]*stringC" ./*.sql
When I first had to use this approach myself, I wrote up the following explanation...
Specifically: I wanted to match "any character, any number of times,
non-greedy (so defer to subsequent explicit patterns), and NOT
MATCHING THE SEQUENCE />".
The last part is what I'm writing to share: "not matching the sequence
/>". This is the first time I've used character sequences combined
with "any character" logic.
My target string:
<img class="photo" src="http://d3gqasl9vmjfd8.cloudfront.net/49c7a10a-4a45-4530-9564-d058f70b9e5e.png" alt="Iron or Gold" />
My first attempt:
<img.*?class="photo".*?src=".*?".*?/>
This worked in online regex testers, but failed for some reason within
my actual Java code. Through trial and error, I found that replacing
every ".?" with "[^<>]?" was successful. That is, instead of
"non-greedy matching of any character", I could use "non-greedy
matching of any character except < or >".
But, I didn't want to use this, since I've seen alt text which
includes these characters. In my particular case, I wanted to use the
character sequence "/>" as the exclusion sequence -- once that
sequence was encountered, stop the "any character" matching.
This brings me to my lesson:
Part 1: Character sequences can be achieved using (?:regex). That is,
use the () parenthesis as normal for a character sequence, but prepend
with "?:" in order to prevent the sequence from being matched as a
target group. Ergo, "(?:/>)" would match "/>", while "(?:/>)*" would
match "/>/>/>/>".
Part 2: Such character sequences can be used in the same manner as
single characters. That is, "[^(?:/>)]*?" will match any character
EXCEPT the sequence "/>", any number of times, non-greedy.
That's pretty much it. The keywords for searching are "non-capturing
groups" and "negative lookahead|lookbehind", and the latter feature
goes much deeper than I've gone so far, with additional flags that I
don't yet grok. But the initial understanding gave me the tool I
needed for my immediate task, and it's a feature that I've wondered
about for awhile -- thus, I figured I'd share the basic introduction
in case any of you were curious about tucking it away in your toolset.
After playing around with the statement provided by the DreadPirateShawn:
stringA[^(?:stringB)]*stringC
I figured out that it is not a truly valid regex. This statement was excluding every character in the given set and not the full string. So I continued digging.
After some googling and testing the pattern, I came up with the following statement, that seems to fit my needs:
stringA\s*\t*(?:(?!stringB).)*\s*\t*stringC
This pattern matches any text except the provided string between 2 specified strings. It also takes into consideration whitespace characters.
There is more testing to be done, but it seems that this pattern perfectly fits my requirements
UPDATE: Here is a final version of the statement that seems to work for me:
grep -lriz "(set feedback on){0,}[ \t]*(?:(?!set feedback off).)*[ \t]*select sysdate from dual" ./*.sql