I'm trying to use a regular expression to find everything except for the data I don't want to replace. I'm then wanting to replace everything except for the found expression.
My match is to find all words that start with "CN=" and ends with 12 characters after. I'm currently using the logic (!?CN=)\w{12}. It finds the two occurrences in the string. However, I'm wanting to find everything but these two occurrences and find everything else and replace everything else with an empty value so that I just have the two CN= word values.
The attached image shows my testing and results. I'm wanting the reverse affect.
image
Thanks,
Fred
You could use .* to capture whatever comes before a match, and add $ (end-of-input) as alternative for CN=\w{12}:
.*?($|CN=\w{12})
Replace with just the captured groups:
$1
Demo on Java regex tester
NB: I did not add the !? as you wrote ...words that start with "CN=". If you really need to keep the exclamation mark before CN=, then you should add it to the regular expression.
Related
I want to use regular expressions to skip until i find a (-)Hyphen and store whatever comes after. I tried a few things but it didn't work out.
This is an example string:
Fall Down Seven Times; Stand Up Eight." -Naoki Higashida
I am just learning regular expressions and want to use them in my project to skip until I encounter different symbols.
*edit 1: this is what I have used so far, with some other stuff I found online.
"(?:[a-zA-Z;.;""]*)[^-][a-zA-Z]*"
Thank you for your help.
You can use .*-(.*). The capture group will contain everything after the hyphen.
.* matches anything
- matches literal hyphen
(.*) matches anything and captures it
Here's a demo.
I'm normally ok with regex but I'm struggling with this.
I have a simple file with two words that start and end a set of data. The data between the words changes but - start and status are always in the same place.
Example :
start
Everything in between
status
I'm trying to work out how to delete (replace) everything between and including start and status
I'm sure I had it working with this at one time
(?i)^start.+?status
set(#replaceAll,$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 :)
The following is in PHP but the regex will also be used in javascript.
Trying to extract repeating patterns from a string
string can be any of the following:
"something arbitrary"
"D123"
"D111|something"
"D197|what.org|when.net"
"D297|who.197d234.whatever|when.net|some other arbitrary string"
I'm currently using the following regex: /^D([0-9]{3})(?:\|([^\|]+))*/
This correctly does not match the first string, matches the second and third correctly. The problem is the third and fourth only match the Dxxx and the last string. I need each of the strings between the '|' to be matched.
I'm hoping to use a regex as it makes it a single step. I realize I could just detect the leading Dxxx then use explode or split as appropriate to break the strings out. I've just gotten stuck on wanting a single regular expression match step.
This same regex may be used in Python as well so just want a generic regex solution.
There is no way to have a dynamic number of capture groups in a regular expression, but if you know some upper limit to how many parts you would have in one string, you can just repeat the pattern that many times:
/^D([0-9]{3})(?:$|\|)(.*?)(?:$|\|)(.*?)(?:$|\|)(.*?)(?:$|\|)(.*?)(?:$|\|)/
So after the initial ^D([0-9]{3})(?:$|\|) you just repeat (.*?)(?:$|\|) as many times as you need it.
When the string has fewer elements, those remaining capture groups will match the empty string.
See regex tester.
Is something like preg_match_all() (the PHP variant of a global match) also acceptable for you?
Then you could use:
^(?|D([0-9]{3})|^.+$|(?!^)\|([^|\n]*)(?=\||$))
This will match everything in a string in different matches, e.g. take your string:
D197|what.org|when.net
It will you then give three matches:
D197
what.org
when.net
Running live: https://regex101.com/r/jL2oX6/4 (Everything in green are your group matches. Ignore what's in blue.)
I have a column of strings, and I want to use a regex to find commas or pipes in every cell, and then make an action. I tried this, but it doesn't work (no syntax error, just doesn't match neither commas nor pipes).
if(value.contains(/(,|\|)/), ...
The funny thing is that the same regex works with the same data in SublimeText. (Yes, I can work it there and then reimport, but I would like to understand what's the difference or what is my mistake).
I'm using Google Refine 2.5.
Since value.match should return captured texts, you need to define a regex with a capture group and check if the result is not null.
Also, pay attention to the regex itself: the string should be matched in its entirety:
Attempts to match the string s in its entirety against the regex pattern p and returns an array of capture groups.
So, add .* before and after the pattern you are looking inside a larger string:
if(value.match(/.*([,|]).*/) != null)
You can use a combination of if and isNonBlank like:
if(isNonBlank(value.match(/your regex/), ...
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.