example:
"today [(["hi"],{"my"})],["ok"],("good")),(["gg"],["fire"])] nice game [(["1"],{"2"})],["3"],("4")),(["5"],["6"])] end."
->
[(["hi"],{"my"})],["ok"],("good")),(["gg"],["fire"])]
[(["1"],{"2"})],["3"],("4")),(["5"],["6"])]
In this case, there is a noise in the middle.
It doesn't work in the usual way
Please give me an idea of extracting the value between parentheses in another way
I love regex, almost as much as I hate them. Writing an effective regex can take a problem from impossible to trivial.
Knowing how to iterate one is very important. Let me teach a man to fish.
Consider using an online regex tool: https://regex101.com/
Input your string:
image of me adding in the text
Then try to Match it. I did with the following \([^()]+\)
That is:
escape the parenthesis you want to match on \( and \) because they are special symbols.
You know you want to match everything that isn't a parenthesis between them. So I use a no-match clause by doing [^...] where the elipsis are all the characters I don't want to match. I then add my parenthesis raw in there [^()]
Finally. I want to have 1 or more of those so I use +.
Thus I have my awnser.
image of me matching the text
edit: cannot post images because of rep
Related
I've been looking around and could not make this happen. I am not totally noob.
I need to get text delimited by (including) START and END that doesn't contain START. Basically I can't find a way to negate a whole word without using advanced stuff.
Example string:
abcSTARTabcSTARTabcENDabc
The expected result:
STARTabcEND
Not good:
STARTabcSTARTabcEND
I can't use backward search stuff. I am testing my regex here: www.regextester.com
Thanks for any advice.
Try this
START(?!.*START).*?END
See it here online on Regexr
(?!.*START) is a negative lookahead. It ensures that the word "START" is not following
.*? is a non greedy match of all characters till the next "END". Its needed, because the negative lookahead is just looking ahead and not capturing anything (zero length assertion)
Update:
I thought a bit more, the solution above is matching till the first "END". If this is not wanted (because you are excluding START from the content) then use the greedy version
START(?!.*START).*END
this will match till the last "END".
START(?:(?!START).)*END
will work with any number of START...END pairs. To demonstrate in Python:
>>> import re
>>> a = "abcSTARTdefENDghiSTARTjlkENDopqSTARTrstSTARTuvwENDxyz"
>>> re.findall(r"START(?:(?!START).)*END", a)
['STARTdefEND', 'STARTjlkEND', 'STARTuvwEND']
If you only care for the content between START and END, use this:
(?<=START)(?:(?!START).)*(?=END)
See it here:
>>> re.findall(r"(?<=START)(?:(?!START).)*(?=END)", a)
['def', 'jlk', 'uvw']
The really pedestrian solution would be START(([^S]|S*S[^ST]|ST[^A]|STA[^R]|STAR[^T])*(S(T(AR?)?)?)?)END. Modern regex flavors have negative assertions which do this more elegantly, but I interpret your comment about "backwards search" to perhaps mean you cannot or don't want to use this feature.
Update: Just for completeness, note that the above is greedy with respect to the end delimiter. To only capture the shortest possible string, extend the negation to also cover the end delimiter -- START(([^ES]|E*E[^ENS]|EN[^DS]|S*S[^STE]|ST[^AE]|STA[^RE]|STAR[^TE])*(S(T(AR?)?)?|EN?)?)END. This risks to exceed the torture threshold in most cultures, though.
Bug fix: A previous version of this answer had a bug, in that SSTART could be part of the match (the second S would match [^T], etc). I fixed this but by the addition of S in [^ST] and adding S* before the non-optional S to allow for arbitrary repetitions of S otherwise.
May I suggest a possible improvement on the solution of Tim Pietzcker?
It seems to me that START(?:(?!START).)*?END is better in order to only catch a START immediately followed by an END without any START or END in between. I am using .NET and Tim's solution would match also something like START END END. At least in my personal case this is not wanted.
[EDIT: I have left this post for the information on capture groups but the main solution I gave was not correct.
(?:START)((?:[^S]|S[^T]|ST[^A]|STA[^R]|STAR[^T])*)(?:END)
as pointed out in the comments would not work; I was forgetting that the ignored characters could not be dropped and thus you would need something such as ...|STA(?![^R])| to still allow that character to be part of END, thus failing on something such as STARTSTAEND; so it's clearly a better choice; the following should show the proper way to use the capture groups...]
The answer given using the 'zero-width negative lookahead' operator "?!", with capture groups, is: (?:START)((?!.*START).*)(?:END) which captures the inner text using $1 for the replace. If you want to have the START and END tags captured you could do (START)((?!.*START).*)(END) which gives $1=START $2=text and $3=END or various other permutations by adding/removing ()s or ?:s.
That way if you are using it to do search and replace, you can do, something like BEGIN$1FINISH. So, if you started with:
abcSTARTdefSTARTghiENDjkl
you would get ghi as capture group 1, and replacing with BEGIN$1FINISH would give you the following:
abcSTARTdefBEGINghiFINISHjkl
which would allow you to change your START/END tokens only when paired properly.
Each (x) is a group, but I have put (?:x) for each of the ones except the middle which marks it as a non-capturing group; the only one I left without a ?: was the middle; however, you could also conceivably capture the BEGIN/END tokens as well if you wanted to move them around or what-have-you.
See the Java regex documentation for full details on Java regexes.
When you address a regex capture, things can get tricky when digits follow the capture. In PCRE, I can write
${1}000
to substitute the capture of Group 1 followed by three zeroes.
Does anyone know the equivalent syntax in Dreamweaver replace operations, if any?
If we had a series of "A"s instead of zeroes, we could use:
$1AAAA
But these:
$10000
${1}0000
do not work.
I believe the regex flavor is ECMAScript. Just cannot find the information.
This may not be addressed in the syntax. If so, that would be good to know.
Thank you!
Edit: I should add that this is not matter of life and death as I have a number of grep tools at my fingertips. I would just like to know.
Dreamweaver's regular expression find and replace is supposed to be based on JavaScript's implementation of RegExp. You should be able to just use $1000 in the replacement text. However, like you've found, the replacement groups ($ + group number) are not properly recognized when the replacement text has digits immediately after the grouping token.
FWIW: I've logged a bug on this at http://adobe.ly/DWwish
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
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)).
Following on from a previous question in which I asked:
How can I use a regular expression to match text that is between two strings, where those two strings are themselves enclosed two other strings, with any amount of text between the inner and outer enclosing strings?
I got this answer:
/outer-start.*?inner-start(.*?)inner-end.*?outer-end/
I would now like to know how to exclude certain strings from the text between the outer enclosing strings and the inner enclosing strings.
For example, if I have this text:
outer-start some text inner-start text-that-i-want inner-end some more text outer-end
I would like 'some text' and 'some more text' not to contain the word 'unwanted'.
In other words, this is OK:
outer-start some wanted text inner-start text-that-i-want inner-end some more wanted text outer-end
But this is not OK:
outer-start some unwanted text inner-start text-that-i-want inner-end some more unwanted text outer-end
Or to explain further, the expression between outer and inner delimiters in the previous answer above should exclude the word 'unwanted'.
Is this easy to match using regexes?
Replace the first and last (but not the middle) .*? with (?:(?!unwanted).)*?. (Where (?:...) is a non-capturing group, and (?!...) is a negative lookahead.)
However, this quickly ends up with corner cases and caveats in any real (instead of example) use, and if you would ask about what you're really doing (with real examples, even if they're simplified, instead of made up examples), you'll likely get better answers.
A better question to ask yourself than "how do I do this with regular expressions?" is "how do I do solve this problem?". In other words, don't get hung up on trying to solve a big problem with regular expressions. If you can solve half the problem with regular expressions, do so, then solve the other half with another regular expression or some other technique.
For example, make a pass over your data getting all matches, ignoring the unwanted text (read: get results both with and without the unwanted text). Then, make a pass over the reduced set of data and weed out those results that have the unwanted text. This sort of a solution is easier to write, easier to understand and easier to maintain over time. And for any problem you're likely to need to solve with this approach it will be sufficiently fast enough.
You can replace .*? with
([^u]|u[^n]|un[^w]|unw[^a]|unwa[^n]|unwan[^t]|unwant[^e]|unwante[^d])*?
This is a solution in "pure" regex; the language you are using might allow you to use some more elegant construct.
You can't easily do that with plain regexes, but some systems such as Perl have extensions that make it easier. One way is to use a negative look-ahead assertion:
/outer-start(?:u(?!nwanted)|[^u])*?inner-start(.*?)inner-end.*?outer-end/
The key is to split up the "unwanted" into ("u" not followed by "nwanted") or (not "u"). That allows the pattern to advance, but will still find and reject all "unwanted" strings.
People may start hating your code if you do much of this though. ;)
Tola, resurrecting this question because it had a fairly simple regex solution that wasn't mentioned. This problem is a classic case of the technique explained in this question to "regex-match a pattern, excluding..."
The idea is to build an alternation (a series of |) where the left sides match what we don't want in order to get it out of the way... then the last side of the | matches what we do want, and captures it to Group 1. If Group 1 is set, you retrieve it and you have a match.
So what do we not want?
First, we want to eliminate the whole outer block if there is unwanted between outer-start and inner-start. You can do it with:
outer-start(?:(?!inner-start).)*?unwanted.*?outer-end
This will be to the left of the first |. It matches a whole outer block.
Second, we want to eliminate the whole outer block if there is unwanted between inner-end and outer-end. You can do it with:
outer-start(?:(?!outer-end).)*?inner-end(?:(?!outer-end).)*?unwanted.*?outer-end
This will be the middle |. It looks a bit complicated because we want to make sure that the "lazy" *? does not jump over the end of a block into a different block.
Third, we match and capture what we want. This is:
inner-start\s*(text-that-i-want)\s*inner-end
So the whole regex, in free-spacing mode, is:
(?xs)
outer-start(?:(?!inner-start).)*?unwanted.*?outer-end # dont want this
| # OR (also don't want that)
outer-start(?:(?!outer-end).)*?inner-end(?:(?!outer-end).)*?unwanted.*?outer-end
| # OR capture what we want
inner-start\s*(text-that-i-want)\s*inner-end
On this demo, look at the Group 1 captures on the right: It contains what we want, and only for the right block.
In Perl and PCRE (used for instance in PHP), you don't even have to look at Group 1: you can force the regex to skip the two blocks we don't want. The regex becomes:
(?xs)
(?: # non-capture group: the things we don't want
outer-start(?:(?!inner-start).)*?unwanted.*?outer-end # dont want this
| # OR (also don't want that)
outer-start(?:(?!outer-end).)*?inner-end(?:(?!outer-end).)*?unwanted.*?outer-end
)
(*SKIP)(*F) # we don't want this, so fail and skip
| # OR capture what we want
inner-start\s*\Ktext-that-i-want(?=\s*inner-end)
See demo: it directly matches what you want.
The technique is explained in full detail in the question and article below.
Reference
How to match (or replace) a pattern except in situations s1, s2, s3...
Article about matching a pattern unless...
Try replacing the last .*? with: (?!(.*unwanted text.*))
Did it work?