I need help with REGEX please.
The 4 and 5 lignes are fine.
But the first, 2nd and 3rd I don't want the "t" be matched.
I tried a lot of things but didn't worked unfortunatly.
So here is the link of what I did :
www.regexr.com/39rtb
/!\ Click the link to see what I see :p
\bt^(t\u0027)|(\bt)
No I don't
just t'oyota
just test
No I don t
t
/!\ I need a code for the 2sd line and an other for the last ! (last line is a t with nothing else !)
Thanks
If you just want to match a single t, assuming your flavor lets you use this syntax, you can enforce the t to be surrounded by whitespace or nothing else like so:
(?<=\s|^)t(?=\s|$)
Demo
This makes use of lookaround expressions.
The splitted version (assuming (?m) - if it doesn't work then try to prepend (?m) to each pattern):
(?!^t$)(?<=\s|^)t(?=\s|$)
And:
^t$
But it doesn't make much sense to separate these cases using regexes. You'd be better off doing the check in your language.
Related
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.
Here's the input string:
loadMedia('mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', 'http://www.something.com/videos/JohnsAwesomeCaption.xml', '/videos/video-splash-image.gif)
With this RegExp: \'.+.xml\'
... we get this:
'mediacontainer1', 'http://www.something.com/videos/JohnsAwesomeVideo.flv', 'http://www.something.com/videos/JohnsAwesomeCaption.xml'
... but I want to extract only this:
http://www.something.com/videos/JohnsAwesomeCaption.xml
Any suggestions? I'm sure this problem has been asked before, but it's difficult to search for. I'll be happy to Accept a solution.
Thanks!
If you want to get everything within quotes that starts with http:
(?<=')http:[^']+(?=')
If you only want those ending with .xml
(?<=')http:[^']+\.xml(?=')
It doesn't select the quotation marks (as you asked)
It's fast!
Fair warning: it only works if the regex engine you're using can handle lookbehind
Knowing the language would be helpful. Basically, you are having a problem because the + quantifier is greedy, meaning it will match the largest part of the string that it can. you need to use a non-greedy quantifier, which will match as little as possible.
We will need to know the language you're in to know what the syntax for the non-greedy quantifier should be.
Here is a perl recipe. Just as a sidenote, instead of .+, you probably want to match [^.]+.xml.
\'.+?.xml\'
should work if your language supports perl-like regexes.
This should work (tested in javascript, but pretty sure it would work in most cases)
'[^']+?\.xml'
it looks for these rules
starts with '
is followed by anything but '
ends in .xml'
you can demo it at http://RegExr.com?2tp6q
in .net this regex works for me:
\'[\w:/.]+\.xml\'
breaking it down:
a ' character
followed by a word character or ':' or '/' or '.' any number of times (which matches the url bit)
followed by '.xml' (which differentiates the sought string from the other urls which it will match without this)
followed by another ' character
I tested it here
Edit
I missed that you don't want the quotes in the result, in which case as has been pointed out you need to use look behind and look ahead to include the quotes in the search, but not in the answer. again in .net:
(?<=')[\w:/.]+\.xml(?=')
but I think the best solution is a combination of those offered already:
(?<=')[^']+\.xml(?=')
which seems the simplest to read, at least to me.
I need a regexp to find strings that start with a specific word then comes colon and whitespace for example
"ErrorID: blabla"
Please help. :(
This should work fine:
^(\w+): (.+)$
First match group will give the first word (e.g. ErrorID), second the rest (e.g. blabla).
Exact implementation would depend on the programming language you use.
This should do what you want:
^ErrorID: .*$
I have a feed in Yahoo Pipes and want to match everything after a question mark.
So far I've figured out how to match the question mark using..
\?
Now just to match everything that is after/follows the question mark.
\?(.*)
You want the content of the first capture group.
Try this:
\?(.*)
The parentheses are a capturing group that you can use to extract the part of the string you are interested in.
If the string can contain new lines you may have to use the "dot all" modifier to allow the dot to match the new line character. Whether or not you have to do this, and how to do this, depends on the language you are using. It appears that you forgot to mention the programming language you are using in your question.
Another alternative that you can use if your language supports fixed width lookbehind assertions is:
(?<=\?).*
With the positive lookbehind technique:
(?<=\?).*
(We're searching for a text preceded by a question mark here)
Input: derpderp?mystring blahbeh
Output: mystring blahbeh
Example
Basically the ?<= is a group construct, that requires the escaped question-mark, before any match can be made.
They perform really well, but not all implementations support them.
\?(.*)$
If you want to match all chars after "?" you can use a group to match any char, and you'd better use the "$" sign to indicate the end of line.
?(.*\n)+
With this you can get everything Even a new line
Check out this site: http://rubular.com/ Basically the site allows you to enter some example text (what you would be looking for on your site) and then as you build the regular expression it will highlight what is being matched in real time.
str.replace(/^.+?\"|^.|\".+/, '');
This is sometimes bad to use when you wanna select what else to remove between "" and you cannot use it more than twice in one string. All it does is select whatever is not in between "" and replace it with nothing.
Even for me it is a bit confusing, but ill try to explain it. ^.+? (not anything OPTIONAL) till first " then | Or/stop (still researching what it really means) till/at ^. has selected nothing until before the 2nd " using (| stop/at). And select all that comes after with .+.
I've got a working regular expression, but I'd like to make it a tad more readable, and I'm far from a regex guru, so I was humbly hoping for some tips.
This is designed to scrape the output of several different compilers, linkers, and other build tools, and is used to build a nice little summery report. It does it's job great, but I'm left feeling like I wrote it in a clunky fashion, and I'd sooner learn than keep it the wrong way.
(.*?)\s?:?\s?(informational|warning|error|fatal error)?\s([A-Z]+[0-9][0-9][0-9][0-9]):\s(.*)$
Which, broken down simply, is as follows:
(.*?) # non-greedily match up until...
\s?:?\s? # we come across a possible " : "
(informational|warning|error|fatal error)? # possibly followed by one of these
\s([A-Z]+[0-9][0-9][0-9][0-9]):\s # but 100% followed by this alphanum
(.*)$ # and then capture the rest
I'm mostly interested in making the 2nd and 4th entry above more... beautiful. For some reason, the regex tester I was using (The Regulator) didn't match plain spaces, so I had to use the \s... but it is not meant to match any other whitespace.
Any schooling will be greatly appreciated.
The easiest way to make a long regex more readable is to use the "free-spacing" (or \x) modifier, which would let you write your regex just like you did in the second block of code -- it makes whitespace ignored. This isn't supported by all engines, however (according to the page linked above, .NET, Java, Perl, PCRE, Python, Ruby and XPath support it).
Note also that in free-spacing mode, you can use [ ] instead of \s if you want to only match a space character (unless you're using Java, in which case you have to use \ , which is an escaped space).
There's not really anything you can do for the second line, if you want each element to be optional independently of the other elements, but the fourth can be shortened:
\s([A-Z]+\d{4}):\s
\d is a shorthand class equivalent to [0-9], and {4} specifies that it should appear exactly four times.
The third line can be slightly shortened as well ((?:…) specifies a non-capturing group):
(informational|warning|(?:fatal )? error)?
From an efficiency standpoint, unless you actually need to capture subpatterns each time you use brackets, you can remove all of them, except for on the third line, where the group is needed for the alternation) -- but that one can be made non-capturing. Putting this all together you'd get:
.*?
\s?:?\s?
(?:informational|warning|(?:fatal )?error)?
\s[A-Z]+\d{4}:\s
.*$
Line 2
I think your regular expression doesn't match with the comment. You probably want this instead:
(\s:\s)?
To make it non-capturing:
(?:\s:\s)?
You should be able to use a literal space instead of \s. This must be a restriction in the tool you are using.
Line 4
[0-9][0-9][0-9][0-9] can be replaced with [0-9]{4}.
In some languages [0-9] is equivalent to \d.
Perhaps you can build the RE from sub-expressions, so that your end RE would look something like this:
/$preamble$possible_colon$keyword$alphanum$trailer/