I'm trying to filter specific files in Java by their names using Regex.
Idea being a lot of files are called SomethingSupport.java, AnotherSupport.java, MoreThingsSupport.java, so as they all have the "Support.java" I was trying to do:
[Support.java]
But of course that's meant for characters so it will filter S,u,p,o, etc... Looking through RegExr I've tried:
(Support.java)
But it takes all "Support.java" occurrences but I'm trying to take ThingsSupport.java, SomethingSupport.java, etc. not Support.java.
Parentheses just group things. There is no difference between what regex "Support.java" and regex "(Support.java)" would match.
Note that . is regexpese for any character, so, e.g. Supportxjava, unlikely as it is that you have a file with that name in your source base, would match too. \. is regexpese for "an actual dot, please".
I think you're looking for the regex .*Support.*\.java. Which means: Absolutely anything (0 or more any character), followed by the string Support, followed by absolutely anything again, followed by .java.
That would find e.g. FooSupportBar.java, Support.java, HelloSupport.java, and SupportHello.java. It wouldn't find anything that doesn't end in .java.
Related
I am in the process of learning Regex and have been stuck on this case. I have a url that can be in two states EXAMPLE 1:
spotify.com/track/1HYcYZCOpaLjg51qUg8ilA?si=Nf5w1q9MTKu3zG_CJ83RWA
OR EXAMPLE 2:
spotify.com/track/1HYcYZCOpaLjg51qUg8ilA
I need to extract the 1HYcYZCOpaLjg51qUg8ilA ID
So far I am using this: (?<=track\/)(.*)(?=\?)? which works well for Example 2 but it includes the ?si=Nf5w1q9MTKu3zG_CJ83RWA when matching with Example 1.
BUT if I remove the ? at the end of the expression then it works for Example 1 but not Example 2! Doesn't that mean that last group (?=\?) is optional and should match?
Where am I going wrong?
Thanks!
I searched a handful of "Questions that may already have your answer" suggestions from SO, and didn't find this case, so I hope asking this is okay!
The capturing group in your regular expression is trying to match anything (.) as much as possible due to the greediness of the quantifier (*).
When you use:
(?<=track\/)(.*)(?=\?)
only 1HYcYZCOpaLjg51qUg8ilA from the first example is captured, as there is no question mark in your second example.
When using:
(?<=track\/)(.*)(?=\??)
You are effectively making the positive lookahead optional, so the capturing group will try to match as much as possible (including the question mark), so that 1HYcYZCOpaLjg51qUg8ilA?si=Nf5w1q9MTKu3zG_CJ83RWA and 1HYcYZCOpaLjg51qUg8ilA are matched, which is not the desired output.
Rather than matching anything, it is perhaps more appropriate for you to match alphanumerical characters \w only.
(?<=track\/)(\w*)(?=\??)
Alternatively, if you are expecting other characters , let's say a hyphen - or a underscore _, you may use a character class.
(?<=track\/)([a-zA-Z0-9_-]*)(?=\??)
Or you might want to capture everything except a question mark ? with a negated character class.
(?<=track\/)([^?]*)(?=\??)
As pointed out by gaganso, a look-behind is not necessary in this situation (or indeed the lookahead), however it is indeed a good idea to start playing around with them. The look-around assertions do not actually consume the characters in the string. As you can see here, the full match for both matches only consists of what is captured by the capture group. You may find more information here.
This should work:
track\/(\w+)
Please see here.
Since track is part of both the strings, and the ID is formed from alphanumeric characters, the above regex which matches the string "track/" and captures the alphanumeric characters after that string, should provide the required ID.
Regex : (\w+(?=\?))|(\w+&)
See the demo for the regex, https://regexr.com/3s4gv .
This will first try to search for word which has '?' just after it and if thats unsuccessful it will fetch the last word.
My string being of the form:
"as.asd.sd fdsfs. dfsd d.sdfsd. sdfsdf sd .COM"
I only want to match against the last segment of whitespace before the last period(.)
So far I am able to capture whitespace but not the very last occurrence using:
\s+(?=\.\w)
How can I make it less greedy?
In a general case, you can match the last occurrence of any pattern using the following scheme:
pattern(?![\s\S]*pattern)
(?s)pattern(?!.*pattern)
pattern(?!(?s:.*)pattern)
where [\s\S]* matches any zero or more chars as many as possible. (?s) and (?s:.) can be used with regex engines that support these constructs so as to use . to match any chars.
In this case, rather than \s+(?![\s\S]*\s), you may use
\s+(?!\S*\s)
See the regex demo. Note the \s and \S are inverse classes, thus, it makes no sense using [\s\S]* here, \S* is enough.
Details:
\s+ - one or more whitespace chars
(?!\S*\s) - that are not immediately followed with any 0 or more non-whitespace chars and then a whitespace.
You can try like so:
(\s+)(?=\.[^.]+$)
(?=\.[^.]+$) Positive look ahead for a dot and characters except dot at the end of line.
Demo:
https://regex101.com/r/k9VwC6/3
"as.asd.sd ffindMyLastOccurrencedsfs. dfindMyLastOccurrencefsd d.sdfsd. sdfsdf sd ..COM"
.*(?=((?<=\S)\s+)).*
replaced by `>\1<`
> <
As a more generalized example
This example defines several needles and finds the last occurrence of either one of them. In this example the needles are:
defined word findMyLastOccurrence
whitespaces (?<=\S)\s+
dots (?<=[^\.])\.+
"as.asd.sd ffindMyLastOccurrencedsfs. dfindMyLastOccurrencefsd d.sdfsd. sdfsdf sd ..COM"
.*(?=(findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+)).*
replaced by `>\1<`
>..<
Explanation:
Part 1 .*
is greedy and finds everything as long as the needles are found. Thus, it also captures all needle occurrences until the very last needle.
edit to add:
in case we are interested in the first hit, we can prevent the greediness by writing .*?
Part 2 (?=(findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+|(?<=**Not**NeedlePart)NeedlePart+))
defines the 'break' condition for the greedy 'find-all'. It consists of several parts:
(?=(needles))
positive lookahead: ensure that previously found everything is followed by the needles
findMyLastOccurrence|(?<=\S)\s+|(?<=[^\.])\.+)|(?<=**Not**NeedlePart)NeedlePart+
several needles for which we are looking. Needles are patterns themselves.
In case we look for a collection of whitespaces, dots or other needleparts, the pattern we are looking for is actually: anything which is not a needlepart, followed by one or more needleparts (thus needlepart is +). See the example for whitespaces \s negated with \S, actual dot . negated with [^.]
Part 3 .*
as we aren't interested in the remainder, we capture it and dont use it any further. We could capture it with parenthesis and use it as another group, but that's out of scope here
SIMPLE SOLUTION for a COMMON PROBLEM
All of the answers that I have read through are way off topic, overly complicated, or just simply incorrect. This question is a common problem that regex offers a simple solution for.
Breaking Down the General Problem
THE STRING
The generalized problem is such that there is a string that contains several characters.
THE SUB-STRING
Within the string is a sub-string made up of a few characters. Often times this is a file extension (i.e .c, .ts, or .json), or a top level domain (i.e. .com, .org, or .io), but it could be something as arbitrary as MC Donald's Mulan Szechuan Sauce. The point it is, it may not always be something simple.
THE BEFORE VARIANCE (Most important part)
The before variance is an arbitrary character, or characters, that always comes just before the sub-string. In this question, the before variance is an unknown amount of white-space. Its a variance because the amount of white-space that needs to be match against varies (or has a dynamic quantity).
Describing the Solution in Reference to the Problem
(Solution Part 1)
Often times when working with regular expressions its necessary to work in reverse.
We will start at the end of the problem described above, and work backwards, henceforth; we are going to start at the The Before Variance (or #3)
So, as mentioned above, The Before Variance is an unknown amount of white-space. We know that it includes white-space, but we don't know how much, so we will use the meta sequence for Any Whitespce with the one or more quantifier.
The Meta Sequence for "Any Whitespace" is \s.
The "One or More" quantifier is +
so we will start with...
NOTE: In ECMAS Regex the / characters are like quotes around a string.
const regex = /\s+/g
I also included the g to tell the engine to set the global flag to true. I won't explain flags, for the sake of brevity, but if you don't know what the global flag does, you should DuckDuckGo it.
(Solution Part 2)
Remember, we are working in reverse, so the next part to focus on is the Sub-string. In this question it is .com, but the author may want it to match against a value with variance, rather than just the static string of characters .com, therefore I will talk about that more below, but to stay focused, we will work with .com for now.
It's necessary that we use a concept here that's called ZERO LENGTH ASSERTION. We need a "zero-length assertion" because we have a sub-string that is significant, but is not what we want to match against. "Zero-length assertions" allow us to move the point in the string where the regular expression engine is looking at, without having to match any characters to get there.
The Zero-Length Assertion that we are going to use is called LOOK AHEAD, and its syntax is as follows.
Look-ahead Syntax: (?=Your-SubStr-Here)
We are going to use the look ahead to match against a variance that comes before the pattern assigned to the look-ahead, which will be our sub-string. The result looks like this:
const regex = /\s+(?=\.com)/gi
I added the insensitive flag to tell the engine to not be concerned with the case of the letter, in other words; the regular expression /\s+(?=\.cOM)/gi
is the same as /\s+(?=\.Com)/gi, and both are the same as: /\s+(?=\.com)/gi &/or /\s+(?=.COM)/gi. Everyone of the "Just Listed" regular expressions are equivalent so long as the i flag is set.
That's it! The link HERE (REGEX101) will take you to an example where you can play with the regular expression if you like.
I mentioned above working with a sub-string that has more variance than .com.
You could use (\s*)(?=\.\w{3,}) for instance.
The problem with this regex, is even though it matches .txt, .org, .json, and .unclepetespurplebeet, the regex isn't safe. When using the question's string of...
"as.asd.sd fdsfs. dfsd d.sdfsd. sdfsdf sd .COM"
as an example, you can see at the LINK HERE (Regex101) there are 3 lines in the string. Those lines represent areas where the sub-string's lookahead's assertion returned true. Each time the assertion was true, a possibility for an incorrect final match was created. Though, only one match was returned in the end, and it was the correct match, when implemented in a program, or website, that's running in production, you can pretty much guarantee that the regex is not only going to fail, but its going to fail horribly and you will come to hate it.
You can try this. It will capture the last white space segment - in the first capture group.
(\s+)\.[^\.]*$
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 source file with literally hundreds of occurrences of strings flecha.jpg and flecha1.jpg, but I need to find occurrences of any other .jpg image (i.e. casa.jpg, moto.jpg, whatever)
I have tried using a regular expression with negative lookbehind, like this:
(?<!flecha|flecha1).jpg
but it doesn't work! Notepad++ simply says that it is an invalid regular expression.
I have tried the regex elsewhere and it works, here is an example so I guess it is a problem with NPP's handling of regexes or with the syntax of lookbehinds/lookaheads.
So how could I achieve the same regex result in NPP?
If useful, I am using Notepad++ version 6.3 Unicode
As an extra, if you are so kind, what would be the syntax to achieve the same thing but with optional numbers (in this case only '1') as a suffix of my string? (even if it doesn't work in NPP, just to know)...
I tried (?<!flecha[1]?).jpg but it doesn't work. It should work the same as the other regex, see here (RegExr)
Notepad++ seems to not have implemented variable-length look-behinds (this happens with some tools). A workaround is to use more than one fixed-length look-behind:
(?<!flecha)(?<!flecha1)\.jpg
As you can check, the matches are the same. But this works with npp.
Notice I escaped the ., since you are trying to match extensions, what you want is the literal .. The way you had, it was a wildcard - could be any character.
About the extra question, unfortunately, as we can't have variable-length look-behinds, it is not possible to have optional suffixes (numbers) without having multiple look-behinds.
Solving the problem of the variable-length-negative-lookbehind limitation in Notepad++
Given here are several strategies for working around this limitation in Notepad++ (or any regex engine with the same limitation)
Defining the problem
Notepad++ does not support the use of variable-length negative lookbehind assertions, and it would be nice to have some workarounds. Let's consider the example in the original question, but assume we want to avoid occurrences of files named flecha with any number of digits after flecha, and with any characters before flecha. In that case, a regex utilizing a variable-length negative lookbehind would look like (?<!flecha[0-9]*)\.jpg.
Strings we don't want to match in this example
flecha.jpg
flecha1.jpg
flecha00501275696.jpg
aflecha.jpg
img_flecha9.jpg
abcflecha556677.jpg
The Strategies
Inserting Temporary Markers
Begin by performing a find-and-replace on the instances that you want to avoid working with - in our case, instances of flecha[0-9]*\.jpg. Insert a special marker to form a pattern that doesn't appear anywhere else. For this example, we will insert an extra . before .jpg, assuming that ..jpg doesn't appear elsewhere. So we do:
Find: (flecha[0-9]*)(\.jpg)
Replace with: $1.$2
Now you can search your document for all the other .jpg filenames with a simple regex like \w+\.jpg or (?<!\.)\.jpg and do what you want with them. When you're done, do a final find-and-replace operation where you replace all instances of ..jpg with .jpg, to remove the temporary marker.
Using a negative lookahead assertion
A negative lookahead assertion can be used to make sure that you're not matching the undesired file names:
(?<!\S)(?!\S*flecha\d*\.jpg)\S+\.jpg
Breaking it down:
(?<!\S) ensures that your match begins at the start of a file name, and not in the middle, by asserting that your match is not preceded by a non-whitespace character.
(?!\S*flecha\d*\.jpg) ensures that whatever is matched does not contain the pattern we want to avoid
\S+\.jpg is what actually gets matched -- a string of non-whitespace characters followed by .jpg.
Using multiple fixed-length negative lookbehinds
This is a quick (but not-so-elegant) solution for situations where the pattern you don't want to match has a small number of possible lengths.
For example, if we know that flecha is only followed by up to three digits, our regex could be:
(?<!flecha)(?<!flecha[0-9])(?<!flecha[0-9][0-9])(?<!flecha[0-9][0-9][0-9])\.jpg
Are you aware that you're only matching (in the sense of consuming) the extension (.jpg)? I would think you wanted to match the whole filename, no? And that's much easier to do with a lookahead:
\b(?!flecha1?\b)\w+\.jpg
The first \b anchors the match to the beginning of the name (assuming it's really a filename we're looking at). Then (?!flecha1?\b) asserts that the name is not flecha or flecha1. Once that's done, the \w+ goes ahead and consumes the name. Then \.jpg grabs the extension to finish off the match.
First timer and relatively inexperienced with RegEx and Notepad++. What I am trying to do is replace everything but the policy numbers in these two firewall session. Mind you, I have a list multiple lists 700+ lines long so I want to replace everything in one pass, leaving just the policy number for each line.
id 1978781/s23,vsys 0,flag 00200440/4000/0003,policy 4332,time 5972, dip 0 module 0
id 1997645/s23,vsys 0,flag 00200440/4000/0003,policy 30562,time 6283, dip 0 module 0
There are thousands of different policy numbers, so a simple search wont do.
I would like my lines to look like this after a replace.
4332
30562
After two hours of trying to learn RegEx for this one problem, I realized this its more involved than I expected, and I need to spend time learning this since its a very powerful tool. This could really save a lot of time, which unfortunately I don't have at the moment. I'm looking forward to learning more about RegEx and appreciate any help or direction you could give me.
Given the fact the lines always look the same you can use the following
^.+policy (\d+).+$
Replace by : $1
The dot is a wild card so , .+ means find everything before the word "policy ". Then find a group of digits (\d+ is for finding digits) and save them (thats what the parenthesis are for in many regex engines). Then find all the characters till the end of the line.
The ^ character means start of line. The $ means end of line.
You can try the following:
Find:
^.*policy ([0-9]+).*$
Replace with:
\1
Why does this work?
The dot matches any character, and the star means "zero or more of" the character preceding it. This means that .* matches everything.
What you want is to match everything before and after the policy and erase it, and keep just the policy number, so between your everything matchers you look for the string "policy xxxxx" where the xxxxx are numbers.
Each term surrounded by parenthesis in your regex is saved to be used in the replacement. I put parenthesis around the number matcher, [0-9]+ and then use what was matched in the repace part with \1. If your regex contains several parenthesized parts, you can get them with \1, \2, \3...
Regexes are really powerful, you should read a tutorial about them to learn what they can offer.