I have such an http-url detector regex:
(?:http|https)(?::\/{2}[\w]+)(?:[\/|\.]?)(?:[^\s<"]*)
It works pretty well for the following url representation:
http://www.acer.com/clearfi/download/
What kind of modification I can do to extract
http://schemas.microsoft.com/office/word/2003/wordml2450
from
Huanghhttp://schemas.microsoft.com/office/word/2003/wordml2450...)()()()()()
?
You can modify it to capture:
group of http stuff
followed by (group of) subdomain stuff
followed by as many as possible groups of:
one point or slash
followed by a group of characters (non-point, non-space, non-", non-<)
(?:http|https)(?:\/{2}[\w]+)([\/|\.][^\s<"\.]+)*
I made capturing groups to visualize the results
I've changed your expression here and there: (.*)(https?:\/{2}[\w]+[\/|\.]?[^\s<"]*)(\.{3}.*) and get only second capturing group from it. See example here: https://regex101.com/r/0viPC5/2
This expression probably can be simplified further but I don't know your exact input and search criteria so let's stick with what you already wrote.
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.
I am beginner in Regular Expression. I am working on a problem where need to match the number that resides in square brackets as a part of parsing for my Sumo Logic query. I need to match just the number '45678'.
2017-08-24 08:55:36,659 INFO [CompanyServiceImpl:XXX] Getting isEducation for company id [45678]
I tried it with above example but it did not work. I came up with [^\[]\d+[^\]] but this solution matches the other numbers in the string such as timestamp.
Above is just an example. There are different id's in square brackets in multiple logs.I need to match all. Not specific to the 45678. I would appreciate if anybody help me with that.
If using capturing group
Well, use \[(\d+)\] the number contained is now in the first capturing group
else
(?<=\[)\d+(?=\])
See demo on regex101
(?<=\[)\d+(?=\])
This captures 1 or more digits between brackets.
Demo
You can use positive lookaheads for that.
\d+(?=\])
You can also test out your regex here: http://regexr.com/. Very useful resource IMO.
So basically I want to reformat a 10 digit number like so:
1234567890 --> (123) 456-7890
A long way to do this would be to have each number be its own capture group and then back-reference each one individually:
'([0-9])([0-9])...([0-9])' --> (\1\2\3) \4\5\6-\7\8\9\10
This seems unnecessary and verbose, but when I try the following
'([0-9]){10}'
There appears to be only one back-reference and its of the last digit in the number.
Is there is a more elegant way to reference each character as its own capture group?
Thanks!
The following pattern will do the job: ^(\d{3})(\d{3})(\d{4})$
^(\d{3}): beginning of the string, then exactly 3 digits
(\d{3}): exactly 3 digits
(\d{4})$: exactly 4 digits, then end of the string.
Then replace by: (\1) \2-\3
Although the other answer with its example regex patterns hopefully shed light on the correct application of capture groups, it does not directly answer the question. If you fail to understand how regular expressions work (capture groups in particular), you may find yourself wanting to do the same thing with a different pattern in the future.
Is there is a more elegant way to reference each character as its own
capture group?
The initial answer is "No", there is no way to reference an individual capture of a single capture group using traditional replacement syntax - regardless of whether it is a single digit or any other capture group. Consider that you indicate a precise number of matches with {10} and it seems perfectly reasonable to be able to access each capture. But what if you had indicated a variable number of matches with + or {,3}? There would be no well-defined way of knowing how many possible captures occurred. If the same regex pattern had had more capture groups following the "repeated" capture group, there would be no way of correctly referencing the later groups. Example: Given the pattern ([a-z])+(\d){3}, the first capture group could match 4 letters one time, then the next time match 11 letters. If you wanted to refer to the captured digits, how would you do that? You could not, since \1, \2, \3, ... would all be reserved for possible capture instances of the first group.
But the inability of basic regular expressions syntax to do what you want does not remove the validity of your question, nor does it necessarily place the solution outside the realm of many regular expression implementations. Various regex implementations (i.e. language syntax and regex libraries) resolve this limitation by facilitating regex matching with various objects for accessing repeated captures. (c# and .Net regex library is one example, like match.Groups[1].Captures[3]) So even though you can't use basic replacement patterns to get want you want, the answer is often "Yes", depending on the specific implementation.
I am trying to extract express route named parameters with regex.
So, for example:
www.test.com/something/:var/else/:var2
I am trying with this regex:
.*\/?([:]+\w+)+
but I am getting only last matched group.
Does anyone knows how to match both :var and :var2.
The first problem is that .* is greedy, and will therefore bypass all matches until the final one is found. This means that the first :var is bypassed.
However, as you are searching for a variable number of capture groups (with thanks to #MichaelTang), I recommend using two regexes in sequence. First, use
^(?:.*?\/?\:\w+)+$
to detect which lines contain colon-elements...
Debuggex Demo
...and then search that line repeatedly for, simply
\/:(\w+)
This places the text post-colon into capture group one.
Debuggex Demo
Here is how you can match both of them:
www.test.com/something/:var/else/:var2'.match(/\:(\w+)/g)
[":var", ":var2"]
I have such txt file:
ххх.prontube.ru
salo.ru
bbb.antichat.ru
yyy.ru
xx.bb.prontube.ru
zzz.com
srfsf.jwbefw.com.ua
Trying to delete all subdomains with such regex:
Find: .+\.((.*?)\.(ru|ua|com\.ua|com|net|info))$
Replace with: \1
Receive:
prontube.ru
salo.ru
antichat.ru
yyy.ru
prontube.ru
zzz.com
com.ua
Why last line becomes com.ua instead of jwbefw.com.ua ?
This works without look around:
Find: [a-zA-Z0-9-.]+\.([a-zA-Z0-9-]+)\.([a-zA-Z0-9-]+)$
Replace: \1\.\2
It finds something with at least 2 periods and only letters, numbers, and dashes following the last two periods; then it replaces it with the last 2 parts. More intuitive, in my opinion.
There's something funny going on with that leading xxx. It doesn't appear to be plain ASCII. For the sake of this question, I'm going to assume that's just something funny with this site and not representative of your real data.
Incorrect
Interestingly, I previously had an incorrect answer here that accumulated a lot of upvotes. So I think I should preserve it:
Find: [a-zA-Z0-9-]+\.([a-zA-Z0-9-]+)\.(.+)$
Replace: \1\.\2
It just finds a host name with at least 2 periods in it, then replaces it with everything after the first dot.
The .+ part is matching as much as possible. Try using .+? instead, and it will capture the least possible, allowing the com.ua option to match.
.+?\.([\w-]*?\.(?:ru|ua|com\.ua|com|net|info))$
This answer still uses the specific domain names that the original question was looking at. As some TLD (top level domains) have a period in them, and you could theoretically have a list including multiple subdomains, whitelisting the TLD in the regex is a good idea if it works with your data set. Both current answers (from 2013) will not handle the difference between "xx.bb.prontube.ru" and "srfsf.jwbefw.com.ua" correctly.
Here is a quick explanation of why this psnig's original regex isn't working as intended:
The + is greedy.
.+ will zip all the way to the right at the end of the line capturing everything,
then work its way backwards (to the left) looking for a match from here:
(ru|ua|com\.ua|com|net|info)
With srfsf.jwbefw.com.ua the regex engine will first fail to match a,
then it will move the token one place to the left to look at "ua"
At that point, ua from the regex (the second option) is a match.
The engine will not keep looking to find "com.ua" because ".ua" met that requirement.
Niet the Dark Absol's answer tells the regex to be "lazy"
.+? will match any character (at least one) and then try to find the next part of the regex. If that fails, it will advance the token, .+ matching one more character and then evaluating the rest of the regex again.
The .+? will eventually consume: srfsf.jwbefw before matching the period, and then matching com.ua.
But the implimentation of ? also creates issues.
Adding in the question mark makes that first .+ lazy, but then causes group1 to match bb.prontube.ru instead of prontube.ru
This is because that first period after the bb will match, then inside group 1 (.*?) will match bb.prontube. before \.(ru|ua|com\.ua|com|net|info))$ matches .ru
To avoid this, change that third group from (.*?) to ([\w-]*?) so it won't capture . only letters and numbers, or a dash.
resulting regex:
.+?\.(([\w-])*?\.(ru|ua|com\.ua|com|net|info))$
Note that you don't need to capture any groups other than the first. Adding ?: makes the TLD options non-capturing.
last change:
.+?\.([\w-]*?\.(?:ru|ua|com\.ua|com|net|info))$
Search what: .+?\.(\w+\.(?:ru|com|com\.au))
Replace with: $1
Look in the picture above, what regex capture referring
It's color the way you will not need a regex explaination anymore ....