I'm struggling to get the right regex to match the following;
content/foo/B6128/8918/foo+bar+foo
OR
content/foo/B6128/8918/foo+bar+foo/randomstringnumsletters
I'm sure this isn't that complicated and I'm nearly there, just can't get it perfected. Here's what I've tried;
content\/(\w+)\/(\w+)\/(\d+)\/([^\/]+[\w]+)\/?(\w*)$
using this online tester: http://regex101.com/r/sB8rR5/2
It still matches a 5th item with this string content/foo/B6128/8918/foo+bar+foo;
And while technically this pattern does match either OR url structures. I don't want it to match the 5th item when there's no randomstringnumsletters present.
After playing around with it for a bit, I do realise some elements are redundant with what I've tried, but I'm not getting anywhere with it...
Just turn the last capturing group into an optional one, and change \w* to \w+ in the last capturing group inorder to prevent null character to be captured by the 5th group.
content\/(\w+)\/(\w+)\/(\d+)\/([^\/]+[\w]+)\/?(\w+)?$
DEMO
Looks like your REAL pattern should be:
content\/((?:\w+\/?)+)
DEMO
or am I wrong? This will match the whole string (after content/) and return it all / delimited. You can parse each variable from there.
You can take each part as an array, then take the part that you need...
DEMO
Related
I have the following string;
Start: 738392E, 6726376N
I extracted 738392 ok using (?<=.art\:\s)([0-9A-Z]*). This gave me a one group match allowing me to extract it as a column value
.
I want to extract 6726376 the same way. Have only one group appear because I am parsing that to a column value.
Not sure why is (?=(art\:\s\s*))(?=[,])*(.*[0-9]*) giving me the entire line after S.
Helping me get it right with an explanation will go along way.
Because you used positive lookaheads. Those just make some assertions, but don't "move the head along".
(?=(art\:\s\s*)) makes sure you're before "art: ...". The next thing is another positive lookahead that you quantify with a star to make it optional. Finally you match anything, so you get the rest of the line in your capture group.
I propose a simpler regex:
(?<=(art\:\s))(\d+)\D+(\d+)
Demo
First we make a positive lookback that makes sure we're after "art: ", then we match two numbers, seperated by non-numbers.
There is no need for you to make it this complicated. Just use something like
Start: (\d+)E, (\d+)N
or
\b\d+(?=[EN]\b)
if you need to match each bit separately.
Your expression (?=(art\:\s\s*))(?=[,])*(.*[0-9]*) has several problems besides the ones already mentioned: 1) your first and second lookahead match at different locations, 2) your second lookahead is quantified, which, in 25 years, I have never seen someone do, so kudos. ;), 3) your capturing group matches about anything, including any line or the empty string.
You match the whole part after it because you use .* which will match until the end of the line.
Note that this part [0-9]* at the end of the pattern does not match because it is optional and the preceding .* already matches until the end of the string.
You could get the match without any lookarounds:
(art:\s)(\d+)[^,]+,\s(\d+)
Regex demo
If you want the matches only, you could make use of the PyPi regex module
(?<=\bStart:(?:\s+\d+[A-Z],)* )\d+(?=[A-Z])
Regex demo (For example only, using a different engine) | Python demo
I tried to get the sub-strings from a string
such like:
test strings:
cat_zoo_New_York_US
dog_zoo_South_Carolina
dolphin_zoo_Montreal_Canada
pokemon_home_d_K2-155
returned sub strings:
cat, New_York
dog, South_Carolina
dolphin, Montreal
pokemon, d
the Regex pattern I have tried is
([\w]+)(?:(_zoo_|_home_))(((?!(_US|_Canada|_K2-155))\w)+)
which I don't think is very concise and it returns other sub-strings besides what I need. Do you have any other suggestions?
Thanks!
Some updates
after #The fourth bird's answer #03/15/2018.
First of all, I like the idea of utilizing both ([^_]+) and the (?:) for different part of the sample strings.But let me extend a little more of the sample strings.
cat_zoo_New_York_US
dog_zoo_South_Carolina
yellow_dolphin_zoo_Montreal_Canada
pokemon_home_d_K2-155
pokemon_home_zoo_d_K2-155
I actually want to use the anchor strings such as 'zoo','home' or 'home_zoo' to separate the characters before and after, together with matching(and discarding) the last part of the country(or whatever specified place ID), which makes this question a bit less general(I like the idea of using _,but let me make it more tricky to learn better).
two questions here
what is the function of (?=) and .* here in
(?=(?:_US|_Canada|_K2-155|$)).*$? It seems if I use
(?:_US|_Canada|_K2-155|$), it is still ok...
since I extended a little bit on the anchor string to let it support
_, I used:
(.*?)(?:_*)(?:home_zoo|zoo|home)(?:_*)(.*?)(?:_*)(?:US|Canada|K2-155|$)
It seems ok, but if I use:
(.*?)(?:_*)(?:home|zoo|home_zoo)(?:_*)(.*?)(?:_*)(?:US|Canada|K2-155|$)
It will firstly match home for the last sample string. Is there a
greedy algorithm to catch this without specify the order of the pattern
string?
Well again, I don't like to make a long list of anchor strings, but I don't have other ideas make it more general without doing so.
Thanks again!
You could try it like this:
^([^_]+)_[^_]+_(.*?)(?=(?:_US|_Canada|_K2-155|$)).*$
This will capture 2 groups. You could for example use this in a replacement with group1, group2.
First capture the first part ending on an underscore in group 1 like cat_. Then match the second part ending with an underscore like zoo_ or home_.
From that point capture in a group until you encounter one of your values using a lookahead (?= or the end of the string.
That would match:
^ Begin of the string
([^_]+) Match in a capturing group not an _ one or more times (group 1)
_[^_]+_ match _ then not an _ one or more times followed by _
(.*?) Capture in a group any character zero or more times greedy (group 2)
(?= Positive lookahead that asserts what is on the right side is
(?: Non capturing group
_US|_Canada|_K2-155|$ your values or end of the string
) Close group
) Close group
.*$ Match any character zero or more times till the end of the string
Edit: After the updated question, perhaps this will suit your requirements:
^(.*?)_(?:home_zoo|zoo|home)(.*?)(?=(?:_US|_Canada|_K2-155|$))
This will match any charcter zero or more times non greedy (.*?), then an underscore and a non capturing group (?:home|zoo|home_zoo) to separate the characters before and after.
Well, I tried a more straightforward approach. If your data is more complex than the sample that you gave above, this may fail. Otherwise, for the above text, it works fine.
Here is the expression that I used:
^([^_]*)_[^_]*_(.*)_.*$
1 23 45 67
Basically what I did was:
Group the first char stream, which does not contain _, starting at the beginning of the line.
Then there is an _ following the above group
Follows an arbitrary length string, which does not have _'s in it
Then comes an _
Group the next arbitrary length string
Comes and _ afterwards
Rest of the string
replace it with \1, \2 (first group, second group).
You can find a fiddle here
If you are using vim, you can also achieve the same thing in vim with the following command:
:%s/^[^_]*_\([^_]*\)_\(.*\)_.*$/\1, \2/g
UPDATE
^([^_]*)_[^_]*_(((?:South_)|(?:New_))*[^_]*)((?:_US)|(?:_Canada)|(?:_K2-155))*$
You can find the new fiddle (here)[https://regex101.com/r/qQ2dE4/273]
What is the difference between this one and the previous one?
Now, I cheat a little, as such that I look for adjectives, which modify the state name, like South_ or New_. You can add more here, like East_, West_, Old_ or whatever if there is a case in your date.
There are cases where country is skipped in data. Plus looks like that last token on the very last line does not follow up a pattern. So, I explicitly listed those options in the expression, like US, Canada etc. You may need to add more exceptional cases in here as well.
I'm trying to extract First and Last name from a string that looks like this:
CN=First\, Last,OU=Standard users,OU=Users,OU=Place,OU=DOMAIN,DC=dfe,DC=stuff,DC=asdf
([^CN=,\\])([a-zA-Z]*)?(?!OU)
My attempt is above, but it obviously doesn't work.
Can anyone point me in the correct direction?
Thanks
You can use this expression:
^CN=(.*?)\\, (.*?),
Live demo. It uses two capturing groups for first and last names with the other static text around them.
CN=(.*)\\, ([^,]*),
This should get your First until the \, and last until the next comma.
You could come up with:
CN=(?P[^,]+),\s*
(?P[^,]+)
See a demo on regex101.com.
In the regex you have tried, the [^CN=,\\] this part will do exactly the reverse of what you need, like will match the characters except CN=,\\ these.
You can use:
^CN=[^,]+,\s+[^,]+
^CN=(.+)\\, (.+?),
This will get you First in the first capture group and Last in the second capture group, assuming everything matches the pattern of line starting with CN=, then first name, then \,, then last name, followed by ,. First and last are will not be limited to only letters though.
When you put the ^ inside of brackets like you attempted, [^CN=,\\], then you are telling Regex to look for any characters except C, N, =, ,, and \.
I have a string like aaa**b***c****ddd, and I want to get a sequence of matched text of pattern [^*]\*+[^*], which should I thank be [a**b, b***c, c***d]. However, when I test this in text editor like vim or emacs, the second (b***c) is not matched.
aaa**b***c***ddd
|--| |---|
first third
|---|
second, which I think should be matched but not
How should I modify the regular expression to match the second?
Yes you can, the trick consists to put all in a capturing group inside a lookahead to allow overlapping results:
(?=([^*]\*+[^*]))
But you can't use this do to replacements since this pattern matches nothing. (or perhaps if you can get the capture group length and the current offset)
EDIT:
it seems to be possible to obtain the capture group length with vim with strlen(submatch(1))
#CommuSoft is correct. One way to approach this problem would be to match the whole string against this regex and then the second time around, you match this regex against the substring that starts at (index_of_first_previous_match + 1) until the end of the string. Hope that is clear.
So if the index of your first match above (a**b) was 2. Then the new substring that you match against the regex the second time should start from index 3 till the end of the string. This will give you the two results.
However, Casimir's answer is much simpler.
I am trying out the quiz from Regex 101
In Task 6, the question is
Oh no! It seems my friends spilled beer all over my keyboard last night and my keys are super sticky now. Some of the time when I press a key, I get two duplicates. Can you pppllleaaaseee help me fix this? Content in bold should be removed.
I have tried this regex
([a-z])(\1{2})
But couldn't get the solution.
The solution for the riddle on that website is:
/(.)\1{2}/g
Since any key on the keyboard can get stuck, so we need to use ..
\1 in the regex means match whatever the 1st capturing group (.) matches.
Replacement is $1 or \1.
The rest of your regex is correct, just that there are unnecessary capturing groups.
Your regex is correct if you want to match exactly three characters. If you want to match at least three, that is
([a-z])(\1{2,})
or
([a-z])(\1\1+)
Since you don't need to capture anything but the first occurence, these are slightly better:
([a-z])\1{2} # your original regex (exactly three occurences)
([a-z])\1{2,}
([a-z])\1\1+
Now, the replacement should be exactly one occurence of the character, and nothing more:
\1
Replace:
(.)\1+
with:
\1
This of course requires that your regex engine suports backreferences... Also, in the replacement part, and according to regex engines, \1 may have to be written as $1.
I'd do it with (\w)(\1+)? but can't find out how to "remove" within the given site...
Best way would be to replace the results of the secound match with empty strings