I've tried a lot to split this string into something i can work with, however my experience isn't enough to reach the goal. Tried first 3 pages on google, which helped but still didn't give me an idea how to properly do this:
I have a string which looks like this:
My Dogs,213,220#Gallery,635,210#Screenshot,219,530#Good Morning,412,408#
The result should be:
MyDogs
213,229
Gallery
635,210
Screenshot
219,530
Good Morning
412,408
Anyone have an idea how to use regex to split the string like shown above?
Given the shared patterns, it seems you're looking for a regex like the following:
[A-Za-z ]+|\d+,\d+
It matches two patterns:
[A-Za-z ]+: any combination of letters and spaces
\d+,\d+: any combination of digits + a comma + any combination of digits
Check the demo here.
If you want a more strict regex, you can include the previous pattern between a lookbehind and a lookahead, so that you're sure that every match is preceeded by either a comma, a # or a start/end of string character.
(?<=^|,|#)([A-Za-z ]+|\d+,\d+)(?=,|#|$)
Check the demo here.
Related
I'm trying to create a regex that matches url strings within normal text. I have this:
http[s]?://[^\s]+
This seems to work well with the exception that if the url is at the end of a sentence it will grab the period as well. For example for this string:
I am typing some text with the url http://something.com/something-?args=someargs. This is another sentence.
it matches:
http://something.com/some-thing?args=someargs.
I would like it to match:
http://something.com/some-thing?args=someargs
Obviously I can't exclude periods because they are in the url previously but I can't figure out how to tell it to exclude the last period if there is one. I could potentially use a negative lookahead for end of line or whitespace, but if it's in the middle of the line (without a period after it) that would leave off the last character of the url.
Most of the ones I have seen online have the same issue that they match the ending dot so maybe it's not possible? I know basic regex but certainly not a genius with it so if someone has a solution I would be very grateful :).
Also, I can do some post-process in this case to remove the dot if I need to, just seems like there should be a Regex solution...
Try this one
http[s]?://[^\s]+[^. ]
I'm having an issue with Regex.
I'm trying to match T0000001 (2, 3 and so on).
However, some of the lines it searches has what I can describe as positioners. These are shown as a question mark, followed by 2 digits, such as ?21.
These positioners describe a new position if the document were to be printed off the website.
Example:
T123?214567
T?211234567
I need to disregard ?21 and match T1234567.
From what I can see, this is not possible.
I have looked everywhere and tried numerous attempts.
All we have to work off is the linked image. The creators cant even confirm the flavour of Regex it is - they believe its Python but I'm unsure.
Regex Image
Update
Unfortunately none of the codes below have worked so far. I thought to test each code in live (Rather than via regex thinking may work different but unfortunately still didn't work)
There is no replace feature, and as mentioned before I'm not sure if it is Python. Appreciate your help.
Do two regex operations
First do the regex replace to replace the positioners with an empty string.
(\?[0-9]{2})
Then do the regex match
T[0-9]{7}
If there's only one occurrence of the 'positioners' in each match, something like this should work: (T.*?)\?\d{2}(.*)
This can be tested here: https://regex101.com/r/XhQXkh/2
Basically, match two capture groups before and after the '?21' sequence. You'll need to concatenate these two matches.
At first, match the ?21 and repace it with a distinctive character, #, etc
\?21
Demo
and you may try this regex to find what you want
(T(?:\d{7}|[\#\d]{8}))\s
Demo,,, in which target string is captured to group 1 (or \1).
Finally, replace # with ?21 or something you like.
Python script may be like this
ss="""T123?214567
T?211234567
T1234567
T1234434?21
T5435433"""
rexpre= re.compile(r'\?21')
regx= re.compile(r'(T(?:\d{7}|[\#\d]{8}))\s')
for m in regx.findall(rexpre.sub('#',ss)):
print(m)
print()
for m in regx.findall(rexpre.sub('#',ss)):
print(re.sub('#',r'?21', m))
Output is
T123#4567
T#1234567
T1234567
T1234434#
T123?214567
T?211234567
T1234567
T1234434?21
If using a replace functionality is an option for you then this might be an approach to match T0000001 or T123?214567:
Capture a T followed by zero or more digits before the optional part in group 1 (T\d*)
Make the question mark followed by 2 digits part optional (?:\?\d{2})?
Capture one or more digits after in group 2 (\d+).
Then in the replacement you could use group1group2 \1\2.
Using word boundaries \b (Or use assertions for the start and the end of the line ^ $) this could look like:
\b(T\d*)(?:\?\d{2})?(\d+)\b
Example Python
Is the below what you want?
Use RegExReplace with multiline tag (m) and enable replace all occurrences!
Pattern = (T\d*)\?\d{2}(\d*)
replace = $1$2
Usage Example:
I have to match an exact string at the end of a url, but not match all other urls that have more characters after that string
I can better explain with example.
I need to match the url having the string 'white' at its end: http//mysite.com/white
But I also need to not match urls having one or more characters postponed to it, like http//mysite.com/white__blue or http//mysite.com/white/yellow or http//mysite.com/white/
How to do that?
Thanks
Regex to match any url*
^(https?:\/\/)?([\da-z\.-]+\.[a-z\.]{2,6}|[\d\.]+)([\/:?=&#]{1}[\da-z\.-]+)*[\/\?]?$
Regex to match a url containing white in the end
^(https?:\/\/)?([\da-z\.-]+\.[a-z\.]{2,6}|[\d\.]+)([\/:?=&#]{1}[\da-z\.-]+)*[\/\?]?white$
You can check the regex here
From regexr.com
It does not match urls(which are not valid anyway) like
httpabrakadabra.co//
http:google.com
http://no-tld-here-folks.a
http://potato.54.211.192.240/
Based on your limited sample inputs, I'd say you could get away with this very minimal pattern:
^http[^\s]+white$
However, depending on what you are truly trying to achieve, what language/function you are implementing this pattern with, and what the full input string looks like, this pattern may need to be refined.
It would be best if you would improve your question to include all of the above relevant information.
I am trying to write a pattern match to only match when a string is not followed by both following patterns. Right now I have a pattern that I've tried to manipulate but I can't seem to get it to match correctly.
Current pattern:
/(address|alias|parents|members|notes|host|name)(?!(\t{5}|\S+))/
I am trying to match when a string is not spaced correctly but not if it is part of a larger word.
For example I want it to match,
host \t{4} something
but not,
hostgroup \t{5} something
In the above example it will match hostgroup and end up separating it into 2 separate words "host" and "group"
Match:
notes \t{4} something
but not,
notes_url \t{5} something
Using my pattern it ends up turning into:
notes \t{5} _url
Hopefully that makes a bit more sense.
I'm not at all clear what you want, but word boundaries will probably do what you ask.
Does this work for you?
/\b(address|alias|parents|members|notes|host|name)\b(?!\t{5})/
Update
Having understood your problem better, does this do what you want?
/\b(address|alias|parents|members|notes|host|name)\b(?!\t{5}(?!\t))/
I have some files like that
15.58.55.ser 16.22.20.ser 16.36.23.ser 16.40.13.ser 16.59.41.ser 17.05.08.ser 17.14.40.ser 18.14.40.ser 18.20.43.ser
I want to replace these filenames with the following format
image_1.ser image_2.ser ....
I don't know how to achieve it.
please give me some advice.
The regex is quite simple:
(?:\d{2}\.){3}ser
It matches two digits \d{2} and a dot \. three times {3}, ending in ser.
You can see from RegExr that is matches all of your test cases.
However, in order to know how to do the replacement, you'd have to specify a language that you're working with.
Try this(If you need Java code)
String regex = "\\.ser";
fileName = "15.58.55.ser";
System.out.println(filename.replaceAll(fileName.split(regex)[0], "image_1"));
This is just for only one entry. If you want to replace multiple files, do it in For loop or whatever