Reputation: 109
I have a URL that contains multiple sequences of numbers I want to capture them all in groups suppose I have the following
https://www.example.com//first/part/54323?key=value
or https://www.example.com/first/12345/second/part/part2/5432?key=value
I tried to use something like that but it only matches one sequence of numbers
(.*\/)([0-9]{4,})(\/.*|$|)
I want to have multiple groups represent different sections if numbers sequence is included
Upvotes: -2
Views: 779
Reputation: 18641
It sounds very straight-forward:
https?:\/\/(?:www\.)?(.*?)\/(\d+)\/(.*?)\/(\d+)(?:\?(.*))?
See regex proof.
EXPLANATION
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http 'http'
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s? 's' (optional (matching the most amount
possible))
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: ':'
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\/ '/'
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\/ '/'
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(?: group, but do not capture (optional
(matching the most amount possible)):
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www 'www'
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\. '.'
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)? end of grouping
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( group and capture to \1:
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.*? any character except \n (0 or more times
(matching the least amount possible))
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) end of \1
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\/ '/'
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( group and capture to \2:
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\d+ digits (0-9) (1 or more times (matching
the most amount possible))
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) end of \2
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\/ '/'
--------------------------------------------------------------------------------
( group and capture to \3:
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.*? any character except \n (0 or more times
(matching the least amount possible))
--------------------------------------------------------------------------------
) end of \3
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\/ '/'
--------------------------------------------------------------------------------
( group and capture to \4:
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\d+ digits (0-9) (1 or more times (matching
the most amount possible))
--------------------------------------------------------------------------------
) end of \4
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(?: group, but do not capture (optional
(matching the most amount possible)):
--------------------------------------------------------------------------------
\? '?'
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( group and capture to \5:
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.* any character except \n (0 or more
times (matching the most amount
possible))
--------------------------------------------------------------------------------
) end of \5
--------------------------------------------------------------------------------
)? end of grouping
Upvotes: 0
Reputation: 2625
The initial .* is Greedy, meaning it tries to match as much as possible. It matched everything up to the last slash "https://www.example.com/first/12345/second/part
". You can modify this behavior by replacing the initial .* with .*?, but that will stop after the first slash, which is also not what you want "https:/
" because there are no digits after those slashes.
But really we need to back up and ask some questions about your pattern. Apparently, you have a preamble you are not interested in, an indefinite number of sequences of 'character string, followed by slash, followed by number string' and then there is the "everything after there are no more slash digit patterns".
The key question is whether the number of char/char/digits combos are indefinite or limited to a definite number like the two pairs in your example. To get the regex parser to return an unbounded number of string-number pairs, you are going to want to turn on the /g (Global) switch so regex will return all matches. That is a problem with the part of your URL at the beginning and end which does not fit your pattern.
I recommend first using a regular expression to divide your URL into three parts, preamble, path, remaining data
. Then you can pass the path string to a second regular expression to parse the pairs - it will be much simpler.
If you do it that way your first expression could be:
^[a-z+.-]+?:\/\/(:www\.)?([^?#]+?)(.*)$
The first part skips over everything through the optional www. and does not capture it because you are not interested in that part. The second part captures everything up to any query or fragment (delimited by ? and #, respectively) and places it in the first capture group. The last part captures the rest of the URL into the the second capture group. In your example that is ?key=value.
Now take your first capture group, which contains the host and the path, and pass it to a second regex with the global flag set (so it processes all pairs repeatedly). This second regex will be:
(.*?)\/([0-9]{4,})\/?
For each match of this string, the parsed values and numbers will be in capture groups 1 & 2.
Upvotes: 0