I'm trying to separate out path elements from a URL in ruby. My URL is:
/cm/api/m_deploymentzones_getDeploymentZones.htm
And using a regex:
/\/([^\/]*)\//
The match is /cm/. I expected to see just cm, given that the / characters are explicitly excluded from the capturing group. What am I doing wrong?
I am also trying to get the next path element with:
/\/[^\/]*\/([^\/]*)\//
And this returns: /cm/api/
Any ideas?
It would be easier to use URI.parse():
require 'uri'
url = '/cm/api/m_deploymentzones_getDeploymentZones.htm'
parts = URI.parse(url).path.split('/') # ["", "cm", "api", "m_deploymentzones_getDeploymentZones.htm"]
Here's a regex matching the path components:
(?<=/).+?(?=/|$)
But you'd be better off just splitting the string on / characters. You don't need regex for this.
Related
I have this URL in a Sinatra-based application:
<li><a href="/blog/<%= blog.title.tr(' ', '-') %>/<%= blog.slug %>"
method="get">Show</a></li>
When I click on it, the URL looks like this:
http://127.0.0.1:9292/blog/A-lovely-day/654790
I am trying to make the last / also a - too, so it will be:
http://127.0.0.1:9292/blog/A-lovely-day-654790
How do I replace it after the URL has been rendered?
Given that you started with:
The slash is not part of the title, but simply the character in red. Replace it with - in the code:
You can specify more than one character to transform
blog.title.tr(" /", "-")
r = /
.* # match any character zero or more times (greedily)
\K # forget all matches so far
\/ # match a forward slash
/x # free-spacing regex definition mode
To return a new string with the replacement:
blog.title.sub(r, '-')
To make the replacement in the existing string:
blog.title.sub!(r, '-')
One could use capture groups in place of \K:
blog.title.sub(/(.*)\/(.*)/, '\1-\2')
Another way to make the replacement in the existing string:
blog.title[blog.title.rindex('/')] = '-'
Here's how I'd go about this:
require 'uri'
title = 'A lovely day'
slug = '654790'
uri = URI.parse('http://127.0.0.1:9292/blog/')
[*title.split, slug].join('-') # => "A-lovely-day-654790"
uri.path += [*title.split, slug].join('-')
uri.to_s # => "http://127.0.0.1:9292/blog/A-lovely-day-654790"
Generate the URL in the controller and only output the variable in the view.
It's always good to use the built-in tools. URI helps when manipulating URLs/URIs, and understands appropriate encoding if necessary.
Also, it's useful to remember that the path is actually a file pathname, so sometimes the File package can be very useful for manipulating/splitting/joining. This wasn't a good example, but it's come in very handy.
'http://127.0.0.1:9292/blog/A-lovely-day/654790'.
sub(/\/(?!.*\/)/,'-') # match a / that is not followed by another /
#=> "http://127.0.0.1:9292/blog/A-lovely-day-654790"
I need to extract a string 'MT/23232' I have written the below code, but
it's not working, Can any one help me here?
'Policy created with MT/1212'
'Policy created with MT/121212'
'Policy created with MT/21212121212'
I have written this code
msg="MT/33235"
id = msg.scan(/MT/\d+/\d+/)[0]
But it's not working for me, Can any one help me to extract this string?
You need to escape the forward slash which exists next to MT in your regex and you don't need to have a forward slash after \d+ . And also i suggest you to add a lookbehind, so that you get a clean result. (?<=\s) Positive lookbehind which asserts that the match must be preceded by a space character.
msg.scan(/(?<=\s)MT\/\d+/)[0]
If you don't care about the preceding character then the below regex would be fine.
msg.scan(/MT\/\d+/)[0]
Example:
> msg = 'Policy created with MT/21212121212'
=> "Policy created with MT/21212121212"
> msg.scan(/(?<=\s)MT\/\d+/)[0]
=> "MT/21212121212"
> msg.match(/(?<=\s)MT\/\d+/)[0]
=> "MT/21212121212"
your_string.scan(/\sMT.*$/).last.strip
If your required substring can be anywhere in the string, then:
your_string.scan(/\bMT\/\d+\b/).last.strip # "\b" is for word boundaries
Or you can specify the acceptable digits this way:
your_string.scan(/\bMT\/[0-9]+\b/).last.strip
Lastly, if the string format is going to remain as you specified, then:
your_string.split.last
http://something.com/bOhxBeD,SyhyTGi,TMDDSIB,U72gx2J,kQTIRy9,7VXgGDw,eSxIcK6,S5oNlnn,WBHHsLk,BdMGd2d,U9kNlsF,cHVyc7Y,D83kaJ5,cLWgdSO,iWtCIF3,ount8L6
I have tried to get the value: bOhxBeD, SyhyTGi and so on. This is what I come up with ( yes fairly simple ) /([a-zA-Z0-9]{7})/, it seems to work with PCRE:
([a-zA-Z0-9]{7})
Debuggex Demo
But when it comes to Ruby, I use it like this :
str.match(/([a-zA-Z0-9]{7})/)
#<MatchData "bOhxBeD" 1:"bOhxBeD">
it doesn't seem to work. Can anyone point out what's wrong with this regex ? Thanks
You need to add word boundary \b inorder to match an exact 7 alphanumeric characters.
\b[a-zA-Z0-9]{7}\b
DEMO
irb(main):006:0> "http://something.com/bOhxBeD,SyhyTGi,TMDDSIB,U72gx2J,kQTIRy9,7VXgGDw,eSxIcK6,S5oNlnn,WBHHsLk,BdMGd2d,U9kNlsF,cHVyc7Y,D83kaJ5,cLWgdSO,iWtCIF3,ount8L6".scan(/\b([a-zA-Z0-9]{7})\b/)
=> [["bOhxBeD"], ["SyhyTGi"], ["TMDDSIB"], ["U72gx2J"], ["kQTIRy9"], ["7VXgGDw"], ["eSxIcK6"], ["S5oNlnn"], ["WBHHsLk"], ["BdMGd2d"], ["U9kNlsF"], ["cHVyc7Y"], ["D83kaJ5"], ["cLWgdSO"], ["iWtCIF3"], ["ount8L6"]]
(?!.*?\/)[a-zA-Z0-9]{7}
Is should be this.Or else it will pick 7 letter words from link as well."somethi" will be in ans.But i guess that is not required.
match only picks up the first match.
You can try the global version of match which is scan.
You can use scan to search string not containing specific characters using [^...]:
str.scan(/[^\/\.\,]+/)[3..-1]
#=> ["bOhxBeD", "SyhyTGi", "TMDDSIB", "U72gx2J", "kQTIRy9", "7VXgGDw", "eSxIcK6", "S5oNlnn", "WBHHsLk", "BdMGd2d", "U9kNlsF", "cHVyc7Y", "D83kaJ5", "cLWgdSO", "iWtCIF3", "ount8L6"]
Update:
If you know that the strings between the comma are always 7 characters, you can use this instead:
str.scan(/[^\/\.\,]{7}/)[1..-1]
it happens because your regexp match just one element which contain 7 chars, nothing more,
as simple solution could be:
str.match(/\/(.*)\z/)[1].split(',')
You could use String#[] and String#split:
str[/.*\/(.*)/,1].split(',')
#=> ["bOhxBeD", "SyhyTGi", "TMDDSIB", "U72gx2J", "kQTIRy9", "7VXgGDw",
# "eSxIcK6", "S5oNlnn", "WBHHsLk", "BdMGd2d", "U9kNlsF", "cHVyc7Y",
# "D83kaJ5", "cLWgdSO", "iWtCIF3", "ount8L6"]
.*\/ in the regex, "greedy" as it is, will consume characters up to and including the last forward slash in the string. Capture group #1 (.*) sucks up the remainder of the string and, due to the presence of ,1, returns it. split(',') then breaks up the string to give you the desired array.
Another way:
str[str[/.*\//].size..-1].split(',')
Suppose I have the following string:
mystring = "start/abc123/end"
How can you splice out the abc123 with something else, while leaving the "/start/" and "/end" elements intact?
I had the following to match for the pattern, but it replaces the entire string. I was hoping to just have it replace the abc123 with 123abc.
mystring.gsub(/start\/(.*)\/end/,"123abc") #=> "123abc"
Edit: The characters between the start & end elements can be any combination of alphanumeric characters, I changed my example to reflect this.
You can do it using this character class : [^\/] (all that is not a slash) and lookarounds
mystring.gsub(/(?<=start\/)[^\/]+(?=\/end)/,"7")
For your example, you could perhaps use:
mystring.gsub(/\/(.*?)\//,"/7/")
This will match the two slashes between the string you're replacing and putting them back in the substitution.
Alternatively, you could capture the pieces of the string you want to keep and interpolate them around your replacement, this turns out to be much more readable than lookaheads/lookbehinds:
irb(main):010:0> mystring.gsub(/(start)\/.*\/(end)/, "\\1/7/\\2")
=> "start/7/end"
\\1 and \\2 here refer to the numbered captures inside of your regular expression.
The problem is that you're replacing the entire matched string, "start/8/end", with "7". You need to include the matched characters you want to persist:
mystring.gsub(/start\/(.*)\/end/, "start/7/end")
Alternatively, just match the digits:
mystring.gsub(/\d+/, "7")
You can do this by grouping the start and end elements in the regular expression and then referring to these groups in in the substitution string:
mystring.gsub(/(?<start>start\/).*(?<end>\/end)/, "\\<start>7\\<end>")
I'm new to Rails, and furthermore to regex. Been looking around, but I'm blocked...
I have a string like this :
Current: http://zs.domain.com/user_images/123456789/imageName_size.ext
Wanted: http://zs.domain.com/user_images/123456789/imageName.ext
I've managed to get to this :
http://a0.twimg.com/profile/1240267050/logo1.png
=> losing all occurrences with
picture.gsub!(/_([a-z0-9-]+)/, '')
or this :
http://a0.twimg.com/profile_images/1240267050/logo1
=> changing only the last occurrence, but losing the extension with
picture.gsub!(/_([a-z0-9-]+)**.(png|gif|jpg|jpeg)**/, '')
You're almost there. The second parameter is the string with which the match will be replaced, and you can re-use matched groups from the match. This will do the trick:
picture.gsub!(/_([a-z0-9-]+).(png|gif|jpg|jpeg)/, '.\2')
To accomodate for the additional conditions, as posed in the comment:
picture.gsub!(/_([^\/]+).(png|gif|jpg|jpeg)/, '.\2')
markijbema's answer will change the string
.../xxx_yyygifzzz/...,
into
.../xxxgifzzz/....
In order to avoid that, you can do this:
picture.gsub!(/_[^\/]+(?=\.[^\.]+\z)/, '')
(?=...) is understood as a context that follows the string, and will not be included in the match.
\z describes the end of the string, so this regexp is safe to use when some intermediate directory includes a string like above.