I have the following lines I have to match
A
Ab
A#
F#7+9d
G3+9d
Gm
Basically I need to match the first letter and the # or the b following. I also need to match anything else (such as the 7+9d or the m).
Here is my code so far but it's not picking up the second part
preg_match('/A-G([A-Z0-9+]?)/i', $start_key, $matches)
Any ideas?
Try this
/^A-G(#|b)?([A-Z0-9+]*)$/i
Try adding a-z and # to the body of the character set, converting your statement into:
preg_match('/A-G([A-Za-z0-9+#]?)/i', $start_key, $matches)
Related
Original string '4.0.0-4.0-M-672092'
How to modify the Original string to "4.0-M-672092" using a one line code.
Any Help is highly appreciated .
Thanks and Regards
The 'split' method works in this case
https://apidock.com/ruby/String/split
'4.0.0-4.0-M-672092'.split('-')[1..-1].join('-')
# => "4.0-M-672092"
Just be careful, in this application is fine, but in long texts this might become unoptimized, since it splits all the string and then joins the array all over again
If you need this in wider texts to be more optimized, you can find the "-" index (which is your split) and use the next position to make a substring
text = '4.0.0-4.0-M-672092'
text[(text.index('-') + 1)..-1]
# => "4.0-M-672092"
But you can't do it in one line, and not finding a split character will result in an error, so use a rescue statement if that is possible to happen
Simplest way:
'4.0.0-4.0-M-672092'.split('-', 2).second
"4.0.0-4.0-M-672092"[/(?<=-).*/]
#=> "4.0-M-672092"
The regular expression reads, "Match zero or more characters other than newlines, as many as possible (.*), provided the match is preceded by a hyphen. (?<=-) is a positive lookbehind. See String#[].
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 have unfortunately wandered into a situation where I need regex using Ruby. Basically I want to match this string after the underscore and before the first parentheses. So the end result would be 'table salt'.
_____ table salt (1) [F]
As usual I tried to fight this battle on my own and with rubular.com. I got the first part
^_____ (Match the beginning of the string with underscores ).
Then I got bolder,
^_____(.*?) ( Do the first part of the match, then give me any amount of words and letters after it )
Regex had had enough and put an end to that nonsense and crapped out. So I was wondering if anyone on stackoverflow knew or would have any hints on how to say my goal to the Ruby Regex parser.
EDIT: Thanks everyone, this is the pattern I ended up using after creating it with rubular.
ingredientNameRegex = /^_+([^(]*)/;
Everything got better once I took a deep breath, and thought about what I was trying to say.
str = "_____ table salt (1) [F]"
p str[ /_{3}\s(.+?)\s+\(/, 1 ]
#=> "table salt"
That says:
Find at least three underscores
and a whitespace character (\s)
and then one or more (+) of any character (.), but as little as possible (?), up until you find
one or more whitespace characters,
and then a literal (
The parens in the middle save that bit, and the 1 pulls it out.
Try this: ^[_]+([^(]*)\(
It will match lines starting with one or more underscores followed by anything not equal to an opening bracket: http://rubular.com/r/vthpGpVr4y
Here's working regex:
str = "_____ table salt (1) [F]"
match = str.match(/_([^_]+?)\(/)
p match[1].strip # => "table salt"
You could use
^_____\s*([^(]+?)\s*\(
^_____ match the underscore from the beginning of string
\s* matches any whitespace character
( grouping start
[^(]+ matches all non ( character at least once
? matches the shortest possible string (non greedy)
) grouping end
\s* matches any whitespace character
\( find the (
"_____ table salt (1) [F]".gsub(/[_]\s(.+)\s\(/, ' >>>\1<<< ')
# => "____ >>>table salt<<< 1) [F]"
It seems to me the simplest regex to do what you want is:
/^_____ ([\w\s]+) /
That says:
leading underscores, space, then capture any combination of word chars or spaces, then another space.
I have a string:
string = %q{<span class="no">2503</span>read_attribute_before_type_cast(<span class="pc">self</span>.class.primary_key)}
In this example I want to match the words 'class' which are not in the tag. Regexp for this:
/\bclass[^=]/
But the problem is that it matches the last letter
/\bclass[^=]/.match(string) => 'class.'
I don't want have a last dot in a result. I've tried this regexp:
/\bclass(?:[^=])/
but still got the same result. How to limit the result to 'class'? Thanks
You are almost correct, but you have an error in your look ahead. Try this:
/\bclass(?!=)/
The regex term (?!=) means the input to the right must not match the character '='
You can take your variable string and extract a subsection using groups:
substring = string[/\b(class)[^=]/, 1]
The brackets around class will set that as the first "group", which is referred to by the 1 as the second parameter in the square brackets.
Assuming your only issue is keeping it from matching span.class.blah, just ignore . as well, so [^=.].