I need to perform to search and replace activity
"{{content}}" => replace. (this to keep same type) regex gsub(/"{{(.*?)}}"/)
"hello {{content}}" => repalce (this to replace from string) regex gsub(/{{(.*?)}}/)
Method that i build is
def fill_in(template)
template.gsub(/\"\{\{(.*?)\}\}\"/) do
"test"
end
end
Tried template.gsub(/\"\{\{(.*?)\}\}\"/).gsub(/\{\{(.*?)\}\}/) do but this is giving
me error
undefined method `gsub' for #<Enumerator:
"{{user_name}}":gsub(/"{{(.*?)}}"/)>
first gsub is priority if it matches that pattern replace based on then if not check for second gsub
template.gsub(/\"\{\{(.*?)\}\}\"/) do
# content will be replaced from the data object
end.gsub(/\"\{\{(.*?)\}\}\"/) do
# content will be replaced from the data object
end
do body for both gsub is same, how to stop this repetition
The gsub with just a regex as a single argument to return an Enumerator, so you won't be able to chain the gsub in this way.
You can combine the two patterns into one:
/(")?\{\{(.*?)\}\}(?(1)"|)/
See the regex demo. Details:
(")? - Capturing group 1 (optional):
\{\{ - a {{ text
(.*?) - Capturing group 2: any zero or more chars other than line break chars, as few as possible (if you need to match line breaks, too, use ((?s:.*?)) instead, or simply add /m flag)
\}\} - a }} string
(?(1)"|) - a conditional construct: if Group 1 matched, match ", else, match an empty string.
In the code, you will need to check if Group 1 matched, and if so, implement one replacement logic, else, use another replacement logic. See the Ruby demo:
def fill_in(template)
template.gsub(/(")?\{\{(.*?)\}\}(?(1)"|)/) {
$~[1] ? "Replacement 1" : "Replacement 2"
}
end
p fill_in('"{{hello}}" and {{hello}}')
# => "Replacement 1 and Replacement 2"
Related
I have a string, which describe some word, I must change ending of it to "sd", if ending == "jk".
For an example, I have word: "lazerjk", I need to get from it "lazersd".
I tried to use method .gsub!, but it doesn't work correctly if we have more than one occurrence of substring "jk" in a word.
String#rindex returns the index of the last occurrence of the given substring
String#[]= can take two integers arguments, first is index where start to replace and second - length of replaced string
You can use them this way:
replaced = "foo"
replacing = "booo"
string = "foo bar foo baz"
string[string.rindex(replaced), replaced.size] = replacing
string
# => "foo bar booo baz"
"jughjkjkjk\njk".sub(/jk$\z/, 'sd')
=> "jughjkjkjk\nsd"
without $ is probably sufficient.
It sounds like you're looking to replace a specific suffix only. If so, I would probably suggest using sub along with an anchored regex (to check for the desired characters only at the end of the string):
string_1 = "lazerjk"
string_2 = "lazerjk\njk"
string_3 = "lazerjkr"
string_1.sub(/jk\z/, "sd")
#=> "lazersd"
string_2.sub(/jk\z/, "sd")
#=> "lazerjk\nsd"
string_3.sub(/jk\z/, "sd")
#=> "lazerjkr"
Or, you could do without a regex at all by using the reverse! method along with a simple conditional statement to sub! only when the suffix is present:
string = "lazerjk"
old_suffix = "jk"
new_suffix = "sd"
string.reverse!.sub!(old_suffix.reverse, new_suffix.reverse).reverse! if string.end_with? (old_suffix)
string
#=> "lazersd"
OR, you could even use a completely different approach. Here's an example using chomp to remove the unwanted suffix and then ljust to pad the desired suffix to the modified string.
string = "lazerjk"
string.chomp("jk").ljust(string.length, "sd")
#=> "lazersd"
Note that the new suffix only gets added if the length of the string was modified with the initial chomp. Otherwise, the string remains unchanged.
If the goal is to substitute the LAST OCCURRENCE (as opposed to suffix only), then this could be accomplished by using sub along with reverse:
string = "jklazerjkm"
old_substring = "jk"
new_substring = "sd"
string.reverse.sub(old_substring.reverse, new_substring.reverse).reverse
#=> "jklazersdm"
Replacing "jk" at the end of a string with something else is straightforward and can be addressed without concern for other instances of "jk" that may be in the string, so I assume that is not what is being asked. Rather, I assume the problem is to replace the last instance of "jk" in a string with "sd".
Here are two solutions that make use of String#sub with a regular expression.
Use a negative lookahead
The idea here is to match "jk" provided it is not followed later in the string by another instance of "jk".
"lajkz\nejkrjklm".sub(/jk(?!.*jk)/m, "sd")
#=> "lajkz\nejkrsdlm"
Capture the part of the string that precedes the last "jk"
The match, if there is one, consists of the front of the string followed by the last "jk", which is replaced by the captured string followed by "sd".
"lajkz\nejkrjklm".sub(/\A(.*)jk/m) { $1 + "sd" }
#=> "lajkz\nejkrsdlm"
The two regular expressions can be written in free-spacing mode to make them self-documenting. The first is the following.
/
jk # match literal
(?! # begin a negative lookahead
.* # match zero or more characters other than line terminators
jk # match literal
) # end negative lookahead
/mx # invoke multiline and free-spacing regex definition modes.
Multiline mode causes . to match any character, including a line terminator.
The second regular expression can be written as follows.
\A # match the beginning of the string
(.*) # match zero or more characters other than line terminators
# and save the match to capture group 1
jk # match literal
/mx # invoke multiline and free-spacing regex definition modes.
Note that in both expressions .* is greedy, meaning that it will match as many characters as possible, including "jk" so long as other requirements of the expression are met, here that the last instance of "jk" in the string is matched.
Here is a different solution:
str = "jughjkjkjk\njk"
pattern = "jk"
replace_with = "sd"
str = str.reverse.sub(pattern.reverse, replace_with.reverse).reverse
I have a name spaced class..
"CommonCar::RedTrunk"
I need to convert it to an underscored string "common_car_red_trunk", but when I use
"CommonCar::RedTrunk".underscore, I get "common_car/red_trunk" instead.
Is there another method to accomplish what I need?
Solutions:
"CommonCar::RedTrunk".gsub(':', '').underscore
or:
"CommonCar::RedTrunk".sub('::', '').underscore
or:
"CommonCar::RedTrunk".tr(':', '').underscore
Alternate:
Or turn any of these around and do the underscore() first, followed by whatever method you want to use to replace "/" with "_".
Explanation:
While all of these methods look basically the same, there are subtle differences that can be very impactful.
In short:
gsub() – uses a regex to do pattern matching, therefore, it's finding any occurrence of ":" and replacing it with "".
sub() – uses a regex to do pattern matching, similarly to gsub(), with the exception that it's only finding the first occurrence (the "g" in gsub() meaning "global"). This is why when using that method, it was necessary to use "::", otherwise a single ":" would have been left. Keep in mind with this method, it will only work with a single-nested namespace. Meaning "CommonCar::RedTrunk::BigWheels" would have been transformed to "CommonCarRedTrunk::BigWheels".
tr() – uses the string parameters as arrays of single character replacments. In this case, because we're only replacing a single character, it'll work identically to gsub(). However, if you wanted to replace "on" with "EX", for example, gsub("on", "EX") would produce "CommEXCar::RedTrunk" while tr("on", "EX") would produce "CEmmEXCar::RedTruXk".
Docs:
https://apidock.com/ruby/String/gsub
https://apidock.com/ruby/String/sub
https://apidock.com/ruby/String/tr
This is a pure-Ruby solution.
r = /(?<=[a-z])(?=[A-Z])|::/
"CommonCar::RedTrunk".gsub(r, '_').downcase
#=> "common_car_red_trunk"
See (the first form of) String#gsub and String#downcase.
The regular expression can be made self-documenting by writing it in free-spacing mode:
r = /
(?<=[a-z]) # assert that the previous character is lower-case
(?=[A-Z]) # assert that the following character is upper-case
| # or
:: # match '::'
/x # free-spacing regex definition mode
(?<=[a-z]) is a positive lookbehind; (?=[A-Z]) is a positive lookahead.
Note that /(?<=[a-z])(?=[A-Z])/ matches an empty ("zero-width") string. r matches, for example, the empty string between 'Common' and 'Car', because it is preceeded by a lower-case letter and followed by an upper-case letter.
I don't know Rails but I'm guessing you could write
"CommonCar::RedTrunk".delete(':').underscore
In my application I've got a procedure which should check if an input is valid or not. You can set up a regex for this input.
But in my case it returns false instead of true. And I can't find the problem.
My code looks like this:
gaps.each_index do | i |
if gaps[i].first.starts_with? "~"
# regular expression
begin
regex = gaps[i].first[1..-1]
# a pipe is used to seperate the regex from the solution-string
if regex.include? "|"
puts "REGEX FOUND ------------------------------------------"
regex = regex.split("|")[0..-2].join("|")
end
reg = Regexp.new(regex, true)
unless reg.match(data[i])
puts "REGEX WRONGGGG -------------------"
#wrong_indexes << i
end
rescue
end
else
# normal string
if data[i].nil? || data[i].strip != gaps[i].first.strip
#wrong_indexes << i
end
end
An example would be:
[[~berlin|berlin]]
The left one before the pipe is the regex and the right one next to the pipe is the correct solution.
This easy input should return true, but it doesn't.
Does anyone see the problem?
Thank you all
EDIT
Somewhere in this lines must be the problem:
if regex.include? "|"
puts "REGEX FOUND ------------------------------------------"
regex = regex.split("|")[0..-2].join("|")
end
reg = Regexp.new(regex, true)
unless reg.match(data[i])
Update: Result without ~
The whole point is that you are initializing regex using the Regexp constructor
Constructs a new regular expression from pattern, which can be either a String or a Regexp (in which case that regexp’s options are propagated, and new options may not be specified (a change as of Ruby 1.8).
However, when you pass the regex (obtained with regex.split("|")[0..-2].join("|")) to the constructor, it is a string, and reg = Regexp.new(regex, true) is getting ~berlin (or /berlin/i) as a literal string pattern. Thus, it actually is searching for something you do not expect.
See, regex= "[[/berlin/i|berlin]]" only finds a *literal /berlin/i text (see demo).
Also, you need to get the pattern from the [[...]], so strip these brackets with regex = regex.gsub(/\A\[+|\]+\z/, '').split("|")[0..-2].join("|").
Note you do not need to specify the case insensitive options, since you already pass true as the second parameter to Regexp.new, it is already case-insensitive.
If you are performing whole word lookup, add word boundaries: regex= "[[\\bberlin\\b|berlin]]" (see demo).
I'm reading some source code at https://github.com/plataformatec/devise and found that line of code:
class_eval <<-URL_HELPERS, __FILE__, __LINE__ + 1
What the params __FILE__ and __LINE__ + 1 does in block declaration (what changes in relation of a string block without these params)?
https://github.com/plataformatec/devise/blob/master/lib/devise/controllers/url_helpers.rb#L47
Thank
Those params belong to the class_eval method, not to the here document. It's common practice to ensure that error, which can raise the evaled code, will be shown with a reference to the current file and with correct line number.
By way of an alternative example showing how HEREDOCs work, the other day in IRB I wrote:
require 'nokogiri'
doc = Nokogiri.XML(<<ENDXML,&:noblanks)
...gobs and gobs of pasted xml...
ENDXML
Even crazier is this legal syntax for passing multiple HEREDOC strings at once:
p( <<END1, <<-END2, <<END3 )
This indented text is part of
the first parameter, END1
END1
And this text is part of param2
whose ending sigil may be indented
END2
and finally this text
is part of
the third parameter
END3
#=> " This indented text is part of\n the first parameter, END1\n"
#=> "And this text is part of param2\n whose ending sigil may be indented\n"
#=> "and finally this text\nis part of\nthe third parameter\n"
Suppose I have string variables like following:
s1="10$"
s2="10$ I am a student"
s3="10$Good"
s4="10$ Nice weekend!"
As you see above, s2 and s4 have white space(s) after 10$ .
Generally, I would like to have a way to check if a string start with 10$ and have white-space(s) after 10$ . For example, The rule should find s2 and s4 in my above case. how to define such rule to check if a string start with '10$' and have white space(s) after?
What I mean is something like s2.RULE? should return true or false to tell if it is the matched string.
---------- update -------------------
please also tell the solution if 10# is used instead of 10$
You can do this using Regular Expressions (Ruby has Perl-style regular expressions, to be exact).
# For ease of demonstration, I've moved your strings into an array
strings = [
"10$",
"10$ I am a student",
"10$Good",
"10$ Nice weekend!"
]
p strings.find_all { |s| s =~ /\A10\$[ \t]+/ }
The regular expression breaks down like this:
The / at the beginning and the end tell Ruby that everything in between is part of the regular expression
\A matches the beginning of a string
The 10 is matched verbatim
\$ means to match a $ verbatim. We need to escape it since $ has a special meaning in regular expressions.
[ \t]+ means "match at least one blank and/or tab"
So this regular expressions says "Match every string that starts with 10$ followed by at least one blank or tab character". Using the =~ you can test strings in Ruby against this expression. =~ will return a non-nil value, which evaluates to true if used in a conditional like if.
Edit: Updated white space matching as per Asmageddon's suggestion.
this works:
"10$ " =~ /^10\$ +/
and returns either nil when false or 0 when true. Thanks to Ruby's rule, you can use it directly.
Use a regular expression like this one:
/10\$\s+/
EDIT
If you use =~ for matching, note that
The =~ operator returns the character position in the string of the
start of the match
So it might return 0 to denote a match. Only a return of nil means no match.
See for example http://www.regular-expressions.info/ruby.html on a regular expression tutorial for ruby.
If you want to proceed to cases with $ and # then try this regular expression:
/^10[\$#] +/