I need to replace certain ascii characters like # and & with their hex representations for a URL which would be 40 and 26 respectively.
How can I do this in ruby? there are also some characters most notably '-' which does not need to be replaced.
require 'uri'
URI.escape str, /[#&]/
Obviously, you can widen the regex with more characters you want to escape. Or, if you want to do a whitelisting approach, you can do, say,
URI.escape str, /[^-\w]/
This is ruby, so there's a mandatory 20 different ways to do it. Here's mine:
>> a = 'one&two%three'
=> "one&two%three"
>> a.gsub(/[&%]/, '&' => '&'.ord, '%' => '%'.ord)
=> "one38two37three"
I'm pretty sure Ruby has this functionality built in for URLs. However, if you want to define some more general translation facility you may use code like the following:
s = "h#llo world"
t = { " " => "%20", "#" => "%40" };
puts s.split(//).map { |c| t[c] || c }.join
Which would output
h%40llo%20world
In the above code, t is a hash defining the mapping from specific characters to their representation. The string is broken into characters and the hash is searched for each character's equivalent.
More generically and easily:
require 'uri'
URI.escape(your_string,Regexp.new("[^#{URI::PATTERN::UNRESERVED}]")
Related
I have a string as given below,
./component/unit
and need to split to get result as component/unit which I will use this as key for inserting hash.
I tried with .split(/.\//).last but its giving result as unit only not getting component/unit.
I think, this should help you:
string = './component/unit'
string.split('./')
#=> ["", "component/unit"]
string.split('./').last
#=> "component/unit"
Your regex was almost fine :
split(/\.\//)
You need to escape both . (any character) and / (regex delimiter).
As an alternative, you could just remove the first './' substring :
'./component/unit'.sub('./','')
#=> "component/unit"
All the other answers are fine, but I think you are not really dealing with a String here but with a URI or Pathname, so I would advise you to use these classes if you can. If so, please adjust the title, as it is not about do-it-yourself-regexes, but about proper use of the available libraries.
Link to the ruby doc:
https://docs.ruby-lang.org/en/2.1.0/URI.html
and
https://ruby-doc.org/stdlib-2.1.0/libdoc/pathname/rdoc/Pathname.html
An example with Pathname is:
require 'pathname'
pathname = Pathname.new('./component/unit')
puts pathname.cleanpath # => "component/unit"
# pathname.to_s # => "component/unit"
Whether this is a good idea (and/or using URI would be cool too) also depends on what your real problem is, i.e. what you want to do with the extracted String. As stated, I doubt a bit that you are really intested in Strings.
Using a positive lookbehind, you could do use regex:
reg = /(?<=\.\/)[\w+\/]+\w+\z/
Demo
str = './component'
str2 = './component/unit'
str3 = './component/unit/ruby'
str4 = './component/unit/ruby/regex'
[str, str2, str3, str4].each { |s| puts s[reg] }
#component
#component/unit
#component/unit/ruby
#component/unit/ruby/regex
Trying to convert output from a rest_client GET to the characters that are represented with escape sequences.
Input: ..."sub_id":"\u0d9c\u8138\u8134\u3f30\u8139\u2b71"...
(which I put in 'all_subs')
Match: m = /sub_id\"\:\"([^\"]+)\"/.match(all_subs.to_str) [1]
Print: puts m.force_encoding("UTF-8").unpack('U*').pack('U*')
But it just comes out the same way I put it in. ie, "\u0d9c\u8138\u8134\u3f30\u8139\u2b71"
However, if I convert a raw string of it:
puts "\u0d9c\u8138\u8134\u3f30\u8139\u2b71".unpack('U*').pack('U*')
The output is perfect as "ග脸脴㼰脹⭱"
What you're getting when you parse the input string is actually this:
m = "\\u0d9c\\u8138\\u8134\\u3f30\\u8139\\u2b71"
Which is not the same as:
"\u0d9c\u8138\u8134\u3f30\u8139\u2b71"
Therefore one option is to eval the string so that ruby applies the codepoints:
puts eval("\"#{m}\"")
=> ග脸脴㼰脹
However note that there are security implications when running eval.
If the string is always like in your example. You could also do something like this, which is safe:
puts m.split("\\u")[1..-1].map { |c| c.to_i(16) }.pack("U*")
=> ග脸脴㼰脹
How can I remove from a string all characters except white spaces, numbers, and some others?
Something like this:
oneLine.gsub(/[^ULDR0-9\<\>\s]/i,'')
I need only: 0-9 l d u r < > <space>
Also, is there a good document about the use of regex in Ruby, like a list of special characters with examples?
The regex you have is already working correctly. However, you do need to assign the result back to the string you're operating on. Otherwise, you're not changing the string (.gsub() does not modify the string in-place).
You can improve the regex a bit by adding a '+' quantifier (so consecutive characters can be replaced in one go). Also, you don't need to escape angle brackets:
oneLine = oneLine.gsub(/[^ULDR0-9<>\s]+/i, '')
A good resource with special consideration of Ruby regexes is the Regular Expressions Cookbook by Jan Goyvaerts and Steven Levithan. A good online tutorial by the same author is here.
Good old String#delete does this without a regular expression. The ^ means 'NOT'.
str = "12eldabc8urp pp"
p str.delete('^0-9ldur<> ') #=> "12ld8ur "
Just for completeness: you don't need a regular expression for this particular task, this can be done using simple string manipulation:
irb(main):005:0> "asdasd123".tr('^ULDRuldr0-9<>\t\r\n ', '')
=> "dd123"
There's also the tr! method if you want to replace the old value:
irb(main):009:0> oneLine = 'UasdL asd 123'
irb(main):010:0> oneLine.tr!('^ULDRuldr0-9<>\t\r\n ', '')
irb(main):011:0> oneLine
=> "UdL d 123"
This should be a bit faster as well (but performance shouldn't be a big concern in Ruby :)
I want to strip leading and trailing quotes, in Ruby, from a string. The quote character will occur 0 or 1 time. For example, all of the following should be converted to foo,bar:
"foo,bar"
"foo,bar
foo,bar"
foo,bar
You could also use the chomp function, but it unfortunately only works in the end of the string, assuming there was a reverse chomp, you could:
'"foo,bar"'.rchomp('"').chomp('"')
Implementing rchomp is straightforward:
class String
def rchomp(sep = $/)
self.start_with?(sep) ? self[sep.size..-1] : self
end
end
Note that you could also do it inline, with the slightly less efficient version:
'"foo,bar"'.chomp('"').reverse.chomp('"').reverse
EDIT: Since Ruby 2.5, rchomp(x) is available under the name delete_prefix, and chomp(x) is available as delete_suffix, meaning that you can use
'"foo,bar"'.delete_prefix('"').delete_suffix('"')
I can use gsub to search for the leading or trailing quote and replace it with an empty string:
s = "\"foo,bar\""
s.gsub!(/^\"|\"?$/, '')
As suggested by comments below, a better solution is:
s.gsub!(/\A"|"\Z/, '')
As usual everyone grabs regex from the toolbox first. :-)
As an alternate I'll recommend looking into .tr('"', '') (AKA "translate") which, in this use, is really stripping the quotes.
Another approach would be
remove_quotations('"foo,bar"')
def remove_quotations(str)
if str.start_with?('"')
str = str.slice(1..-1)
end
if str.end_with?('"')
str = str.slice(0..-2)
end
end
It is without RegExps and start_with?/end_with? are nicely readable.
It frustrates me that strip only works on whitespace. I need to strip all kinds of characters! Here's a String extension that will fix that:
class String
def trim sep=/\s/
sep_source = sep.is_a?(Regexp) ? sep.source : Regexp.escape(sep)
pattern = Regexp.new("\\A(#{sep_source})*(.*?)(#{sep_source})*\\z")
self[pattern, 2]
end
end
Output
'"foo,bar"'.trim '"' # => "foo,bar"
'"foo,bar'.trim '"' # => "foo,bar"
'foo,bar"'.trim '"' # => "foo,bar"
'foo,bar'.trim '"' # => "foo,bar"
' foo,bar'.trim # => "foo,bar"
'afoo,bare'.trim /[aeiou]/ # => "foo,bar"
Assuming that quotes can only appear at the beginning or end, you could just remove all quotes, without any custom method:
'"foo,bar"'.delete('"')
I wanted the same but for slashes in url path, which can be /test/test/test/ (so that it has the stripping characters in the middle) and eventually came up with something like this to avoid regexps:
'/test/test/test/'.split('/').reject(|i| i.empty?).join('/')
Which in this case translates obviously to:
'"foo,bar"'.split('"').select{|i| i != ""}.join('"')
or
'"foo,bar"'.split('"').reject{|i| i.empty?}.join('"')
Regexs can be pretty heavy and lead to some funky errors. If you are not dealing with massive strings and the data is pretty uniform you can use a simpler approach.
If you know the strings have starting and leading quotes you can splice the entire string:
string = "'This has quotes!'"
trimmed = string[1..-2]
puts trimmed # "This has quotes!"
This can also be turned into a simple function:
# In this case, 34 is \" and 39 is ', you can add other codes etc.
def trim_chars(string, char_codes=[34, 39])
if char_codes.include?(string[0]) && char_codes.include?(string[-1])
string[1..-2]
else
string
end
end
You can strip non-optional quotes with scan:
'"foo"bar"'.scan(/"(.*)"/)[0][0]
# => "foo\"bar"
I need to print escaped characters to a binary file using Ruby. The main problem is that slashes need the whole byte to escape correctly, and I don't know/can't create the byte in such a way.
I am creating the hex value with, basically:
'\x' + char
Where char is some 'hex' value, such as 65. In hex, \x65 is the ASCII character 'e'.
Unfortunately, when I puts this sequence to the file, I end up with this:
\\x65
How do I create a hex string with the properly escaped value? I have tried a lot of things, involving single or double quotes, pack, unpack, multiple slashes, etc. I have tried so many different combinations that I feel as though I understand the problem less now then I did when I started.
How?
You may need to set binary mode on your file, and/or use putc.
File.open("foo.tmp", "w") do |f|
f.set_encoding(Encoding::BINARY) # set_encoding is Ruby 1.9
f.binmode # only useful on Windows
f.putc "e".hex
end
Hopefully this can give you some ideas even if you have Ruby <1.9.
Okay, if you want to create a string whose first byte
has the integer value 0x65, use Array#pack
irb> [0x65].pack('U')
#=> "e"
irb> "e"[0]
#=> 101
10110 = 6516, so this works.
If you want to create a literal string whose first byte is '\',
second is 'x', third is '6', and fourth is '5', then just use interpolation:
irb> "\\x#{65}"
#=> "\\x65"
irb> "\\x65".split('')
#=> ["\\", "x", "6", "5"]
If you have the hex value and you want to create a string containing the character corresponding to that hex value, you can do:
irb(main):002:0> '65'.hex.chr
=> "e"
Another option is to use Array#pack; this can be used if you need to convert a list of numbers to a single string:
irb(main):003:0> ['65'.hex].pack("C")
=> "e"
irb(main):004:0> ['66', '6f', '6f'].map {|x| x.hex}.pack("C*")
=> "foo"