How can I replace double-quotes to "e; in a string?
This is what I have tried:
1.9.3-p362 :009 > a = "\"That's it\", she said."
=> "\"That's it\", she said."
1.9.3-p362 :010 > a.tr('"', ""e;")
=> "&That's it&, she said."
As you see instead of "es; I only get &, any idea?
use gsub instead
a.gsub(/\"/, ""e;")
# without regex as noted by hirolau
a.gsub("\"", ""e;")
# => ""e;That's it"e;, she said."
Related
#!/usr/bin/ruby
puts "Please enter the path-name of the directory:"
p = STDIN.gets
isdir = File.directory?(p)
puts "#{isdir} #{p}"
it always return me a false! even though I know the user input is a directory. I think (p) is not working as a parameter. So i think its saying that p is not a directory not the user input for example "/usr/bin/". any help?
Using p = STDIN.gets '\n' was getting appended. Instead you can use gets.chomp. Also you need to use File.expand_path. Check the example below.
# My irb
1.9.3-p545 :002 > p = gets.chomp
~/.ssh
=> "~/.ssh"
1.9.3-p545 :003 > File.directory?(p)
=> false
1.9.3-p545 :004 > File.exists? File.expand_path(p)
=> true
The p value is not strictly equal to what you expect it to be. It contains \n at the end:
# in my irb:
1.9.3p392 :010 > p = STDIN.gets
/home/
=> "/home/\n"
1.9.3p392 :011 > isdir = File.directory?(p)
=> false
1.9.3p392 :012 > isdir = File.directory?(p.strip)
=> true
The strip method:
Strips entire range of Unicode whitespace from the right and left of the string.
Source: http://apidock.com/rails/ActiveSupport/Multibyte/Chars/strip
I'm following a tutorial on regular expressions in ruby, but a method called regex that seems not working
Tutorial
re = /(\w*)\s(\w*),\s?([\w\s]*)/
match1 = str1.match re
match2 = str2.match re
match1.regex # => wsw,s[ws] (this is IRB's unique way of showing regular expressions; it will still work normally)
My console
The regex method the method throws an error
1.9.3-p547 :033 > re = /(\w*)\s(\w*),\s?([\w\s]*)/
=> /(\w*)\s(\w*),\s?([\w\s]*)/
1.9.3-p547 :034 > match1 = str1.match re
=> #<MatchData "Joe Schmo, Plumber" 1:"Joe" 2:"Schmo" 3:"Plumber">
1.9.3-p547 :035 > match2 = str2.match re
=> #<MatchData "Stephen Harper, Prime Minister" 1:"Stephen" 2:"Harper" 3:"Prime Minister">
1.9.3-p547 :036 > match1.regex
NoMethodError: undefined method `regex' for #<MatchData "Joe Schmo, Plumber" 1:"Joe" 2:"Schmo" 3:"Plumber">
from (irb):36
from /home/fernando/.rvm/rubies/ruby-1.9.3-p547/bin/irb:12:in `<main>'
1.9.3-p547 :037 >
I think it should be
match1.regexp
with a final 'p'
I am using Rails 4.0.0 with Ruby 2.0.0 p247. I am writing an URL regexp matcher but I have no idea why it does not work:
2.0.0-p247 :033 > REGEXP = %r{\Ahttps:\/\/#{ Rails.configuration.aws[:bucket] }\.s3(-#{Rails.configuration.aws[:region]}|)\.amazonaws\.com\/(?<path>uploads\/.+\/(?<filename>.+))\?.+\z}.freeze
=> /\Ahttps:\/\/test-gem\.s3(-eu-west-1|)\.amazonaws\.com\/(?<path>uploads\/.+\/(?<filename>.+))\?.+\z/
2.0.0-p247 :034 > url = "https://test-gem.s3.amazonaws.com/uploads/2alrg16mvx6r-29590d114fb3257846c1a03330418da9/3031674-poster-p-1-for-25.jpg"
=> "https://test-gem.s3.amazonaws.com/uploads/2alrg16mvx6r-29590d114fb3257846c1a03330418da9/3031674-poster-p-1-for-25.jpg"
2.0.0-p247 :035 > REGEXP.match(url)
=> nil
But when I try to debug in things like Rubular, it does work. Any idea? Thanks!
Remove \?.+ in the end if your regexp
Might be a bug with Ruby 2.0.0. I'm using 2.1.3 and it works like you'd expect.
> r = /\Ahttps:\/\/test-gem\.s3(\A-eu-west-1\z|)\.amazonaws\.com\/(?<path>uploads\/.+\/(?<filename>.+))\z/
=> /\Ahttps:\/\/test-gem\.s3(\A-eu-west-1\z|)\.amazonaws\.com\/(?<path>uploads\/.+\/(?<filename>.+))\z/
> r.match("https://test-gem.s3.amazonaws.com/uploads/2alrg16mvx6r-29590d114fb3257846c1a03330418da9/3031674-poster-p-1-for-25.jpg")
=> #<MatchData
"https://test-gem.s3.amazonaws.com/uploads/2alrg16mvx6r-29590d114fb3257846c1a03330418da9/3031674-poster-p-1-for-25.jpg"
path:"uploads/2alrg16mvx6r-29590d114fb3257846c1a03330418da9/3031674-poster-p-1-for-25.jpg"
filename:"3031674-poster-p-1-for-25.jpg">
Context
I have Rails (4.0.1 + Ruby 2.0.0) connected to a PostgreSQL database filled with strings like "€ 289,95". The values have been scraped from a website using Nokogiri. I want to convert the strings to floating points.
What I've tried
Rails console:
listing = Listing.find(1)
=> #<Listing id: 1, title: #, subtitle: #, name: #, price: "€ 289,95", url: #, created_at: #, updated_at: #>
listing_price = listing.price
=> "€ 289,95"
listing_price_1 = listing_price.gsub(/,/, ".")
=> "€ 289.95"
listing_price_2 = listing_price_1.gsub(/€\s/, "")
=> "€ 289.95"
listing_price_3 = listing_price_2.to_f
=> 0.0
Problem
The code works in irb but doesn't work in the rails console.
What I want to know
How to convert a string "€ 289,95" to a float "289.95" in Rails?
The step where your technique is failing is when trying to strip away € and the space from € 289.95 with the regexp /€\s/, but this is not matching, leaving the string unchanged.
The space character in € 289,95 is probably a non-breaking space (U+00A0) rather than a “normal” space, and would be used in the web page so that the € and the value are not separated.
In Ruby the non-breaking space is not matched by \s in a regexp, so your call to gsub doesn’t replace anything:
2.0.0p353 :001 > s = "€\u00a0289.95"
=> "€ 289.95"
2.0.0p353 :002 > s.gsub(/€\s/, "")
=> "€ 289.95"
Non-breaking space is matched by the POSIX bracket expression [[:space:]], or by the character property \{Blank}:
2.0.0p353 :003 > s.gsub /€[[:space:]]/, ""
=> "289.95"
2.0.0p353 :004 > s.gsub /€\p{Blank}/, ""
=> "289.95"
So if you wanted a more specific regexp than in the other answer you could use one of these.
"€ 289,95".sub(/\A\D+/, "").sub(",", ".").to_f
# => 289.95
listing.price.delete('€ ') # => "289,95"
listing.price.delete('€ ').tr(',', '.') # => "289.95"
listing.price.delete('€ ').tr(',', '.').to_f # => 289.95
String's 'delete' method is good for removing all occurrences of the target strings.
and 'tr' method takes a string of characters to search for, and a string of characters used to replace them.
Better probably than the accepted answer is:
"€ 289,95"[/[\d,.]+/].tr ',', '.'
I have a string like
"car\"
which I will be storing in postgres db. I want to remove the backslash from the string before saving. Is there a way I can do that in ruby or in postgres? When I try to remove it in ruby, it considers the quote after the backslash as an escape character.
See following code:
1.9.3p125 :022 > s = "cat\\"
=> "cat\\"
1.9.3p125 :023 > puts s
cat\
=> nil
1.9.3p125 :024 > s.chomp("\\")
=> "cat"
1.9.3p125 :025 >
People don't do this much, but Ruby's String class supports:
irb(main):002:0> str = 'car\\'
=> "car\\"
irb(main):003:0> str[/\\$/] = ''
=> ""
irb(main):004:0> str
=> "car"
It's a conditional search for a trailing '\', and replacement with an empty string.
To remove a trailing backslash:
"car\\".gsub!(/\\$/, "")
Note that the backslash has to be escaped itself with a backslash.
puts '"car\"'.gsub(/\\(")?$/, '\1')
that will do it,
but, is the trailing slash always at the en followed by a quote?
See what says the
str.dump
operation, and then try to operate on that.