I'm currently learning Ruby, and am enjoying most everything except a small string comparason issue.
answer = gets()
if (answer == "M")
print("Please enter how many numbers you'd like to multiply: ")
elsif (answer. == "A")
print("Please enter how many numbers you'd like to sum: ")
else
print("Invalid answer.")
print("\n")
return 0
end
What I'm doing is I'm using gets() to test whether the user wants to multiply their input or add it (I've tested both functions; they work), which I later get with some more input functions and float translations (which also work).
What happens is that I enter A and I get "Invalid answer."The same happens with M.
What is happening here? (I've also used .eql? (sp), that returns bubcus as well)
gets returns the entire string entered, including the newline, so when they type "M" and press enter the string you get back is "M\n". To get rid of the trailing newline, use String#chomp, i.e replace your first line with answer = gets.chomp.
The issue is that Ruby is including the carriage return in the value.
Change your first line to:
answer = gets().strip
And your script will run as expected.
Also, you should use puts instead of two print statements as puts auto adds the newline character.
your answer is getting returned with a carriage return appended. So input "A" is never equal to "A", but "A(return)"
You can see this if you change your reject line to print("Invalid answer.[#{answer}]"). You could also change your comparison to if (answer.chomp == ..)
I've never used gets put I think if you hit enter your variable answer will probably contain the '\n' try calling .chomp to remove it.
Add a newline when you check your answer...
answer == "M\n"
answer == "A\n"
Or chomp your string first: answer = gets.chomp
Related
Good morning,
I'm starting on Ruby, wanted to create a small tool that fetches my public IP and sends it over by email. I'm stumbling on a basic problem with a string comparison and an if/else block that won't process.
Code is quite simple (see below). The problem comes at the string comparison at line 21. What I'd want is that whenever the IP address changed from what was recorded in a file, the new IP overwrites the one in the file, and other actions (not in the current code) will ensue.
It looks like that sometimes the comparison is not executed, thus the statements following the if do not execute.
I'd like to understand why is that. Line 21 has been changed at times for if (!oldIP == ip) to if (oldIP != ip), same result. I also suspect the return value to be ignored (dead code path ?) sometimes.
Here's the code
#!/usr/bin/env ruby
require "net/http"
puts "\e[H\e[2J"
def FetchIPAddress()
oldIP = ""
if File::exists?('/tmp/wanipaddress.txt')
iFile = File.open('/tmp/wanipaddress.txt')
oldIP = iFile.read()
iFile.close()
end
oFile = File.open('/tmp/wanipaddress.txt', "w+")
ip = Net::HTTP.get(URI("https://api.ipify.org"))
puts "old = " + oldIP
puts "new = " + ip
if (!oldIP == ip)
puts "changed"
oFile.puts ip
oFile.close()
else
ip = "unchanged"
puts "unchanged"
end
return ip
end
Really, I do see some erratic behaviour here; I suspect it's just me being a newbie with Ruby.
Your file likely contains a line break.
Try this
if old_ip.chomp != ip.chomp
...
end
chomp removes a trailing linebreak.
Best use p to print values for debugging, this will escape whitespace and thus make trailing linebreaks visible. You should never use puts for debugging.
And here is why !a == b will never ever work.
!a == b is the same as a.!().==(b) and executed as follows
first the ! method is called on object a, which returns false for a string
then ==(b) method is called on the resulting boolean
and since a boolean is never equal to a string the comparison will always fail
The problem with this line (if (!oldIP == ip)) is pretty simple - what you want it to do is check whether the oldIP is different from the new IP.
What you do instead is take oldIP, check whether it's true or false, then negate it (that's what the ! does), then compare it to ip. Since oldIP is a String, and thus always true, which gets negated to false, and ip is (I'm guessing) a String, it will always be true, your line essentially reads if (false == true).
To solve this problem, you could use the != comparison operator, like if oldIP != ip, or, if you like the negation, if !(oldIP == ip).
I have 4 chars, first one is letter 'L' for example, the other two are numbers and the last one is letter again, all of them are separated by one space. User is entering them in the Ruby console. I need to check that they are separated by one space and don't have other weird characters and that there is nothing after the last letter.
So if a user enters for example gets.chomp = 'L 5 7 A', I need to check that everything is ok and separated by only one space and return input[1], input[2], input[3]. How can I do that? Thanks.
You can do something like this:
puts "Enter string"
input = gets.chomp
r = /^(L)\s(\d)\s(\d)\s([A-Z])$/
matches = input.match r
puts matches ? "inputs: #{$1}, #{$2}, #{$3}, #{$4}" : "input-format incorrect"
Here $1 is the first capture, similarly for $2, $3 etc. If you want to store the result in an array you can use:
matches = input.match(r).to_a
then the first element is the entire match, followed by each capture.
Try
/^\w\s(\d)\s(\d)\s(\w)$/
Rubular is a good sandbox site for experimenting with and debugging regexes.
I am creating a webserver that people connect to using telnet.
Some telnet-clients send data per-line, but others send each and every character as soon as it is typed. In the server, I cache these until a newline is typed. (And therefore we have a full line)
However, backspace characters, (\x08), are also sent and in the string. I would like to implement the backspace behaviour by replacing a character followed by a backspace with nothing.
However, what if a user uses backspace to remove more than one letter at a time?
is it possible to construct a regexp that removes an N amount of characters when followed by an N amount of backspaces?
For completeness sake, I'm using Ruby 2.2 .
Thanks.
You can achive that with a recursive regex with a subroutine call:
str = "\bNewqaz\b\b\b car!!!\b\b\b."
puts str.gsub(/^[\b]+|([^\b](\g<1>)*[\b])/, '').gsub(/[\b]/,'B')
# => New car.
Note that .gsub(/[\b]/,'B') part is used for demo only, to show there are no backspace symbols left. See IDEONE demo.
An alternative approach is
loop do
str = str.gsub(Regexp.new("^\b+|[^\b]\b"), "")
break if !str.include?("\b")
end
See IDEONE demo, turning \bNewqaz\b\b\b car!!!\b\b\b. to New car.:
str = "\bNewqaz\b\b\b car!!!\b\b\b."
puts str.gsub(/\x08/, "B")
loop do
str = str.gsub(Regexp.new("^\b+|[^\b]\b"), "")
break if !str.include?("\b")
end
puts str
A friend of mine is trying to explain to me the answer to this problem:
Define a method binary_multiple_of_4?(s) that takes a string and returns true if the string represents a binary number that is a multiple of 4.
However, his example he gave is this:
if (s) == "0"
return true
end
if /^[01]*(00)$/.match(s) #|| /^0$/.match(s)
return true
else
return false
end
It works, because the software we use says there were no errors, but I don't understand why, or what /^ means, and how it's used.
If you could also explain the /^0$/.match(s), that would be great too.
Thanks!
what he is doing is using regular expressions, see: http://www.tutorialspoint.com/ruby/ruby_regular_expressions.htm
To break it down, there is a pattern that is matched inside the slashes /pattern/ and every character means something. ^ means start of the line [01] means match a 0 or a 1, * means match the previous thing ([01]) zero or more times, and (00) means match 00, and $ means match the end of the line.
If you want to know what /^0$/ matches, you should definitely try to figure it out based on the information in my post or the link I provided. Here's the answer though (hover to view):
It matches the beginning of the line, zero, the end of a line.
Here's my regular expression that I have for this. I'm in Ruby, which — if I'm not mistaken — uses POSIX regular expressions.
regex = /(?:\n^)(\*[\w+ ?]+\*)\n/
Here's my goal: I want to split a string with a regex that is *delimited by asterisks*, including those asterisks. However: I only want to split by the match if it is prefaced with a newline character (\n), or it's the start of the whole string. This is the string I'm working with.
"*Friday*\nDo not *break here*\n*But break here*\nBut again, not this"
My regular expression is not splitting properly at the *Friday* match, but it is splitting at the *But break here* match (it's also throwing in a here split). My issue is somewhere in the first group, I think: (?:\n^) — I know it's wrong, and I'm not entirely sure of the correct way to write it. Can someone shed some light? Here's my complete code.
regex = /(?:\n^)(\*[\w+ ?]+\*)\n/
str = "*Friday*\nDo not *break here*\n*But break here*\nBut again, not this"
str.split(regex)
Which results in this:
>>> ["*Friday*\nDo not *break here*", "*But break here*", "But again, not this"]
I want it to be this:
>>> ["*Friday*", "Do not *break here*", "*But break here*", "But again, not this"]
Edit #1: I've updated my regex and result. (2011/10/18 16:26 CST)
Edit #2: I've updated both again. (16:32 CST)
What if you just add a '\n' to the front of each string. That simplifies the processing quite a bit:
regex = /(?:\n)(\*[\w+ ?]+\*)\n/
str = "*Friday*\nDo not *break here*\n*But break here*\nBut again, not this"
res = ("\n"+str).split(regex)
res.shift if res[0] == ""
res
=> [ "*Friday*", "Do not *break here*",
"*But break here*", "But again, not this"]
We have to watch for the initial extra match but it's not too bad. I suspect someone can shorten this a bit.
Groups 1 & 2 of the regex below :
(?:\A|\\n)(\*.*?\*)|(?:\A|\\n)(.*?)(?=\\n|\Z)
Will give you your desired output. I am no ruby expert so you will have to create the list yourself :)
Why not just split at newlines? From your example, it looks that's what you're really trying to do.
str.split("\n")