I got stuck in another situation: our users enter a text to be stored in a variable. The condition for that text is it can be allowed to enter only 25 characters, Now I have to write a regular expression which will check the condition, kindly help me out in this.
I think you could just use the String#length method...
http://ruby-doc.org/core-1.9.3/String.html#method-i-length
Example:
text = 'The quick brown fox jumps over the lazy dog.'
puts text.length > 25 ? 'Too many characters' : 'Accepted'
Ruby provides a built-in function for checking the length of a string. Say it's called s:
if s.length <= 25
# We're OK
else
# Too long
end
Instead of using a regular expression, just check if string.length > 25
Verification, do not forget the to_s
def nottolonng?(value)
if value.to_s.length <=8
return true
else
return false
end
end
You could take any of the answers above that use the string.length method and replace it with string.size.
They both work the same way.
if string.size <= 25
puts "No problem here!"
else
puts "Sorry too long!"
end
https://ruby-doc.org/core-2.4.0/String.html#method-i-size
Related
I'm trying to make users input a positive integer(>=1). If the user inputs a negative integer or 0, the program gonna notice and say you didn't put a valid integer, please try again. my program works so far until I try to input a string. My program break after I input a string. Does anybody know how to fix this?
I tried elsif !barsofsoap.match (/\A[-+]?[0-9]*.?[0-9]+\Z/)
puts "You didn't enter an integer, please put a positive integer", but it didn't work
loop do
puts "How many soaps are you looking to purchase? (Please put an integer)"
x = gets.chomp
barsofsoap = Integer(x) rescue false
if barsofsoap >= 1
break
else
puts "You didn't put an integer, please put an integer and try again."
end
end
I hope the program will not break if the user inputs a string.
The reason why your current code fails is because you use false as rescue value. You can't compare false >= 1. Sticking close to your original code you might wrap the comparison into a begin/end block.
loop do
puts "How many soaps are you looking to purchase? (Please put an integer)"
x = gets.chomp
begin
barsofsoap = Integer(x)
if barsofsoap >= 1
break
end
rescue ArgumentError => _error
puts "You didn't put an integer, please put an integer and try again."
end
end
Or simplifying the code:
loop do
puts "How many soaps are you looking to purchase? (Please put an integer)"
break if (barsofsoap = gets.to_i).positive?
puts "Invalid input. Provide a positive integer."
end
The above code doesn't detect if the input is a string or an integer. By replacing Integer(x) with x.to_i it would simply return 0 if invalid. Since you want the user to provide an positive integer it still requests them to provide a new integer.
nil.to_i #=> 0
''.to_i #=> 0
'asdf'.to_i #=> 0
'12.34'.to_i #=> 12
Like shown above the string '12.34' will produce the value 12. Depending on your requirements you might not want to use this value and instead mark it as invalid input. In this case you could use a regex to check the provided input.
loop do
puts "How many soaps are you looking to purchase? (Please put an integer)"
input = gets
break barsofsoap = input.to_i if input.match?(/\A[1-9]\d*\Z/)
puts "Invalid input. Provide a positive integer."
end
The regex /\A[1-9]\d*\Z/ will match the following:
\A Matches start of string.
[1-9] Matches one of the digits 1-9.
\d* Matches zero or more digits 0-9.
\Z Matches end of string. If the string ends with a newline, it matches just before the newline.
You try this: num.positive?
see here: https://ruby-doc.org/core-2.6/Numeric.html#method-i-positive-3F
Also you can try convert the text to an integer and validate by doing this:
num.is_a?(Integer)
https://ruby-doc.org/core-2.6/Object.html#method-i-is_a-3F
Or even better since input is always text, you can use a regex to make sure only numbers are entered: "12345" =~ /\A\d+\Z/
How about try this:
loop do
puts "How many soaps are you looking to purchase? (Please put an integer)"
barsofsoap = gets.chomp
if barsofsoap.match(/^[\d]+$/).nil?
puts "You didn't put an integer, please put an integer and try again."
elsif barsofsoap.to_i.positive?
break
else
puts "You didn't put an integer, please put an integer and try again."
end
end
Other option, using n.to_i.to_s == n to check the input is an Integer and moving the loop into a method:
def get_input_integer
loop do
puts "Enter a positive Integer"
n = gets.chomp
if n.to_i.to_s == n
return n.to_i if n.to_i.positive?
puts "It's not positive"
else
puts "It's not an integer"
end
end
end
x = get_input_integer
I am trying to halt input from a user when their input is 42.
The correct answer on the website I'm working on is:
while line = gets
break if (/42/ =~ line)
x << line
end
The code I tried that does not work is:
while line = gets.chomp
break if (line == 42)
x << line
end
Why is this the case? Am I missing some limitations to what I can use in my if statement?
The problem is that 42 is an integer, but line is a string:
1.9.3p392 :001 > "42" == 42
=> false
So it's never the case that your if statement is getting triggered, because it's comparing two different kinds of things. Matching with a Regex fixes it, though it's looking for "42" to appear anywhere in the input (e.g. "3427"). I think what you meant to say was
while line = gets.chomp
break if (line == "42")
x << line
end
In other words, break when the input is a string with the characters 4 and 2 in it.
I suspect it's because you're comparing a number to a string. The example uses a regular expression it appears. "42" == 42 will give you false in ruby.
<< is a method(Append) on Array or String class objects. But your x not holding any referencing to such objects. Thus you are getting undefined local variable or method 'x' for main:Object (NameError).
Try this instead(by fixing local variable x to hold a practical object and converting line value to Fixnum object:
x = "hi"
while line = gets.chomp
break if (line.to_i == 42)
x << line
end
This program will help you to halt input from a user when their input is 42.
until (line = gets.chomp).to_i == 42
x << line
end
This of course bypasses the if statement you were asking about.
Your limitation for the if is based solely on the fact that you are comparing a string that will always be a string to a number, and this will never be equal. (as others have mentioned)
So we must reconsider the conditional statement. In this case, I considered it "out of place" and moved the comparison to the 'while' loop, and then inverted it to an 'until' statement, to be able to positively express the condition to end the loop on. Whenever I see a 'break' in a loop, I try to get rid of that smell, as the condition to leave a loop should really be expressed in the loop condition if possible.
I hope this helps.
I'm trying to validate that the input a user gives my program via gets is an integer. is_a?(Integer) does not work, as far as i can tell, because gets gets a string from the user, so it will always return false even if the user enters an valid integer (in string form). One would think I could simply use to_i on the input and be done with it, but that raises another issue - "75akjfas".to_i results in 75. So if I relied on to_i to solve my problems, anything starting with numbers will work.
How do I cleanly validate that the value is an integer only, and not a combination of numbers and letters? Or do I need to resort to regex for this?
print "Enter an integer (or try to sneak by something other): "
puts Integer(gets) rescue puts "Hey, that's not an integer!"
How about s.to_i.to_s == s? I'd prefer regex however.
Using regex you could do it like this:
class String
def integer?
!!(self =~ /^[-+]?[0-9]+$/)
end
end
You could use Integer() but you need to be careful with hex, binary, or octal input. http://ruby-doc.org/core-2.3.1/Kernel.html#method-i-Integer
def valid_integer?(string)
begin
!!Integer(string)
rescue ArgumentError
false
end
end
Check this code example for how to use the checked string input by the user
puts "Enter a number: "
user_input = gets.chomp
check = (user_input.to_i.to_s == user_input)
while (!check ) do
puts("Wrong Input, Pls, Enter a number: " )
user_input = gets.chomp
check = (user_input.to_i.to_s == user_input)
end
if user_input.to_i < 0
puts "Number is negative"
elsif user_input.to_i > 0
puts "Number is positve"
else
puts "Number is Zero"
end
Ruby can do it without esoteric solutions:
Integer is a integer
1970.is_a?Integer
true
String is not a integer
"1970".is_a?Integer
false
String to integer is a integer
"1970".to_i.is_a?Integer
true
I need to check to see if a string contains at least one number in it using Ruby (and I assume some sort of regex?).
How would I do that?
You can use the String class's =~ method with the regex /\d/ as the argument.
Here's an example:
s = 'abc123'
if s =~ /\d/ # Calling String's =~ method.
puts "The String #{s} has a number in it."
else
puts "The String #{s} does not have a number in it."
end
Alternatively, without using a regex:
def has_digits?(str)
str.count("0-9") > 0
end
if /\d/.match( theStringImChecking ) then
#yep, there's a number in the string
end
Rather than use something like "s =~ /\d/", I go for the shorter s[/\d/] which returns nil for a miss (AKA false in a conditional test) or the index of the hit (AKA true in a conditional test). If you need the actual value use s[/(\d)/, 1]
It should all work out the same and is largely a programmer's choice.
!s[/\d/].nil?
Can be a standalone function -
def has_digits?(s)
return !s[/\d/].nil?
end
or ... adding it to the String class makes it even more convenient -
class String
def has_digits?
return !self[/\d/].nil?
end
end
I have a string, say '123', and I want to convert it to the integer 123.
I know you can simply do some_string.to_i, but that converts 'lolipops' to 0, which is not the effect I have in mind. I want it to blow up in my face when I try to convert something invalid, with a nice and painful Exception. Otherwise, I can't distinguish between a valid 0 and something that just isn't a number at all.
EDIT: I was looking for the standard way of doing it, without regex trickery.
Ruby has this functionality built in:
Integer('1001') # => 1001
Integer('1001 nights')
# ArgumentError: invalid value for Integer: "1001 nights"
As noted in answer by Joseph Pecoraro, you might want to watch for strings that are valid non-decimal numbers, such as those starting with 0x for hex and 0b for binary, and potentially more tricky numbers starting with zero that will be parsed as octal.
Ruby 1.9.2 added optional second argument for radix so above issue can be avoided:
Integer('23') # => 23
Integer('0x23') # => 35
Integer('023') # => 19
Integer('0x23', 10)
# => #<ArgumentError: invalid value for Integer: "0x23">
Integer('023', 10) # => 23
This might work:
i.to_i if i.match(/^\d+$/)
Also be aware of the affects that the current accepted solution may have on parsing hex, octal, and binary numbers:
>> Integer('0x15')
# => 21
>> Integer('0b10')
# => 2
>> Integer('077')
# => 63
In Ruby numbers that start with 0x or 0X are hex, 0b or 0B are binary, and just 0 are octal. If this is not the desired behavior you may want to combine that with some of the other solutions that check if the string matches a pattern first. Like the /\d+/ regular expressions, etc.
Another unexpected behavior with the accepted solution (with 1.8, 1.9 is ok):
>> Integer(:foobar)
=> 26017
>> Integer(:yikes)
=> 26025
so if you're not sure what is being passed in, make sure you add a .to_s.
I like Myron's answer but it suffers from the Ruby disease of "I no longer use Java/C# so I'm never going to use inheritance again". Opening any class can be fraught with danger and should be used sparingly, especially when it's part of Ruby's core library. I'm not saying don't ever use it, but it's usually easy to avoid and that there are better options available, e.g.
class IntegerInString < String
def initialize( s )
fail ArgumentError, "The string '#{s}' is not an integer in a string, it's just a string." unless s =~ /^\-?[0-9]+$/
super
end
end
Then when you wish to use a string that could be a number it's clear what you're doing and you don't clobber any core class, e.g.
n = IntegerInString.new "2"
n.to_i
# => 2
IntegerInString.new "blob"
ArgumentError: The string 'blob' is not an integer in a string, it's just a string.
You can add all sorts of other checks in the initialize, like checking for binary numbers etc. The main thing though, is that Ruby is for people and being for people means clarity. Naming an object via its variable name and its class name makes things much clearer.
I had to deal with this in my last project, and my implementation was similar, but a bit different:
class NotAnIntError < StandardError
end
class String
def is_int?
self =~ /^-?[0-9]+$/
end
def safe_to_i
return self.to_i if is_int?
raise NotAnIntError, "The string '#{self}' is not a valid integer.", caller
end
end
class Integer
def safe_to_i
return self
end
end
class StringExtensions < Test::Unit::TestCase
def test_is_int
assert "98234".is_int?
assert "-2342".is_int?
assert "02342".is_int?
assert !"+342".is_int?
assert !"3-42".is_int?
assert !"342.234".is_int?
assert !"a342".is_int?
assert !"342a".is_int?
end
def test_safe_to_i
assert 234234 == 234234.safe_to_i
assert 237 == "237".safe_to_i
begin
"a word".safe_to_i
fail 'safe_to_i did not raise the expected error.'
rescue NotAnIntError
# this is what we expect..
end
end
end
someString = "asdfasd123"
number = someString.to_i
if someString != number.to_s
puts "oops, this isn't a number"
end
Probably not the cleanest way to do it, but should work.
Re: Chris's answer
Your implementation let's things like "1a" or "b2" through. How about this instead:
def safeParse2(strToParse)
if strToParse =~ /\A\d+\Z/
strToParse.to_i
else
raise Exception
end
end
["100", "1a", "b2", "t"].each do |number|
begin
puts safeParse2(number)
rescue Exception
puts "#{number} is invalid"
end
end
This outputs:
100
1a is invalid
b2 is invalid
t is invalid