How do I remove carriage returns with Ruby?
RubyRegexRuby Problem Overview
I thought this code would work, but the regular expression doesn't ever match the \r\n. I have viewed the data I am reading in a hex editor and verified there really is a hex D and hex A pattern in the file.
I have also tried the regular expressions /\xD\xA/m and /\x0D\x0A/m but they also didn't match.
This is my code right now:
lines2 = lines.gsub( /\r\n/m, "\n" )
if ( lines == lines2 )
print "still the same\n"
else
print "made the change\n"
end
In addition to alternatives, it would be nice to know what I'm doing wrong (to facilitate some learning on my part). :)
Ruby Solutions
Solution 1 - Ruby
Use String#strip
> Returns a copy of str with leading and trailing whitespace removed.
e.g
" hello ".strip #=> "hello"
"\tgoodbye\r\n".strip #=> "goodbye"
Using gsub
string = string.gsub(/\r/," ")
string = string.gsub(/\n/," ")
Solution 2 - Ruby
Generally when I deal with stripping \r or \n, I'll look for both by doing something like
lines.gsub(/\r\n?/, "\n");
I've found that depending on how the data was saved (the OS used, editor used, Jupiter's relation to Io at the time) there may or may not be the newline after the carriage return. It does seem weird that you see both characters in hex mode. Hope this helps.
Solution 3 - Ruby
If you are using Rails, there is a squish
method
"\tgoodbye\r\n".squish => "goodbye"
"\tgood \t\r\nbye\r\n".squish => "good bye"
Solution 4 - Ruby
What do you get when you do puts lines
? That will give you a clue.
By default File.open
opens the file in text mode, so your \r\n
characters will be automatically converted to \n
. Maybe that's the reason lines
are always equal to lines2
. To prevent Ruby from parsing the line ends use the rb
mode:
C:> copy con lala.txt a file with many lines ^ZC:> irb irb(main):001:0> text = File.open('lala.txt').read => "a\nfile\nwith\nmany\nlines\n" irb(main):002:0> bin = File.open('lala.txt', 'rb').read => "a\r\nfile\r\nwith\r\nmany\r\nlines\r\n" irb(main):003:0>
But from your question and code I see you simply need to open the file with the default modifier. You don't need any conversion and may use the shorter File.read
.
Solution 5 - Ruby
modified_string = string.gsub(/\s+/, ' ').strip
Solution 6 - Ruby
lines2 = lines.split.join("\n")
Solution 7 - Ruby
"still the same\n".chomp
or
"still the same\n".chomp!
http://www.ruby-doc.org/core-1.9.3/String.html#method-i-chomp
Solution 8 - Ruby
How about the following?
irb(main):003:0> my_string = "Some text with a carriage return \r"
=> "Some text with a carriage return \r"
irb(main):004:0> my_string.gsub(/\r/,"")
=> "Some text with a carriage return "
irb(main):005:0>
Or...
irb(main):007:0> my_string = "Some text with a carriage return \r\n"
=> "Some text with a carriage return \r\n"
irb(main):008:0> my_string.gsub(/\r\n/,"\n")
=> "Some text with a carriage return \n"
irb(main):009:0>
Solution 9 - Ruby
I think your regex is almost complete - here's what I would do:
lines2 = lines.gsub(/[\r\n]+/m, "\n")
In the above, I've put \r and \n into a class (that way it doesn't matter in which order they might appear) and added the "+" qualifier (so that "\r\n\r\n\r\n" would also match once, and the whole thing replaced with "\n")
Solution 10 - Ruby
Just another variant:
lines.delete(" \n")
Solution 11 - Ruby
Why not read the file in text mode, rather than binary mode?
Solution 12 - Ruby
lines.map(&:strip).join(" ")
Solution 13 - Ruby
You can use this :
my_string.strip.gsub(/\s+/, ' ')
Solution 14 - Ruby
def dos2unix(input)
input.each_byte.map { |c| c.chr unless c == 13 }.join
end
remove_all_the_carriage_returns = dos2unix(some_blob)