Ruby Which regular expression pattern to replace \n\n to \n string

Use squeeze function of String class

[18] pry(main)> "Java \n\n c# \n\n c/c++".squeeze("\n")
 => "Java \n c# \n c/c++"

However, it returns a new string where runs of the same character that occur in this set are replaced by a single character, So

 [18] pry(main)> "Java \n\n\n\n\n c# \n\n\n\n\n c/c++".squeeze("\n")
 => "Java \n c# \n c/c++"

You probably want to eliminate triple and so forth occurencies as well. In such a case the best option is to use the match counter:

#                                  ⇓⇓⇓⇓
'Java \n\n c# \n\n c/c++'.gsub /\\n{1,}/, '\n'

In this particular case, “one or more” has s syntactic sugar for it:

#                                  ⇓
'Java \n\n c# \n\n c/c++'.gsub /\\n+/, '\n'

If you are using Ruby2, there is \R match to match any combination of \r and \n.

To eliminate exactly two occurencies, one might use:

#                                  ⇓⇓⇓
'Java \n\n c# \n\n c/c++'.gsub /\\n{2}/, '\n'

And, finally, there is a function to remove multiple occurencies of \n from the string using named matches and backreferences:

def singlify s
  s.gsub /(?<sym>\\n)\g<sym>+/, '\k<sym>'
end

singlify 'Java \n\n c# \n\n c/c++'
# Java \n c# \n c/c++'

'Java \n\n c# \n\n c/c++'.gsub! /\\n\\n/, '\n'