I have a script written in ruby. I need to remove any duplicate newlines (e.g.)
\n
\n
\n
to
\n
My current attempt worked (or rather not) using
str.gsub!(/\n\n/, "\n")
Which gave me no change to the output. What am I doing wrong?
This works for me:
#!/usr/bin/ruby
$s = "foo\n\n\nbar\nbaz\n\n\nquux";
puts $s
$s.gsub!(/[\n]+/, "\n");
puts $s
Use the more idiomatic String#squeeze instead of gsub.
str = "a\n\n\nb\n\n\n\n\n\nc"
str.squeeze("\n") # => "a\nb\nc"
You need to match more than one newline up to an infinite amount. Your code example will work with just a minor tweak:
str.gsub!(/\n+/, "\n")
For example:
str = "this\n\n\nis\n\n\n\n\na\ntest"
str.gsub!(/\n+/, "\n") # => "this\nis\na\ntest"
are you sure it shouldn't be /\n\n\n/, "\n" that what you seem to be wanting in your question above.
also, are you sure it's not doing a windows new-line "\r\n"?
EDIT: Additional info
Per Comment
"The amount of newlines can change. Different lines have between 2 and 5 newlines."
if you only want to hit the 2-5 lines try this
/\n{2,5}/, "\n"
Simply splitting and recombining the lines will give the desired result
>> "one\ntwo\n\nthree\n".split.join("\n")
=> "one\ntwo\nthree"
Edit: I just noticed this will replace ALL whitespace substrings with newlines, e.g.
>> "one two three\n".split.join("\n")
=> "one\ntwo\nthree"
First check that this is what you want!
Simply calling split will also trim out all of your whitespace.
You need to pass \n to split
>> "one ok \ntwo\n\nthree\n".split(/\n+/).join("\n")
=> "one ok \ntwo\nthree"
Additionally, also works with
spaces on blank lines
n number of back to back blank lines
str.gsub! /\n^\s*\n/, "\n\n"
where,
\n is of course newline
\s is space
denotes 1 or more spaces along when used after \s
Try This It Worked for me:
s = test\n\n\nbar\n\n\nfooo
s.gsub("\n\n", '')
Ruby needs the backslashes escaped differently than you have provided.
str.sub!("\\\\n+\\\\n","\\\\n")
http://www.ruby-forum.com/topic/176239
Related
I wants to echo \n like in this way using ruby language:
`echo "array = text.split(/\n/)" > demo`
and demo file should contain output in this way:
array = text.split(/\n/)
But instead to above i am getting in this way, \n produce new line in real which i dont want:
array = text.split(/
/)
Please let me know any options, I already tried backslash and ^ but none of this useful.
If you want to stick with regexps, double-backslash your \n
`echo "array = text.split(/\\n/)" > demo`
You could use split("\n") instead of a regexp.
There are 6 backslash needed here, Like this:
`echo "array = text.split(/\\\\\\n/)" > demo`
This produce excellent results.
I have a string
"So on and so forth $5.99"
I would like to extract everything after the $ until the end of the line.
/$ finds the character $. How do I select the rest of the string? I know it's something \z but I can't get the syntax right.
In regexp $ represents the end of the line.
So in your case you need \$.*$ To include your escaped $ and everything (.*) up until the end of the line $.
No, /$ does not match that character. You need to escape it \ to match a literal.
string = "So on and so forth $5.99"
result = string.match(/\$(.*)$/)
puts result[1] #=> "5.99"
If you want to capture everything after the $, you'll want:
/\$(.*)\z/
See http://rubular.com/r/T4fR1SEl3j
str = "This\n is a sample text for test"
str.scan(/\S.{0,15}\S(?=\s|$)|\S+/)
# => ["This", "is a sample text", "for test"]
Here, it splits when the newline (\n) is present. I actually want the output as,
["This\n is a", "sample text for", "test"]
How can I achieve that?
Use the /m modifier which allows the dot to match newlines:
str.scan(/\S.{0,15}\S(?=\s|\z)|\S+/m)
Also, I suggest you use \z instead of $ because $ matches the end of a line; \z is the only way to force Ruby to match the end of the string. It doesn't matter in this example, but it's a good habit to get into. Ruby differs from all other regex flavors in these two points.
This question is kinda simple (don't be so harsh with me), but I can't get a code-beautiful solution. I have the following code:
ARGF.each_line do |line|
arguments = line.split(',')
arguments.each do |task|
puts "#{task} result"
end
end
It simply read from the standard input numbers. I use it this way:
echo "1,2,3" | ruby prog.rb
The output desired is
1 result
2 result
3 result
But the actual output is
1 result
2 result
3
result
It seems like there's a newline character introduced. I'm skipping something?
Each line ends in a newline character, so splitting on commas in your example means that the last token is 3\n. Printing this prints 3 and then a newline.
Try using
arguments = line.chomp.split(',')
To remove the trailing newlines before splitting.
Your stdin input includes a trailing newline character. Try calling line.chomp! as the first instruction in your each_line block.
cud any body tell me how this expression works
output = "#{output.gsub(/grep .*$/,'')}"
before that opearation value of ouptput is
"df -h | grep /mnt/nand\r\n/dev/mtdblock4 248.5M 130.7M 117.8M 53% /mnt/nand\r\n"
but after opeartion it comes
"df -h | \n/dev/mtdblock4 248.5M 248.5M 130.7M 117.8M 53% /mnt/nand\r\n "
plzz help me
Your expression is equivalent to:
output.gsub!(/grep .*$/,'')
which is much easier to read.
The . in the regular expression matches all characters except newline by default. So, in the string provided, it matches "grep /mnt/nand", and will substitute a blank string for that. The result is the provided string, without the matched substring.
Here is a simpler example:
"hello\n\n\nworld".gsub(/hello.*$/,'') => "\n\n\nworld"
In both your provided regex, and the example above, the $ is not necessary. It is used as an anchor to match the end of a line, but since the pattern immediately before it (.*) matches everything up to a newline, it is redundant (but does not cause harm).
Since gsub returns a string, your first line is exactly the same as
output = output.gsub(/grep .*$/, '')
which takes the string and removes any occurance of the regexp pattern
/grep .*$/
i.e. all parts of the string that start with 'grep ' until the end of the string or a line break.
There's a good regexp tester/reference here. This one matches the word "grep", then a space, then any number of characters until the next line-break (\r or \n). "." by itself means any character, and ".*" together means any number of them, as many as possible. "$" means the end of a line.
For the '$', see here http://www.regular-expressions.info/reference.html
".*$" means "take every character from the end of the string" ; but the parser will interpret the "\n" as the end of a line, so it stops here.