How to keep the Carriage Return characters in string which returned by regular expression

Ruby
1

How can make below regular expression treat the Carriage Return (\r)
like normal character? i.e. let the “puts $1” output Who uses these
\ranyways? To escape the \r in str is one way, I’d like to know is there
a way to let regular expression just treat the \r as normal characters?

str = “Who uses these \ranyways?”
str =~ /(.*)/
puts $1

2

Ying Sun wrote in post #1166006:

How can make below regular expression treat the Carriage Return (\r)
like normal character? i.e. let the “puts $1” output Who uses these
\ranyways? To escape the \r in str is one way, I’d like to know is there
a way to let regular expression just treat the \r as normal characters?

str = “Who uses these \ranyways?”
str =~ /(.*)/
puts $1

On *nix it is a normal character:

irb(main):003:0> s=“Who uses these \ranyways?”
=> “Who uses these \ranyways?”
irb(main):004:0> puts s
anyways? these
=> nil
irb(main):005:0> s[/./]
=> “Who uses these \ranyways?”
irb(main):006:0> s[/./m]
=> “Who uses these \ranyways?”

On what platform are you? If on Windows or Mac maybe you have to use
regex option /m like I did in the last example.

3

Hi Robert,

Thanks for your replay, my platform is Windows7, I added /m option, but
the result still same.

Robert K. wrote in post #1166015:

Ying Sun wrote in post #1166006:

How can make below regular expression treat the Carriage Return (\r)
like normal character? i.e. let the “puts $1” output Who uses these
\ranyways? To escape the \r in str is one way, I’d like to know is there
a way to let regular expression just treat the \r as normal characters?

str = “Who uses these \ranyways?”
str =~ /(.*)/
puts $1

On *nix it is a normal character:

irb(main):003:0> s=“Who uses these \ranyways?”
=> “Who uses these \ranyways?”
irb(main):004:0> puts s
anyways? these
=> nil
irb(main):005:0> s[/./]
=> “Who uses these \ranyways?”
irb(main):006:0> s[/./m]
=> “Who uses these \ranyways?”

On what platform are you? If on Windows or Mac maybe you have to use
regex option /m like I did in the last example.

4

Robert K. wrote in post #1166218:

Ying Sun wrote in post #1166096:

Thanks for your replay, my platform is Windows7, I added /m option, but
the result still same.

Result of what and what is the result? Can you copy paste an IRB
session from the console like I did?

Hi Robert,

Sorry for the delay, IRB as below:

irb(main):001:0> str = “Who uses these \ranyways?”
=> “Who uses these \ranyways?”
irb(main):002:0> str =~ /(.*)/m
=> 0
irb(main):003:0> puts $1
anyways? these
=> nil
irb(main):004:0>

5

This has nothing to do with matching.

irb(main):006:0> str = “Who uses these \ranyways?”
=> “Who uses these \ranyways?”
irb(main):007:0> str =~ /(.*)/m
=> 0
irb(main):008:0> $1
=> “Who uses these \ranyways?”
irb(main):009:0> puts $1
anyways? these
=> nil

6

Ying Sun wrote in post #1166096:

Thanks for your replay, my platform is Windows7, I added /m option, but
the result still same.

Result of what and what is the result? Can you copy paste an IRB
session from the console like I did?