Ruby正则表达式行尾

Question

I am trying to find the variables in a string, eg

"%0" can not be found. %1 Please try again %2

I need to know how each variable ends (space, period, end of line) cause I will check for the existence of same variable in the translated version of this string. Text comes from a CSV and strings do not end with a line break.

I am able to capture them all except the ones at the end of a string with:

reg = /[%@!][^\\s]+[\\s\\.\\z$]+/

I thought either $ or \\z should match end of line but that does not seem to work. How can I capture %2 in the above scenario? (again, there is no line break at the end)

Answer 1

$ matches end-of-line, but not when used inside brackets like that. Writing [$] is how you would look for the normal dollar-sign character '$'.

If the string you are searching is the exact string you listed above, try

reg = /^"(.*)" can not be found[.] (.*) Please try again (.*)$/
error_string =~ reg

Your three matching results will be stored in the special variables $1 , $2 , and $3 .

Answer 2

Okay, I solved it with a different approach. Using a positive lookahead works as the character class is not needed

/[%@!][\w]+(?=\s|\z|\.|\W)/

For the example string, this returns:

%0

%1

%2