红宝石参数化正则表达式

Question

I have a string like "{some|words|are|here}" or "{another|set|of|words}"

So in general the string consists of an opening curly bracket,words delimited by a pipe and a closing curly bracket.

What is the most efficient way to get the selected word of that string ?

I would like do something like this:

@my_string = "{this|is|a|test|case}"
@my_string.get_column(0) # => "this"
@my_string.get_column(2) # => "is"
@my_string.get_column(4) # => "case"

What should the method get_column contain ?

Answer 1

So this is the solution I like right now:

class String
  def get_column(n)
    self =~ /\A\{(?:\w*\|){#{n}}(\w*)(?:\|\w*)*\}\Z/ && $1
  end
end

We use a regular expression to make sure that the string is of the correct format, while simultaneously grabbing the correct column.

Explanation of regex:

• \\A is the beginnning of the string and \\Z is the end, so this regex matches the enitre string.
• Since curly braces have a special meaning we escape them as \\{ and \\} to match the curly braces at the beginning and end of the string.
• next, we want to skip the first n columns - we don't care about them.
  • A previous column is some number of letters followed by a vertical bar, so we use the standard \\w to match a word-like character (includes numbers and underscore, but why not) and * to match any number of them. Vertical bar has a special meaning, so we have to escape it as \\| . Since we want to group this, we enclose it all inside non-capturing parens (?:\\w*\\|) (the ?: makes it non-capturing).
  • Now we have n of the previous columns, so we tell the regex to match the column pattern n times using the count regex - just put a number in curly braces after a pattern. We use standard string substition, so we just put in {#{n}} to mean "match the previous pattern exactly n times.
• the first non skipped column after that is the one we care about, so we put that in capturing parens: (\\w*)
• then we skip the rest of the columns, if any exist: (?:\\|\\w*)* .

Capturing the column puts it into $1 , so we return that value if the regex matched. If not, we return nil, since this String has no n th column.

In general, if you wanted to have more than just words in your columns (like "{a phrase or two|don't forget about punctuation!|maybe some longer strings that have\\na newline or two?}" ), then just replace all the \\w in the regex with [^|{}] so you can have each column contain anything except a curly-brace or a vertical bar.

---

Here's my previous solution

class String
  def get_column(n)
    raise "not a column string" unless self =~ /\A\{\w*(?:\|\w*)*\}\Z/
    self[1 .. -2].split('|')[n]
  end
end

We use a similar regex to make sure the String contains a set of columns or raise an error. Then we strip the curly braces from the front and back (using self[1 .. -2] to limit to the substring starting at the first character and ending at the next to last), split the columns using the pipe character (using .split('|') to create an array of columns), and then find the n'th column (using standard Array lookup with [n] ).

I just figured as long as I was using the regex to verify the string, I might as well use it to capture the column.