How do I split an atom in Parslet? - ruby

Question

I'm building an SQL-like query language. I would like to be able to handle lists of items delimited by commas. I have successfully achieved this with this code:
class QueryParser < Parslet::Parser
rule(:space) { match('\s').repeat(1) }
rule(:space?) { space.maybe }
rule(:delimiter) { space? >> str(',') >> space? }
rule(:select) { str('SELECT') >> space? }
rule(:select_value) { str('*') | match('[a-zA-Z]').repeat(1) }
rule(:select_arguments) do
space? >>
(select_value >> (delimiter >> select_value).repeat).maybe.as(:select) >>
space?
end
rule(:from) { str('FROM') >> space? }
rule(:from_arguments) { match('[a-zA-Z]').repeat(1).as(:from) >> space? }
rule(:query) { select >> select_arguments >> from >> from_arguments }
root(:query)
end
Where something like SELECT id,name,fork FROM forks correctly outputs the {:select=>"id,name,fork"#7, :from=>"forks"#25} tree.
Now, instead of messing around with this later, I would like to be able to convert the SELECT arguments (id,name,fork in this case) into an Array. I can do this by running 'id,name,fork'.split ','. I cannot get the Parslet transformer to do this for me when applied. This my code for my query transformer:
class QueryTransformer < Parslet::Transform
rule(select: simple(:args)) { args.split(',') }
end
When applied like so:
QueryTransformer.new.apply(
QueryParser.new.parse('SELECT id,name,fork FROM forks')
)
The result is the same as when I didn't apply it: {:select=>"id,name,fork"#7, :from=>"forks"#25}.
The value I was hoping :select to be is an Array like this ["id","name","fork"].
My question is: how do I split the value of :select into an Array using transformers?

Answer 1

You need to put "as(:xxx)" on whatever part of the parse tree you want to be able to play with later.
Here I changed your rule(:select_value) to remember the values as a :value
rule(:select_value) { (str('*') | match('[a-zA-Z]').repeat(1)).as(:value) }
Now your parser outputs :
{:select=>[{:value=>"id"#7}, {:value=>"name"#10}, {:value=>"fork"#15}], :from=>"forks"#25}
Which is easy to transform using:
class QueryTransformer < Parslet::Transform
rule(:value => simple(:val)) { val }
end
Then you get:
{:select=>["id"#7, "name"#10, "fork"#15], :from=>"forks"#25}
So in full the code is as follows :-
require 'parslet'
class QueryParser < Parslet::Parser
rule(:space) { match('\s').repeat(1) }
rule(:space?) { space.maybe }
rule(:delimiter) { space? >> str(',') >> space? }
rule(:select) { str('SELECT') >> space? }
rule(:select_value) { (str('*') | match('[a-zA-Z]').repeat(1)).as(:value) }
rule(:select_arguments) do
space? >>
(select_value >> (delimiter >> select_value).repeat).maybe.as(:select) >>
space?
end
rule(:from) { str('FROM') >> space? }
rule(:from_arguments) { match('[a-zA-Z]').repeat(1).as(:from) >> space? }
rule(:query) { select >> select_arguments >> from >> from_arguments }
root(:query)
end
puts QueryParser.new.parse('SELECT id,name,fork FROM forks')
# => {:select=>[{:value=>"id"#7}, {:value=>"name"#10}, {:value=>"fork"#15}], :from=>"forks"#25}
class QueryTransformer < Parslet::Transform
rule(:value => simple(:val)) { val }
end
puts QueryTransformer.new.apply(
QueryParser.new.parse('SELECT id,name,fork FROM forks')
)
# => {:select=>["id"#7, "name"#10, "fork"#15], :from=>"forks"#25}