Merge pull request #211 from quinnj/jq/fix_nested_joins

For nested joins, there was an issue where calls to the inner most so…

src/query_translation.jl

@@ -124,7 +124,6 @@ function query_expression_translation_phase_1(qe)
 			deleteat!(sub_query.args[end].args,6)
 
 			translate_query(sub_query)
-
 			length(sub_query.args)==1 || throw(QueryException("@group ... into subquery too long", sub_query))
 
 			qe[1] = :( @from $x in $(sub_query.args[1]) )
@@ -387,13 +386,31 @@ function replace_transparent_identifier_in_anonym_func(ex::Expr, names_to_put_in
 	end
 end
 
+# translates (:a, :((b.c).d)) to :(((a.b).c).d)
+function shift_access!(sym::Symbol, ex::Expr)
+	while ex.args[1] isa Expr
+		ex = ex.args[1]
+	end
+	ex.args[1] = Expr(:., sym, QuoteNode(ex.args[1]))
+	return
+end
+
 function find_names_to_put_in_scope(ex::Expr)
 	names = []
 	for child_ex in ex.args[2:end]
 		if isa(child_ex,Expr) && child_ex.head==:transparentidentifier
 			child_names = find_names_to_put_in_scope(child_ex)
 			for child_name in child_names
-				push!(names, (Expr(:., ex.args[1], QuoteNode(child_name[1])), child_name[2]))
+				c = child_name[1]
+				if c isa Symbol
+					xx = (Expr(:., ex.args[1], QuoteNode(c)), child_name[2])
+				elseif c.args[1] isa Symbol
+					xx = (Expr(:., Expr(:., ex.args[1], QuoteNode(c.args[1])), c.args[2]), child_name[2])
+				else
+					shift_access!(ex.args[1], c)
+					xx = (c, child_name[2])
+				end
+				push!(names, xx)
 			end
 		elseif isa(child_ex, Symbol)
 			push!(names,(ex.args[1],child_ex))

test/runtests.jl

@@ -32,6 +32,14 @@ end
         @test !Query.iscall(:(map(1,2,3,4)), :map, 3)
     end
 
+    @testset "shift_access!" begin
+        ex = Expr(:., :c, QuoteNode(:d))
+        Query.shift_access!(:b, ex)
+        @test ex == :((b.c).d)
+        Query.shift_access!(:a, ex)
+        @test ex == :(((a.b).c).d)
+    end
+
 source_df = DataFrame(name=["John", "Sally", "Kirk"], age=[23., 42., 59.], children=[3,5,2])
 
 q = @from i in source_df begin
@@ -455,6 +463,18 @@ end
 @test q["Sally"]==5
 @test q["Kirk"]==2
 
+q = @from i in source_df begin
+    @let j = i.name
+    @let k = i.children
+    @let l = i.age
+    @select {a=j, b=k, c=l}
+    @collect DataFrame
+end
+
+@test q.a == ["John", "Sally", "Kirk"]
+@test q.b == [3, 5, 2]
+@test q.c == [23., 42., 59.]
+
 @test @count(source_df)==3
 @test @count(source_df, i->i.children>3)==1