snowflake: PIVOT on derived table factors (apache#1027)

Author: lustefaniak

src/parser/mod.rs

@@ -6737,9 +6737,20 @@ impl<'a> Parser<'a> {
             // `parse_derived_table_factor` below will return success after parsing the
             // subquery, followed by the closing ')', and the alias of the derived table.
             // In the example above this is case (3).
-            return_ok_if_some!(
+            if let Some(mut table) =
                 self.maybe_parse(|parser| parser.parse_derived_table_factor(NotLateral))
-            );
+            {
+                while let Some(kw) = self.parse_one_of_keywords(&[Keyword::PIVOT, Keyword::UNPIVOT])
+                {
+                    table = match kw {
+                        Keyword::PIVOT => self.parse_pivot_table_factor(table)?,
+                        Keyword::UNPIVOT => self.parse_unpivot_table_factor(table)?,
+                        _ => unreachable!(),
+                    }
+                }
+                return Ok(table);
+            }
+
             // A parsing error from `parse_derived_table_factor` indicates that the '(' we've
             // recently consumed does not start a derived table (cases 1, 2, or 4).
             // `maybe_parse` will ignore such an error and rewind to be after the opening '('.

tests/sqlparser_snowflake.rs

@@ -1140,3 +1140,10 @@ fn parse_division_correctly() {
         "SELECT tbl1.field / tbl2.field FROM tbl1 JOIN tbl2 ON tbl1.id = tbl2.entity_id",
     );
 }
+
+#[test]
+fn parse_pivot_of_table_factor_derived() {
+    snowflake().verified_stmt(
+        "SELECT * FROM (SELECT place_id, weekday, open FROM times AS p) PIVOT(max(open) FOR weekday IN (0, 1, 2, 3, 4, 5, 6)) AS p (place_id, open_sun, open_mon, open_tue, open_wed, open_thu, open_fri, open_sat)"
+    );
+}