Support lambda operator and FILTER() in Snowflake

Fixes #789
sql-formatter-org · Oct 21, 2024 · 5d377ac · 5d377ac
1 parent 8672ca4
commit 5d377ac
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diff --git a/src/languages/snowflake/snowflake.formatter.ts b/src/languages/snowflake/snowflake.formatter.ts
@@ -330,6 +330,8 @@ export const snowflake: DialectOptions = {
       '=>',
       // Assignment https://docs.snowflake.com/en/sql-reference/snowflake-scripting/let
       ':=',
+      // Lambda: https://docs.snowflake.com/en/user-guide/querying-semistructured#lambda-expressions
+      '->',
     ],
     propertyAccessOperators: [':'],
   },

diff --git a/src/languages/snowflake/snowflake.functions.ts b/src/languages/snowflake/snowflake.functions.ts
@@ -184,6 +184,7 @@ export const functions: string[] = [
   'EXTRACT',
   'EXTRACT_SEMANTIC_CATEGORIES',
   'FACTORIAL',
+  'FILTER',
   'FIRST_VALUE',
   'FLATTEN',
   'FLOOR',

diff --git a/test/snowflake.test.ts b/test/snowflake.test.ts
@@ -204,4 +204,12 @@ describe('SnowflakeFormatter', () => {
         items;
     `);
   });
+
+  // Issue #771
+  it('supports lambda expressions', () => {
+    expect(format(`SELECT FILTER(my_arr, a -> a:value >= 50);`)).toBe(dedent`
+      SELECT
+        FILTER(my_arr, a -> a:value >= 50);
+    `);
+  });
 });