如何使用SQL SELECT填充缺失的日期并保持值的总和？

CREATE TABLE dates (date_value DATE);

INSERT INTO dates (date_value)
SELECT DATE('2022-01-01') + INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY
FROM (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS a
CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS b
CROSS JOIN (SELECT 0 AS a UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9) AS c
WHERE DATE('2022-01-01') + INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY <= '2022-01-31';

SELECT d.date_value, COALESCE(t.value_column, 0) AS value_column
FROM dates d
LEFT JOIN data_table t ON d.date_value = t.date_column;

SELECT date_value, SUM(value_column) AS total_value
FROM (
    SELECT d.date_value, COALESCE(t.value_column, 0) AS value_column
    FROM dates d
    LEFT JOIN data_table t ON d.date_value = t.date_column
) subquery
GROUP BY date_value;