首先,我想提出您原始查询的问题
主要问题是 1) 使用 LEFT JOIN ; 2) 使用 col != 0
以下是如何修改它以真正捕获两个表的所有差异
运行您的原始查询并低于一个 - 希望您会看到差异
#standardSQL
SELECT key1, key2
FROM
(
SELECT
IFNULL(table1.key1, table2.key1) key1,
IFNULL(table1.key2, table2.key2) key2,
table1.column1 - table2.column1 AS col1,
table1.col2 - table2.col2 AS col2
FROM `table1` AS table1
FULL OUTER JOIN `table2` AS table2
ON table1.key1 = table2.key1
AND table1.key2 = table2.key2
)
WHERE IFNULL(col1, 1) != 0
OR IFNULL(col2, 1) != 0
或者您可以尝试针对虚拟数据运行原始版本和更高版本以查看差异
#standardSQL
WITH `table1` AS (
SELECT 1 key1, 1 key2, 1 column1, 2 col2 UNION ALL
SELECT 2, 2, 3, 4 UNION ALL
SELECT 3, 3, 5, 6
), `table2` AS (
SELECT 1 key1, 1 key2, 1 column1, 29 col2 UNION ALL
SELECT 2, 2, 3, 4 UNION ALL
SELECT 4, 4, 7, 8
)
SELECT key1, key2
FROM
(
SELECT
IFNULL(table1.key1, table2.key1) key1,
IFNULL(table1.key2, table2.key2) key2,
table1.column1 - table2.column1 AS col1,
table1.col2 - table2.col2 AS col2
FROM `table1` AS table1
FULL OUTER JOIN `table2` AS table2
ON table1.key1 = table2.key1
AND table1.key2 = table2.key2
)
WHERE IFNULL(col1, 1) != 0
OR IFNULL(col2, 1) != 0
其次,下面将高度简化您的整体查询
#standardSQL
SELECT
IFNULL(table1.key1, table2.key1) key1,
IFNULL(table1.key2, table2.key2) key2
FROM `table1` AS table1
FULL OUTER JOIN `table2` AS table2
ON table1.key1 = table2.key1
AND table1.key2 = table2.key2
WHERE TO_JSON_STRING(table1) != TO_JSON_STRING(table2)
您可以使用与上述相同的虚拟数据示例对其进行测试
注意:在此解决方案中,您不需要选择特定列 - 它只是比较所有列!但是如果您只需要比较特定的列 - 您仍然需要像下面的示例中那样挑选它们
#standardSQL
SELECT
IFNULL(table1.key1, table2.key1) key1,
IFNULL(table1.key2, table2.key2) key2
FROM `table1` AS table1
FULL OUTER JOIN `table2` AS table2
ON table1.key1 = table2.key1
AND table1.key2 = table2.key2
WHERE TO_JSON_STRING((table1.column1, table1.col2)) != TO_JSON_STRING((table2.column1, table2.col2))