用于比较来自数据库的两个实例的数据的 PSQL 查询

Question

我有两个来自不同日子的同一个数据库实例。某一天的所有 table 称为 tableA*，另一天的所有 tableB*。我想比较数据以查看发生了什么变化。我想 select 所有不完全匹配的行。因此，例如，如果 tables tableA1 和 tableB1 中的一个值不同，我想 select 来自 table A 的相应行并标记它作为 'new' 和来自 table B 并将其标记为 'deleted'。我试过这样的查询：

SELECT 'new', ta1.name, ta2.name, ta3.name, ta4.name, ta5.name
FROM tableA1 ta1
LEFT JOIN tableA2 ta2 ON ta1.ta2_id = ta2.id
LEFT JOIN tableA3 ta3 ON ta1.ta3_id = ta3.id
LEFT JOIN tableA4 ta4 ON ta1.ta4_id = ta4.id
LEFT JOIN tableA5 ta5 ON ta5.ta1_id = ta1.id WHERE NOT EXISTS
(SELECT tb1.name, tb2.name, tb3.name, tb4.name, tb5.name
FROM tableB1 tb1
LEFT JOIN tableB2 tb2 ON tb1.tb2_id = tb2.id
LEFT JOIN tableB3 tb3 ON tb1.tb3_id = tb3.id
LEFT JOIN tableB4 tb4 ON tb1.tb4_id = tb4.id
LEFT JOIN tableB5 tb5 ON tb5.tb1_id = tb1.id WHERE 
tb1.name = ta1.name AND 
tb2.name = ta2.name AND 
tb3.name = ta3.name AND 
tb4.name = ta4.name AND 
tb5.name = ta5.name)
UNION
SELECT 'deleted', tb1.name, tb2.name, tb3.name, tb4.name, tb5.name
FROM tableB1 tb1
LEFT JOIN tableB2 tb2 ON tb1.tb2_id = tb2.id
LEFT JOIN tableB3 tb3 ON tb1.tb3_id = tb3.id
LEFT JOIN tableB4 tb4 ON tb1.tb4_id = tb4.id
LEFT JOIN tableB5 tb5 ON tb5.tb1_id = tb1.id WHERE NOT EXISTS
(SELECT ta1.name, ta2.name, ta3.name, ta4.name, ta5.name
FROM tableA1 ta1
LEFT JOIN tableA2 ta2 ON ta1.ta2_id = ta2.id
LEFT JOIN tableA3 ta3 ON ta1.ta3_id = ta3.id
LEFT JOIN tableA4 ta4 ON ta1.ta4_id = ta4.id
LEFT JOIN tableA5 ta5 ON ta5.ta1_id = ta1.id WHERE 
tb1.name = ta1.name AND 
tb2.name = ta2.name AND 
tb3.name = ta3.name AND 
tb4.name = ta4.name AND 
tb5.name = ta5.name)

希望如果我创建相同的结构并比较所有值，我会得到预期的结果。即使数据库相同，我也会得到很多行 selected.

Answer 1

找到问题了。当比较两个 NULL 值时，结果为 FALSE，查询本身应该没问题。所以我应该添加条件来检查值是否为 NULL。

用于比较来自数据库的两个实例的数据的 PSQL 查询

PSQL query to comapre data from two instances of a database

database-comparison

psql