問(wèn)題描述

同一個(gè)數(shù)據(jù)庫(kù)，本地51張表和遠(yuǎn)程49張表，有差異數(shù)據(jù)表。如何通過(guò)一條SQL來(lái)快速找出這些表的名字。

SQL:

USE performance_schema;SELECT t1.OBJECT_SCHEMA,t1.OBJECT_NAME,t2.OBJECT_SCHEMA,t2.OBJECT_NAMEFROM `table_io_waits_summary_by_table` t1 RIGHT JOIN `table_io_waits_summary_by_table` t2 ON t1.OBJECT_NAME = t2.OBJECT_NAMEWHERE t1.OBJECT_SCHEMA=’db1_local’ AND t2.OBJECT_SCHEMA=’db2_remote’;

結(jié)果集只有49張，無(wú)法羅列出差異的表明。使用了 LEFT OUTER JOIN 還是一樣。

驗(yàn)證是存在差異的：

SELECT OBJECT_NAMEFROM table_io_waits_summary_by_table WHERE OBJECT_SCHEMA=’db1_local’ AND OBJECT_NAME NOT IN (SELECT OBJECT_NAME FROM table_io_waits_summary_by_table WHERE OBJECT_SCHEMA=’db2_remote’ )

問(wèn)題解答

試試這個(gè)：

USE performance_schema;SELECT t1.*FROM `table_io_waits_summary_by_table` t1 LEFT JOIN `table_io_waits_summary_by_table` t2 ON t1.OBJECT_NAME = t2.OBJECT_NAME AND t2.OBJECT_SCHEMA=’db2_remote’WHERE t1.OBJECT_SCHEMA=’db1_local’ AND t2.OBJECT_NAME IS NULL;

其實(shí)你的第一個(gè)SQL只要將對(duì)t2的限制提到連接條件中就行了，將t2.OBJECT_SCHEMA=’db2_remote’寫(xiě)在where條件里面RIGHT JOIN就變成了INNER JOIN ~