显示所有记录从表A中跳过表B中的匹配记录

Question

select r.TableReservationId, d.TableId, d.TableName, d.StoreId
from dining_table d
  LEFT join dining_table_reservation r on d.TableId=r.TableId
where r.Status=0
  and r.StartTime not BETWEEN '2015-10-15 17:40:18' AND '2015-10-15 17:40:18'      

这个只显示匹配记录，但我想要来自dining_table的所有记录，如果记录与条件匹配，则应跳过此记录

Answer 1

select r.TableReservationId, d.TableId, d.TableName, d.StoreId
from dining_table d
  LEFT join dining_table_reservation r on d.TableId=r.TableId
  AND r.Status=0
  and r.StartTime not BETWEEN '2015-10-15 17:40:18' AND '2015-10-15 17:40:18'  

当LEFT JOIN时，如果将这些右侧表条件放在WHERE子句中，它将作为常规内连接执行。转而使用ON子句！

Answer 2

所以你想要所有的记录，表格中没有共同的值使用NOT IN和你当前的结果

SELECT * FROM dining_table where TableId NOT IN
(select r.TableReservationId, d.TableId, d.TableName, d.StoreId
from dining_table d
  LEFT join dining_table_reservation r on d.TableId=r.TableId
where r.Status=0
  and r.StartTime not BETWEEN '2015-10-15 17:40:18' AND '2015-10-15 17:40:18');