mysql - 复合索引顺序 MySQL 查询

Question

我正在尝试优化以下 mysql 查询。运行时间约为 2.5 秒。我已经阅读了复合索引，但我希望有人可以帮助我理解如何使用包含多个联接、许多 where 条件(包括日期范围)、分组依据和顺序依据的查询对复合索引进行排序计算值。我是否缺少有用的复合索引？有没有更有效的方法可以从这些表中提取数据？非常感谢任何帮助!

SELECT 
branch.name AS branch,
SUM(appointment.status =  'completed') AS `Completed`, 
SUM(appointment.status =  'cancelled') AS `Cancelled`, 
SUM(appointment.status =  'not completed') AS `Not Completed`, 
SUM(appointment.status != 'rescheduled') AS `Total`
FROM rep
JOIN customer ON rep.id = customer.rep_id
JOIN office ON rep.office_id = office.id
JOIN appointment ON customer.id = appointment.customer_id
JOIN branch ON office.branch_id = branch.id
WHERE rep.active= 1 
AND rep.group IN (1,2,3) 
AND rep.deleted = 0 
AND customer.saved = 0 
AND (customer.rep_id != appointment.closed_by OR appointment.closed_by IS NULL)
AND customer.rep_id != 0 
AND customer.deleted = 0
AND office.visible = 1 
AND office.deleted = 0 
AND appointment.date >= '2016-12-01'
AND appointment.date < '2017-11-30' 
AND appointment.current = 1 
GROUP BY branch.id
ORDER BY Completed

这是解释输出:

id: 1
select_type: simple
table: office
type: ref
possible_keys: PRIMARY, deleted_branchID_name, deleted_visible 
key: deleted_visible
key_len: 5
ref: const,const
rows:  73
Extra: Using index condition; Using temporary; Using filesort

id: 1
select_type: simple
table: branch
type: eq_ref
possible_keys: PRIMARY
key: PRIMARY
key_len: 4
ref: office.branch_id
rows: 1
Extra: NULL

id: 1
select_type:  simple
table: rep
type: ref
possible_keys: PRIMARY, group_id, office_id, active_deleted
key: office_id
key_len: 5
ref: office.id
rows:  57 
Extra: Using index condition; Using where

id:  1
select_type:  simple
table:  customer
type: ref
possible_keys: PRIMARY, rep_id
key: rep_id
key_len: 4
ref: rep.id
rows:  61 
Extra: Using where

id: 1
select_type: simple
table: appointment
type: ref
possible_keys: date, customer_id, closedByID_date, isCurrent_date
key: customer_id
key_len: 4
ref: customer.id
rows: 1
Extra: Using where

Answer 1

一个建议

删除条件周围无用的 ()

  LEFT JOIN customer ON rep.id = customer.rep_id
  LEFT JOIN office ON rep.office_id = office.id
  LEFT JOIN appointment ON customer.id = appointment.customer_id
  LEFT JOIN branch ON office.branch_id = branch.id