我会尝试以下方法:
首先,请确保以下表和列上都有索引(括号中的每组列都应是一个单独的索引):
table1 : (subscribe, CDate)
(CU_id)
table2 : (O_cid)
(O_ref)
table3 : (I_oref)
(I_pid)
table4 : (P_id)
(P_cat)
table5 : (C_id, store)
SELECT DISTINCT t1.first_name, t1.last_name, t1.email FROM
(SELECT CU_id, t1.first_name, t1.last_name, t1.email
FROM table1
WHERE subscribe = 1 AND
CDate >= $startDate AND
CDate <= $endDate) AS t1
INNER JOIN table2 AS t2
ON t1.CU_id = t2.O_cid
INNER JOIN table3 AS t3
ON t2.O_ref = t3.I_oref
INNER JOIN table4 AS t4
ON t3.I_pid = t4.P_id
INNER JOIN (SELECT C_id FROM table5 WHERE store = 2) AS t5
ON t4.P_cat = t5.C_id
我希望在这里,第一个子选择将大大减少要考虑加入的行数,希望使以后的加入工作更少。同上表5中第二个子选择的推理。
无论如何,请弄乱它。我的意思是,最终它只是一个SELECT-您不能用它真正伤害任何东西。检查由每个不同排列生成的计划,并尝试找出每个计划的优缺点。
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