Mysql-巧用join來優化sql
0. 準備相關表來進行接下來的測試
相關建表語句請看：https://github.com/YangBaohust/my_sql
user1表，取經組 +----+-----------+-----------------+---------------------------------+ | id | user_name | comment | mobile | +----+-----------+-----------------+---------------------------------+ | 1 | 唐僧 | 旃檀功德佛 | 138245623,021-382349 | | 2 | 孫悟空 | 鬥戰勝佛 | 159384292,022-483432,+86-392432 | | 3 | 豬八戒 | 淨壇使者 | 183208243,055-8234234 | | 4 | 沙僧 | 金身羅漢 | 293842295,098-2383429 | | 5 | NULL | 白龍馬 | 993267899 | +----+-----------+-----------------+---------------------------------+ user2表，悟空的朋友圈 +----+--------------+-----------+ | id | user_name | comment | +----+--------------+-----------+ | 1 | 孫悟空 | 美猴王 | | 2 | 牛魔王 | 牛哥 | | 3 | 鐵扇公主 | 牛夫人 | | 4 | 菩提老祖 | 葡萄 | | 5 | NULL | 晶晶 | +----+--------------+-----------+ user1_kills表，取經路上殺的妖怪數量 +----+-----------+---------------------+-------+ | id | user_name | timestr | kills | +----+-----------+---------------------+-------+ | 1 | 孫悟空 | 2013-01-10 00:00:00 | 10 | | 2 | 孫悟空 | 2013-02-01 00:00:00 | 2 | | 3 | 孫悟空 | 2013-02-05 00:00:00 | 12 | | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | | 6 | 豬八戒 | 2013-02-07 00:00:00 | 17 | | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | | 8 | 沙僧 | 2013-01-10 00:00:00 | 3 | | 9 | 沙僧 | 2013-01-22 00:00:00 | 9 | | 10 | 沙僧 | 2013-02-11 00:00:00 | 5 | +----+-----------+---------------------+-------+ user1_equipment表，取經組裝備 +----+-----------+--------------+-----------------+-----------------+ | id | user_name | arms | clothing | shoe | +----+-----------+--------------+-----------------+-----------------+ | 1 | 唐僧 | 九環錫杖 | 錦斕袈裟 | 僧鞋 | | 2 | 孫悟空 | 金箍棒 | 梭子黃金甲 | 藕絲步雲履 | | 3 | 豬八戒 | 九齒釘耙 | 僧衣 | 僧鞋 | | 4 | 沙僧 | 降妖寶杖 | 僧衣 | 僧鞋 | +----+-----------+--------------+-----------------+-----------------+
1. 使用left join優化not in子句
例子：找出取經組中不屬於悟空朋友圈的人
+----+-----------+-----------------+-----------------------+ | id | user_name | comment | mobile | +----+-----------+-----------------+-----------------------+ | 1 | 唐僧 | 旃檀功德佛 | 138245623,021-382349 | | 3 | 豬八戒 | 淨壇使者 | 183208243,055-8234234 | | 4 | 沙僧 | 金身羅漢 | 293842295,098-2383429 | +----+-----------+-----------------+-----------------------+
not in寫法：
select * from user1 a where a.user_name not in (select user_name from user2 where user_name is not null);
left join寫法：
首先看通過user_name進行連線的外連線資料集
select a.*, b.* from user1 a left join user2 b on (a.user_name = b.user_name);
+----+-----------+-----------------+---------------------------------+------+-----------+-----------+ | id | user_name | comment | mobile | id | user_name | comment | +----+-----------+-----------------+---------------------------------+------+-----------+-----------+ | 2 | 孫悟空 | 鬥戰勝佛 | 159384292,022-483432,+86-392432 | 1 | 孫悟空 | 美猴王 | | 1 | 唐僧 | 旃檀功德佛 | 138245623,021-382349 | NULL | NULL | NULL | | 3 | 豬八戒 | 淨壇使者 | 183208243,055-8234234 | NULL | NULL | NULL | | 4 | 沙僧 | 金身羅漢 | 293842295,098-2383429 | NULL | NULL | NULL | | 5 | NULL | 白龍馬 | 993267899 | NULL | NULL | NULL | +----+-----------+-----------------+---------------------------------+------+-----------+-----------+
可以看到a表中的所有資料都有顯示，b表中的資料只有b.user_name與a.user_name相等才顯示，其餘都以null值填充，要想找出取經組中不屬於悟空朋友圈的人，只需要在b.user_name中加一個過濾條件b.user_name is null即可。
select a.* from user1 a left join user2 b on (a.user_name = b.user_name) where b.user_name is null;
+----+-----------+-----------------+-----------------------+ | id | user_name | comment | mobile | +----+-----------+-----------------+-----------------------+ | 1 | 唐僧 | 旃檀功德佛 | 138245623,021-382349 | | 3 | 豬八戒 | 淨壇使者 | 183208243,055-8234234 | | 4 | 沙僧 | 金身羅漢 | 293842295,098-2383429 | | 5 | NULL | 白龍馬 | 993267899 | +----+-----------+-----------------+-----------------------+
看到這裡發現結果集中還多了一個白龍馬，繼續新增過濾條件a.user_name is not null即可。
select a.* from user1 a left join user2 b on (a.user_name = b.user_name) where b.user_name is null and a.user_name is not null;
2. 使用left join優化標量子查詢
例子：檢視取經組中的人在悟空朋友圈的暱稱
+-----------+-----------------+-----------+ | user_name | comment | comment2 | +-----------+-----------------+-----------+ | 唐僧 | 旃檀功德佛 | NULL | | 孫悟空 | 鬥戰勝佛 | 美猴王 | | 豬八戒 | 淨壇使者 | NULL | | 沙僧 | 金身羅漢 | NULL | | NULL | 白龍馬 | NULL | +-----------+-----------------+-----------+
子查詢寫法：
select a.user_name, a.comment, (select comment from user2 b where b.user_name = a.user_name) comment2 from user1 a;
left join寫法：
select a.user_name, a.comment, b.comment comment2 from user1 a left join user2 b on (a.user_name = b.user_name);
3. 使用join優化聚合子查詢
例子：查詢出取經組中每人打怪最多的日期
+----+-----------+---------------------+-------+ | id | user_name | timestr | kills | +----+-----------+---------------------+-------+ | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | | 9 | 沙僧 | 2013-01-22 00:00:00 | 9 | +----+-----------+---------------------+-------+
聚合子查詢寫法：
select * from user1_kills a where a.kills = (select max(b.kills) from user1_kills b where b.user_name = a.user_name);
join寫法：
首先看兩表自關聯的結果集，為節省篇幅，只取豬八戒的打怪資料來看
select a.*, b.* from user1_kills a join user1_kills b on (a.user_name = b.user_name) order by 1;
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+ | id | user_name | timestr | kills | id | user_name | timestr | kills | +----+-----------+---------------------+-------+----+-----------+---------------------+-------+ | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | 6 | 豬八戒 | 2013-02-07 00:00:00 | 17 | | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | | 6 | 豬八戒 | 2013-02-07 00:00:00 | 17 | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | | 6 | 豬八戒 | 2013-02-07 00:00:00 | 17 | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | | 6 | 豬八戒 | 2013-02-07 00:00:00 | 17 | 6 | 豬八戒 | 2013-02-07 00:00:00 | 17 | | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | 6 | 豬八戒 | 2013-02-07 00:00:00 | 17 | | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | +----+-----------+---------------------+-------+----+-----------+---------------------+-------+
可以看到當兩表通過user_name進行自關聯，只需要對a表的所有欄位進行一個group by，取b表中的max(kills)，只要a.kills=max(b.kills)就滿足要求了。sql如下
select a.* from user1_kills a join user1_kills b on (a.user_name = b.user_name) group by a.id, a.user_name, a.timestr, a.kills having a.kills = max(b.kills);
4. 使用join進行分組選擇
例子：對第3個例子進行升級，查詢出取經組中每人打怪最多的前兩個日期
+----+-----------+---------------------+-------+ | id | user_name | timestr | kills | +----+-----------+---------------------+-------+ | 3 | 孫悟空 | 2013-02-05 00:00:00 | 12 | | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | | 5 | 豬八戒 | 2013-01-11 00:00:00 | 20 | | 7 | 豬八戒 | 2013-02-08 00:00:00 | 35 | | 9 | 沙僧 | 2013-01-22 00:00:00 | 9 | | 10 | 沙僧 | 2013-02-11 00:00:00 | 5 | +----+-----------+---------------------+-------+
在oracle中，可以通過分析函式來實現
select b.* from (select a.*, row_number() over(partition by user_name order by kills desc) cnt from user1_kills a) b where b.cnt <= 2;
很遺憾，上面sql在mysql中報錯ERROR 1064 (42000): You have an error in your SQL syntax; 因為mysql並不支援分析函式。不過可以通過下面的方式去實現。
首先對兩表進行自關聯，為了節約篇幅，只取出孫悟空的資料
select a.*, b.* from user1_kills a join user1_kills b on (a.user_name=b.user_name and a.kills<=b.kills) order by a.user_name, a.kills desc;
+----+-----------+---------------------+-------+----+-----------+---------------------+-------+ | id | user_name | timestr | kills | id | user_name | timestr | kills | +----+-----------+---------------------+-------+----+-----------+---------------------+-------+ | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | | 3 | 孫悟空 | 2013-02-05 00:00:00 | 12 | 3 | 孫悟空 | 2013-02-05 00:00:00 | 12 | | 3 | 孫悟空 | 2013-02-05 00:00:00 | 12 | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | | 1 | 孫悟空 | 2013-01-10 00:00:00 | 10 | 1 | 孫悟空 | 2013-01-10 00:00:00 | 10 | | 1 | 孫悟空 | 2013-01-10 00:00:00 | 10 | 3 | 孫悟空 | 2013-02-05 00:00:00 | 12 | | 1 | 孫悟空 | 2013-01-10 00:00:00 | 10 | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | | 2 | 孫悟空 | 2013-02-01 00:00:00 | 2 | 1 | 孫悟空 | 2013-01-10 00:00:00 | 10 | | 2 | 孫悟空 | 2013-02-01 00:00:00 | 2 | 3 | 孫悟空 | 2013-02-05 00:00:00 | 12 | | 2 | 孫悟空 | 2013-02-01 00:00:00 | 2 | 4 | 孫悟空 | 2013-02-12 00:00:00 | 22 | | 2 | 孫悟空 | 2013-02-01 00:00:00 | 2 | 2 | 孫悟空 | 2013-02-01 00:00:00 | 2 | +----+-----------+---------------------+-------+----+-----------+---------------------+-------+
從上面的表中我們知道孫悟空打怪前兩名的數量是22和12，那麼只需要對a表的所有欄位進行一個group by，對b表的id做個count，count值小於等於2就滿足要求，sql改寫如下：
select a.* from user1_kills a join user1_kills b on (a.user_name=b.user_name and a.kills<=b.kills) group by a.id, a.user_name, a.timestr, a.kills having count(b.id) <= 2;
5. 使用笛卡爾積關聯實現一列轉多行
例子：將取經組中每個電話號碼變成一行
原始資料：
+-----------+---------------------------------+ | user_name | mobile | +-----------+---------------------------------+ | 唐僧 | 138245623,021-382349 | | 孫悟空 | 159384292,022-483432,+86-392432 | | 豬八戒 | 183208243,055-8234234 | | 沙僧 | 293842295,098-2383429 | | NULL | 993267899 | +-----------+---------------------------------+
想要得到的資料：
+-----------+-------------+ | user_name | mobile | +-----------+-------------+ | 唐僧 | 138245623 | | 唐僧 | 021-382349 | | 孫悟空 | 159384292 | | 孫悟空 | 022-483432 | | 孫悟空 | +86-392432 | | 豬八戒 | 183208243 | | 豬八戒 | 055-8234234 | | 沙僧 | 293842295 | | 沙僧 | 098-2383429 | | NULL | 993267899 | +-----------+-------------+
可以看到唐僧有兩個電話，因此他就需要兩行。我們可以先求出每人的電話號碼數量，然後與一張序列表進行笛卡兒積關聯，為了節約篇幅，只取出唐僧的資料
select a.id, b.* from tb_sequence a cross join (select user_name, mobile, length(mobile)-length(replace(mobile, ',', ''))+1 size from user1) b order by 2,1;
+----+-----------+---------------------------------+------+ | id | user_name | mobile | size | +----+-----------+---------------------------------+------+ | 1 | 唐僧 | 138245623,021-382349 | 2 | | 2 | 唐僧 | 138245623,021-382349 | 2 | | 3 | 唐僧 | 138245623,021-382349 | 2 | | 4 | 唐僧 | 138245623,021-382349 | 2 | | 5 | 唐僧 | 138245623,021-382349 | 2 | | 6 | 唐僧 | 138245623,021-382349 | 2 | | 7 | 唐僧 | 138245623,021-382349 | 2 | | 8 | 唐僧 | 138245623,021-382349 | 2 | | 9 | 唐僧 | 138245623,021-382349 | 2 | | 10 | 唐僧 | 138245623,021-382349 | 2 | +----+-----------+---------------------------------+------+
a.id對應的就是第幾個電話號碼，size就是總的電話號碼數量，因此可以加上關聯條件(a.id <= b.size)，將上面的sql繼續調整
select b.user_name, replace(substring(substring_index(b.mobile, ',', a.id), char_length(substring_index(mobile, ',', a.id-1)) + 1), ',', '') as mobile from tb_sequence a cross join (select user_name, concat(mobile, ',') as mobile, length(mobile)-length(replace(mobile, ',', ''))+1 size from user1) b on (a.id <= b.size);
6. 使用笛卡爾積關聯實現多列轉多行
例子：將取經組中每件裝備變成一行
原始資料：
+----+-----------+--------------+-----------------+-----------------+ | id | user_name | arms | clothing | shoe | +----+-----------+--------------+-----------------+-----------------+ | 1 | 唐僧 | 九環錫杖 | 錦斕袈裟 | 僧鞋 | | 2 | 孫悟空 | 金箍棒 | 梭子黃金甲 | 藕絲步雲履 | | 3 | 豬八戒 | 九齒釘耙 | 僧衣 | 僧鞋 | | 4 | 沙僧 | 降妖寶杖 | 僧衣 | 僧鞋 | +----+-----------+--------------+-----------------+-----------------+
想要得到的資料：
+-----------+-----------+-----------------+ | user_name | equipment | equip_mame | +-----------+-----------+-----------------+ | 唐僧 | arms | 九環錫杖 | | 唐僧 | clothing | 錦斕袈裟 | | 唐僧 | shoe | 僧鞋 | | 孫悟空 | arms | 金箍棒 | | 孫悟空 | clothing | 梭子黃金甲 | | 孫悟空 | shoe | 藕絲步雲履 | | 沙僧 | arms | 降妖寶杖 | | 沙僧 | clothing | 僧衣 | | 沙僧 | shoe | 僧鞋 | | 豬八戒 | arms | 九齒釘耙 | | 豬八戒 | clothing | 僧衣 | | 豬八戒 | shoe | 僧鞋 | +-----------+-----------+-----------------+
union的寫法：
select user_name, 'arms' as equipment, arms equip_mame from user1_equipment
union all
select user_name, 'clothing' as equipment, clothing equip_mame from user1_equipment
union all
select user_name, 'shoe' as equipment, shoe equip_mame from user1_equipment
order by 1, 2;
join的寫法：
首先看笛卡爾資料集的效果，以唐僧為例
select a.*, b.* from user1_equipment a cross join tb_sequence b where b.id <= 3;
+----+-----------+--------------+-----------------+-----------------+----+ | id | user_name | arms | clothing | shoe | id | +----+-----------+--------------+-----------------+-----------------+----+ | 1 | 唐僧 | 九環錫杖 | 錦斕袈裟 | 僧鞋 | 1 | | 1 | 唐僧 | 九環錫杖 | 錦斕袈裟 | 僧鞋 | 2 | | 1 | 唐僧 | 九環錫杖 | 錦斕袈裟 | 僧鞋 | 3 | +----+-----------+--------------+-----------------+-----------------+----+
使用case對上面的結果進行處理
select user_name,
case when b.id = 1 then 'arms'
when b.id = 2 then 'clothing'
when b.id = 3 then 'shoe' end as equipment,
case when b.id = 1 then arms end arms,
case when b.id = 2 then clothing end clothing,
case when b.id = 3 then shoe end shoe
from user1_equipment a cross join tb_sequence b where b.id <=3;
+-----------+-----------+--------------+-----------------+-----------------+ | user_name | equipment | arms | clothing | shoe | +-----------+-----------+--------------+-----------------+-----------------+ | 唐僧 | arms | 九環錫杖 | NULL | NULL | | 唐僧 | clothing | NULL | 錦斕袈裟 | NULL | | 唐僧 | shoe | NULL | NULL | 僧鞋 | +-----------+-----------+--------------+-----------------+-----------------+
使用coalesce函式將多列資料進行合併
select user_name,
case when b.id = 1 then 'arms'
when b.id = 2 then 'clothing'
when b.id = 3 then 'shoe' end as equipment,
coalesce(case when b.id = 1 then arms end,
case when b.id = 2 then clothing end,
case when b.id = 3 then shoe end) equip_mame
from user1_equipment a cross join tb_sequence b where b.id <=3 order by 1, 2;
7. 使用join更新過濾條件中包含自身的表
例子：把同時存在於取經組和悟空朋友圈中的人，在取經組中把comment欄位更新為"此人在悟空的朋友圈"
我們很自然地想到先查出user1和user2中user_name都存在的人，然後更新user1表，sql如下
update user1 set comment = '此人在悟空的朋友圈' where user_name in (select a.user_name from user1 a join user2 b on (a.user_name = b.user_name));
很遺憾，上面sql在mysql中報錯：ERROR 1093 (HY000): You can't specify target table 'user1' for update in FROM clause，提示不能更新目標表在from子句的表。
那有沒有其它辦法呢？我們可以將in的寫法轉換成join的方式
select c.*, d.* from user1 c join (select a.user_name from user1 a join user2 b on (a.user_name = b.user_name)) d on (c.user_name = d.user_name);
+----+-----------+--------------+---------------------------------+-----------+ | id | user_name | comment | mobile | user_name | +----+-----------+--------------+---------------------------------+-----------+ | 2 | 孫悟空 | 鬥戰勝佛 | 159384292,022-483432,+86-392432 | 孫悟空 | +----+-----------+--------------+---------------------------------+-----------+
然後對join之後的檢視進行更新即可
update user1 c join (select a.user_name from user1 a join user2 b on (a.user_name = b.user_name)) d on (c.user_name = d.user_name) set c.comment = '此人在悟空的朋友圈';
再檢視user1，可以看到user1已修改成功
select * from user1;
+----+-----------+-----------------------------+---------------------------------+ | id | user_name | comment | mobile | +----+-----------+-----------------------------+---------------------------------+ | 1 | 唐僧 | 旃檀功德佛 | 138245623,021-382349 | | 2 | 孫悟空 | 此人在悟空的朋友圈 | 159384292,022-483432,+86-392432 | | 3 | 豬八戒 | 淨壇使者 | 183208243,055-8234234 | | 4 | 沙僧 | 金身羅漢 | 293842295,098-2383429 | | 5 | NULL | 白龍馬 | 993267899 | +----+-----------+-----------------------------+---------------------------------+
8. 使用join刪除重複資料
首先向user2表中插入兩條資料
insert into user2(user_name, comment) values ('孫悟空', '美猴王');
insert into user2(user_name, comment) values ('牛魔王', '牛哥');
例子：將user2表中的重複資料刪除，只保留id號大的
+----+--------------+-----------+ | id | user_name | comment | +----+--------------+-----------+ | 1 | 孫悟空 | 美猴王 | | 2 | 牛魔王 | 牛哥 | | 3 | 鐵扇公主 | 牛夫人 | | 4 | 菩提老祖 | 葡萄 | | 5 | NULL | 晶晶 | | 6 | 孫悟空 | 美猴王 | | 7 | 牛魔王 | 牛哥 | +----+--------------+-----------+
首先檢視重複記錄
select a.*, b.* from user2 a join (select user_name, comment, max(id) id from user2 group by user_name, comment having count(*) > 1) b on (a.user_name=b.user_name and a.comment=b.comment) order by 2;
+----+-----------+-----------+-----------+-----------+------+ | id | user_name | comment | user_name | comment | id | +----+-----------+-----------+-----------+-----------+------+ | 1 | 孫悟空 | 美猴王 | 孫悟空 | 美猴王 | 6 | | 6 | 孫悟空 | 美猴王 | 孫悟空 | 美猴王 | 6 | | 2 | 牛魔王 | 牛哥 | 牛魔王 | 牛哥 | 7 | | 7 | 牛魔王 | 牛哥 | 牛魔王 | 牛哥 | 7 | +----+-----------+-----------+-----------+-----------+------+
接著只需要刪除(a.id < b.id)的資料即可
delete a from user2 a join (select user_name, comment, max(id) id from user2 group by user_name, comment having count(*) > 1) b on (a.user_name=b.user_name and a.comment=b.comment) where a.id < b.id;
檢視user2，可以看到重複資料已經被刪掉了
select * from user2;
+----+--------------+-----------+ | id | user_name | comment | +----+--------------+-----------+ | 3 | 鐵扇公主 | 牛夫人 | | 4 | 菩提老祖 | 葡萄 | | 5 | NULL | 晶晶 | | 6 | 孫悟空 | 美猴王 | | 7 | 牛魔王 | 牛哥 | +----+--------------+-----------+
總結：給大家就介紹到這裡，大家有興趣可以多造點資料，然後比較不同的sql寫法在執行時間上的區別。本文例子取自於慕課網《sql開發技巧