Join執行原理解析
0 目錄
- JOIN語句的執行順序
- INNER/LEFT/RIGHT/FULL JOIN的區別
- ON和WHERE的區別
1 概述
一個完整的SQL語句中會被拆分成多個子句,子句的執行過程中會產生虛擬表(vt),但是結果只返回最後一張虛擬表。從這個思路出發,我們試著理解一下JOIN查詢的執行過程並解答一些常見的問題。
如果之前對不同JOIN的執行結果沒有概念,可以結合這篇文章往下看
2 JOIN的執行順序
以下是JOIN查詢的通用結構
SELECT <row_list> FROM <left_table> <inner|left|right> JOIN <right_table> ON <join condition> WHERE <where_condition>
它的執行順序如下(SQL語句裡第一個被執行的總是FROM子句):
- FROM:對左右兩張表執行笛卡爾積,產生第一張表vt1。行數為n*m(n為左表的行數,m為右表的行數
- ON:根據ON的條件逐行篩選vt1,將結果插入vt2中
- JOIN:新增外部行,如果指定了LEFT JOIN(LEFT OUTER JOIN),則先遍歷一遍左表的每一行,其中不在vt2的行會被插入到vt2,該行的剩餘欄位將被填充為NULL,形成vt3;如果指定了RIGHT JOIN也是同理。但如果指定的是INNER JOIN,則不會新增外部行,上述插入過程被忽略,vt2=vt3(所以INNER JOIN的過濾條件放在ON
- WHERE:對vt3進行條件過濾,滿足條件的行被輸出到vt4
- SELECT:取出vt4的指定欄位到vt5
下面用一個例子介紹一下上述聯表的過程(這個例子不是個好的實踐,只是為了說明join語法)
3 舉例
建立一個使用者資訊表:
CREATE TABLE `user_info` ( `userid` int(11) NOT NULL, `name` varchar(255) NOT NULL, UNIQUE `userid` (`userid`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4
再建立一個使用者餘額表:
CREATE TABLE `user_account` (
`userid` int(11) NOT NULL,
`money` bigint(20) NOT NULL,
UNIQUE `userid` (`userid`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4
隨便匯入一些資料:
select * from user_info;
+--------+------+
| userid | name |
+--------+------+
| 1001 | x |
| 1002 | y |
| 1003 | z |
| 1004 | a |
| 1005 | b |
| 1006 | c |
| 1007 | d |
| 1008 | e |
+--------+------+
8 rows in set (0.00 sec)
select * from user_account;
+--------+-------+
| userid | money |
+--------+-------+
| 1001 | 22 |
| 1002 | 30 |
| 1003 | 8 |
| 1009 | 11 |
+--------+-------+
4 rows in set (0.00 sec)
一共8個使用者有使用者名稱,4個使用者的賬戶有餘額。
取出userid為1003的使用者姓名和餘額,SQL如下:
SELECT i.name, a.money
FROM user_info as i
LEFT JOIN user_account as a
ON i.userid = a.userid
WHERE a.userid = 1003;
第一步:執行FROM子句對兩張表進行笛卡爾積操作
笛卡爾積操作後會返回兩張表中所有行的組合,左表user_info有8行,右表user_account有4行,生成的虛擬表vt1就是8*4=32行:
SELECT * FROM user_info as i LEFT JOIN user_account as a ON 1;
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1001 | 22 |
| 1003 | z | 1001 | 22 |
| 1004 | a | 1001 | 22 |
| 1005 | b | 1001 | 22 |
| 1006 | c | 1001 | 22 |
| 1007 | d | 1001 | 22 |
| 1008 | e | 1001 | 22 |
| 1001 | x | 1002 | 30 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1002 | 30 |
| 1004 | a | 1002 | 30 |
| 1005 | b | 1002 | 30 |
| 1006 | c | 1002 | 30 |
| 1007 | d | 1002 | 30 |
| 1008 | e | 1002 | 30 |
| 1001 | x | 1003 | 8 |
| 1002 | y | 1003 | 8 |
| 1003 | z | 1003 | 8 |
| 1004 | a | 1003 | 8 |
| 1005 | b | 1003 | 8 |
| 1006 | c | 1003 | 8 |
| 1007 | d | 1003 | 8 |
| 1008 | e | 1003 | 8 |
| 1001 | x | 1009 | 11 |
| 1002 | y | 1009 | 11 |
| 1003 | z | 1009 | 11 |
| 1004 | a | 1009 | 11 |
| 1005 | b | 1009 | 11 |
| 1006 | c | 1009 | 11 |
| 1007 | d | 1009 | 11 |
| 1008 | e | 1009 | 11 |
+--------+------+--------+-------+
32 rows in set (0.00 sec)
第二步:執行ON子句過濾掉不滿足條件的行
ON i.userid = a.userid 過濾之後vt2如下:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1003 | 8 |
+--------+------+--------+-------+
第三步:JOIN 新增外部行
LEFT JOIN會將左表未出現在vt2的行插入進vt2,每一行的剩餘欄位將被填充為NULL,RIGHT JOIN同理
本例中用的是LEFT JOIN,所以會將左表user_info剩下的行都添上 生成表vt3:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1003 | 8 |
| 1004 | a | NULL | NULL |
| 1005 | b | NULL | NULL |
| 1006 | c | NULL | NULL |
| 1007 | d | NULL | NULL |
| 1008 | e | NULL | NULL |
+--------+------+--------+-------+
第四步:WHERE條件過濾
WHERE a.userid = 1003 生成表vt4:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1003 | z | 1003 | 8 |
+--------+------+--------+-------+
第五步:SELECT
SELECT i.name, a.money 生成vt5:
+------+-------+
| name | money |
+------+-------+
| z | 8 |
+------+-------+
虛擬表vt5作為最終結果返回給客戶端
介紹完聯表的過程之後,我們看看常用JOIN的區別
4 INNER/LEFT/RIGHT/FULL JOIN的區別
- INNER JOIN...ON...: 返回 左右表互相匹配的所有行(因為只執行上文的第二步ON過濾,不執行第三步 新增外部行)
- LEFT JOIN...ON...: 返回左表的所有行,若某些行在右表裡沒有相對應的匹配行,則將右表的列在新表中置為NULL
- RIGHT JOIN...ON...: 返回右表的所有行,若某些行在左表裡沒有相對應的匹配行,則將左表的列在新表中置為NULL
INNER JOIN
拿上文的第三步新增外部行來舉例,若LEFT JOIN替換成INNER JOIN,則會跳過這一步,生成的表vt3與vt2一模一樣:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1003 | 8 |
+--------+------+--------+-------+
RIGHT JOIN
若LEFT JOIN替換成RIGHT JOIN,則生成的表vt3如下:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1003 | 8 |
| NULL | NULL | 1009 | 11 |
+--------+------+--------+-------+
因為user_account(右表)裡存在userid=1009這一行,而user_info(左表)裡卻找不到這一行的記錄,所以會在第三步插入以下一行:
| NULL | NULL | 1009 | 11 |
FULL JOIN
上文引用的文章中提到了標準SQL定義的FULL JOIN,這在mysql裡是不支援的,不過我們可以通過LEFT JOIN + UNION + RIGHT JOIN 來實現FULL JOIN:
SELECT *
FROM user_info as i
RIGHT JOIN user_account as a
ON a.userid=i.userid
union
SELECT *
FROM user_info as i
LEFT JOIN user_account as a
ON a.userid=i.userid;
他會返回如下結果:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1003 | 8 |
| NULL | NULL | 1009 | 11 |
| 1004 | a | NULL | NULL |
| 1005 | b | NULL | NULL |
| 1006 | c | NULL | NULL |
| 1007 | d | NULL | NULL |
| 1008 | e | NULL | NULL |
+--------+------+--------+-------+
ps:其實我們從語義上就能看出LEFT JOIN和RIGHT JOIN沒什麼差別,兩者的結果差異取決於左右表的放置順序,以下內容摘自mysql官方文件:
RIGHT JOIN works analogously to LEFT JOIN. To keep code portable across databases, it is recommended that you use LEFT JOIN instead of RIGHT JOIN.
所以當你糾結使用LEFT JOIN還是RIGHT JOIN時,儘可能只使用LEFT JOIN吧
5 ON和WHERE的區別
上文把JOIN的執行順序瞭解清楚之後,ON和WHERE的區別也就很好理解了。
舉例說明:
SELECT *
FROM user_info as i
LEFT JOIN user_account as a
ON i.userid = a.userid and i.userid = 1003;
SELECT *
FROM user_info as i
LEFT JOIN user_account as a
ON i.userid = a.userid where i.userid = 1003;
第一種情況LEFT JOIN在執行完第二步ON子句後,篩選出滿足i.userid = a.userid and i.userid = 1003的行,生成表vt2,然後執行第三步JOIN子句,將外部行新增進虛擬表生成vt3即最終結果:
vt2:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1003 | z | 1003 | 8 |
+--------+------+--------+-------+
vt3:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | NULL | NULL |
| 1002 | y | NULL | NULL |
| 1003 | z | 1003 | 8 |
| 1004 | a | NULL | NULL |
| 1005 | b | NULL | NULL |
| 1006 | c | NULL | NULL |
| 1007 | d | NULL | NULL |
| 1008 | e | NULL | NULL |
+--------+------+--------+-------+
而第二種情況LEFT JOIN在執行完第二步ON子句後,篩選出滿足i.userid = a.userid的行,生成表vt2;再執行第三步JOIN子句新增外部行生成表vt3;然後執行第四步WHERE子句,再對vt3表進行過濾生成vt4,得的最終結果:
vt2:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1003 | 8 |
+--------+------+--------+-------+
vt3:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1001 | x | 1001 | 22 |
| 1002 | y | 1002 | 30 |
| 1003 | z | 1003 | 8 |
| 1004 | a | NULL | NULL |
| 1005 | b | NULL | NULL |
| 1006 | c | NULL | NULL |
| 1007 | d | NULL | NULL |
| 1008 | e | NULL | NULL |
+--------+------+--------+-------+
vt4:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1003 | z | 1003 | 8 |
+--------+------+--------+-------+
如果將上例的LEFT JOIN替換成INNER JOIN,不論將條件過濾放到ON還是WHERE裡,結果都是一樣的,因為INNER JOIN不會執行第三步新增外部行
SELECT *
FROM user_info as i
INNER JOIN user_account as a
ON i.userid = a.userid and i.userid = 1003;
SELECT *
FROM user_info as i
INNER JOIN user_account as a
ON i.userid = a.userid where i.userid = 1003;
返回結果都是:
+--------+------+--------+-------+
| userid | name | userid | money |
+--------+------+--------+-------+
| 1003 | z | 1003 | 8 |
+--------+------+--------+-------+