刷Leecode時遇到的MySQL知識點整理html
1. case ... when ... then ...[when ... then ...] else ... endjava
https://blog.csdn.net/helloxiaozhe/article/details/78124138sql
2. limit offset網絡
兩者一般與order by聯合使用以求取第幾大第幾小的記錄。函數
使用區別：假設要返回某一字段的第2-4大的數據性能
limit 3 offset 1 --讀取三條，從第二條開始讀。讀取數量在前，偏移量在後學習
limit 1,3 --從第二條（偏移量從0開始）開始讀，讀取三條。讀取數量在後，偏移量在前測試
select distinct salary from employee order by salary desc limit 3 offset 1 select distinct salary from employee order by salary desc limit 1,3
此外，使用limit,offset能夠實現分頁查詢,pageNumber:頁碼，numberPerPage：每頁數據量優化
select * from table_name [過濾條件] limit pageNumber offset (pageNumber-1)*numberPerPage select * from table_name [過濾條件] limit (pageNumber-1)*numberPerPage,pageNumber
3. IFNULL(expr1,expr2)spa
若是expr1不是NULL，IFNULL()返回expr1，不然它返回expr2。IFNULL()返回一個數字或字符串值
對於2的第一個例子，更嚴謹的寫法是(當查不到數據時返回null，參考https://leetcode-cn.com/problems/second-highest-salary)，這個也是提醒咱們要考慮嚴謹。
select ifnull((select distinct salary from employee order by salary desc limit 1,3),null) as conSalary
4. MySQL Date 函數
http://www.w3school.com.cn/sql/sql_dates.asp
DATE_ADD(date,INTERVAL expr type) : 函數向日期添加指定的時間間隔，例date_add('2019-03-31',interval 1 day)
DATE_SUB(date,INTERVAL expr type) : 函數從日期減去指定的時間間隔，例SELECT OrderId,DATE_SUB(OrderDate,INTERVAL 2 DAY) AS OrderPayDate FROM Orders
DATEDIFF(date1,date2) ： 函數返回兩個日期之間的天數，例SELECT DATEDIFF('2008-12-30','2008-12-29') AS DiffDate 返回1，要是第一個參數小於第二個參數就返回負值
EXTRACT(unit FROM date) ： 函數用於返回日期/時間的單獨部分，好比年、月、日、小時、分鐘等等。
CURTIME()： 函數返回當前的時間。
CURDATE()： 函數返回當前的日期。
NOW()： 函數返回當前的日期和時間。
DATE(date)： 函數返回日期或日期/時間表達式的日期部分
DATE_FORMAT(date,format)：函數用於以不一樣的格式顯示日期/時間數據。
5. join、inner join、left join、right join、full join
參考http://www.w3school.com.cn/sql/sql_join.asp
join和inner join相同，full join至關於left join、right join查詢的並集
6.判斷字段奇偶的方式
https://blog.csdn.net/zhazhagu/article/details/80452473
7.group by，order by，having，聚合函數等做用於結果集上
8.錯誤
You can't specify target table 'person' for update in FROM clause
https://zhidao.baidu.com/question/68619324.html
不能對同一個表select的同時進行更新操做，例以下面的select t.id from就是避免對同一表操做
Every derived table must have its own alias
子查詢必須有別名，例以下面的別名t必須有
例：題目來自於https://leetcode-cn.com/problems/delete-duplicate-emails/
Unknown column 't.id' in 'field list'
min(id)默認列名就是min(id)，因此要記得加別名id
delete from person where id not in ( select t.id from ( select min(p.id) id from person p group by p.email) t )
9.distinct c1,c2,...
過濾掉c1,c2,....都相同的數據記錄，即判斷多列同時不重複
10. and,or優化查詢
https://www.cnblogs.com/ouhouki/p/9661927.html
11.MySQL中實現rank排名查詢
https://blog.csdn.net/justry_deng/article/details/80597916
例子：https://leetcode-cn.com/problems/rank-scores/
select score, convert(case when @preS = score then @curR when @preS := score then @curR := @curR+1 when @preS = 0 then @curR := @curR+1 end,unsigned integer) as rank from Scores , (select @curR:=0, @preS:=NULL ) r order by score desc --------------------------------------- SELECT Score, (SELECT count(DISTINCT score) FROM Scores WHERE score >= s.score) AS Rank FROM Scores s ORDER BY Score DESC ;
第二種性能不如第一種
12.自定義函數
https://leetcode-cn.com/problems/nth-highest-salary/
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT BEGIN set n = n-1; RETURN ( # Write your MySQL query statement below. select t.salary from(select distinct salary from employee order by salary desc limit 1 offset n) t ); END
13. case when then或者if交換座位問題
https://leetcode-cn.com/problems/exchange-seats/
--------------------------------方法一---------------------------------- select t.id, case when t.id%2=0 then (select s.student from seat s where s.id = t.id-1 ) when t.id=(select max(id) from seat) then student when t.id%2=1 then (select s.student from seat s where s.id = t.id+1 ) end as student from seat t ----------------------------------方法二---------------------------------- select if(mod(id,2)=1 and id = (select count(1) from seat),id,if(mod(id,2)=1,id+1,id-1)) id,student from seat order by id ----------------------------------方法三---------------------------------- SELECT (CASE WHEN MOD(id,2) = 1 AND id = (SELECT COUNT(*) FROM seat) THEN id WHEN MOD(id,2) = 1 THEN id+1 ElSE id-1 END) AS id, student FROM seat ORDER BY id;
方法二三，性能至關，if性能稍快於case（可能受網絡影響），都比第一種快了一倍，第一種由於兩個子查詢耗時。
14. 重命名不能出如今where後面嗎？
（下面會報Unknown column 'c' in 'where clause'）
https://leetcode-cn.com/problems/consecutive-numbers/
select num, convert(( case when @pre=num then @n:=@n+1 when @pre:=num then @n:=1 end ),unsigned int) as c from logs ,(select @n:=0,@pre:=null)r where c>=3（去掉「where c>=3」就正確了） -------------------------------------------正確答案------------------------------------------------------------------------- select distinct t.num as ConsecutiveNums from (select num, convert(( case when @pre=num then @n:=@n+1 when @pre:=num then @n:=1 end ),unsigned int) as c from logs ,(select @n:=0,@pre:=null)r)t where t.c>=3
注意點：
①變量前的@必定不要忘帶，「:」不要忘寫
②別名c不能在當前where中使用，會報c找不到
③distinct過濾掉重複的（[1,3],[1,4],[1,5],[1,6],[1,7]，若是不過濾的話會輸出[1,1,1,1,1]）
15.聚合函數只能select被聚合的字段以及分組字段，其餘字段查詢，會發生錯位，由於其它字段與被聚合字段和分組字段是多對一的關係。
16. 派生表關聯，多表關聯
題目來自https://leetcode-cn.com/problems/department-highest-salary/
有Employee，Department 兩張表和由Employee生成的派生表，共有3張表，每張表取一個字段。必定要讓三個表互相關聯到，否則會出現數據錯亂
三表關聯where測試比join關係性能好點，不知原理爲什麼？
select d.name Department , e.name Employee, m.salary
from Employee e,Department d,
(select max(salary) salary,departmentid from employee group by departmentid)m
where e.Salary = m.Salary and d.id=m.departmentid and e.DepartmentId =d.id
------------------
select d.name Department, e.name Employee, m.salary
from
(select max(salary) salary,departmentid from employee group by departmentid) m
join Employee e on e.Salary = m.Salary
join Department d on d.id=m.departmentid and d.id = e.departmentid
要增強學習，作到知其然，也要知其因此然。
17.round(x,d)：x保留d位小數
爲了獲得而努力
2019-04-01
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