如何使这个 sql 查询与旧版本的 mysql 兼容，例如 v.5.5 及更高版本？答案

【问题标题】：How to make this sql query compatible for older verisons of mysql such v.5.5 and above?如何使这个 sql 查询与旧版本的 mysql 兼容，例如 v.5.5 及更高版本？
【发布时间】：2021-04-02 20:25:25
【问题描述】：

我有这个查询，它适用于 MySQL 8.0，但不适用于 MySQL 5.5 等以下版本。当我在 Web 服务器上收到此错误时如何使其兼容：#1064 - 您的 SQL 语法有错误；检查与您的 MySQL 服务器版本相对应的手册，以在第 11 行的 '(partition by pid order by age' 附近使用正确的语法

SELECT p.id,  p.pid, p.name, 
       MAX(CASE WHEN t.dr = 1 THEN t.name END) as name1,
       MAX(CASE WHEN t.dr = 1 THEN age END) as age1,
       MAX(CASE WHEN t.dr = 2 THEN t.name END) as name2,
         MAX(CASE WHEN t.dr = 2 THEN t.age END) as age2,
            MAX(CASE WHEN t.dr = 3 THEN t.name END) as name3,
         MAX(CASE WHEN t.dr = 3 THEN t.age END) as age3,
             MAX(CASE WHEN t.dr = 4 THEN t.name END) as name4,
         MAX(CASE WHEN t.dr = 4 THEN t.age END) as age4
FROM Table1 p
join (select id , pid, name, age,  DENSE_RANK() OVER (partition by pid order by age) as dr
from Table2) t on p.pid= t.pid
group by p.id, p.pid, p.name

https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=d324e7654218dae3d1168df2bf529099

【问题讨论】：

标签： mysql sql

【解决方案1】：

 select version();

version()
5.5.62

架构：

 CREATE TABLE Table1
     (id int , name varchar(15) , pid int   )

 INSERT INTO Table1
  values   (1, 'Jack', 1521), 
  (2, 'Steve',1522)

 CREATE TABLE Table2
     (id  int, pid int, name varchar(15), age int);
 
 INSERT INTO Table2
    
 VALUES
     (1, 1521, 'John', 12),
     (2, 1521, 'Maria', 7),
     (3, 1521, 'Larry', 3),
     (4, 1522, 'Harry', 5)

查询：

 SELECT p.id,  p.pid, p.name, 
        MAX(CASE WHEN t.dr = 1 THEN t.name END) as name1,
        MAX(CASE WHEN t.dr = 1 THEN age END) as age1,
        MAX(CASE WHEN t.dr = 2 THEN t.name END) as name2,
          MAX(CASE WHEN t.dr = 2 THEN t.age END) as age2,
             MAX(CASE WHEN t.dr = 3 THEN t.name END) as name3,
          MAX(CASE WHEN t.dr = 3 THEN t.age END) as age3,
              MAX(CASE WHEN t.dr = 4 THEN t.name END) as name4,
          MAX(CASE WHEN t.dr = 4 THEN t.age END) as age4
 FROM Table1 p
 join (select id , pid, name, age,  (select count(*) from Table2 t2
 where  t.pid=t2.pid and t2.age<=t.age) as dr
 from Table2 t) t on p.pid= t.pid
 group by p.id, p.pid, p.name

输出：

id	pid	name	name1	age1	name2	age2	name3	age3	name4	age4
1	1521	Jack	Larry	3	Maria	7	John	12	null	null
2	1522	Steve	Harry	5	null	null	null	null	null	null

db小提琴here

【讨论】：

【解决方案2】：

一种方法是用相关子查询替换DENSE_RANK()：

FROM Table1 p JOIN
     (SELECT t2.*,
             (SELECT COUNT(DISTINCT tt2.age)
              FROM table2 tt2
              WHERE tt2.pid = t2.pid AND
                    tt2.age <= t2.age
             ) as dr
      FROM Table2 t2
     ) t 
     ON p.pid = t.pid

Here 是一个 dbfiddle。

【讨论】：