Question

解决！谢谢你的帮助！

如何修复此查询

所以Pet的表格字段是

Pet_Code，名称，Employee_code，类型，性别

Employees表的字段是

Employee_code，姓氏，将First_Name，部

这是我尝试但不起作用的地方：

SELECT First_Name,Last_Name 
FROM employees E,pets P 
WHERE P.Employee_code=E.Employee_code AND COUNT(Type='C') > Count(Type='D');

Answer 1

我想你可以试试这个：

SELECT 
Employee_code,
Last_Name,
First_Name,
SUM(CASE WHEN Type = 'C' THEN 1 ELSE 0 END) AS CAT,
SUM(CASE WHEN Type = 'D' THEN 1 ELSE 0 END) AS DOG
FROM Employees_TABLE E
JOIN PET_TABLE P ON E.Employee_code = P.Employee_code
GROUP BY Employee_code 
HAVING SUM(CASE WHEN Type = 'C' THEN 1 ELSE 0 END) > SUM(CASE WHEN Type = 'D' THEN 1 ELSE 0 END)

Answer 2

好的，你可以试试这个：

SELECT DISTINCT
 E.Employee_code, E.Last_Name, E.Department,
 (SELECT COUNT(*) FROM pet WHERE Employee_code = E.Employee_code AND Type = 'C') AS cats,
 (SELECT COUNT(*) FROM pet WHERE Employee_code = E.Employee_code AND Type = 'D') AS dogs
FROM employee E
 LEFT JOIN pet P ON E.Employee_code = P.Employee_code
GROUP BY E.Employee_code
HAVING cats > dogs

它就像前一个，但更简单的是

Answer 3

也许你可以试试这个：

SELECT e.First_Name, e.Last_Name
FROM employees e INNER JOIN pets p ON e.Employee_code = p.Employee_code
WHERE 
    (SELECT COUNT(*) 
     FROM pets p1
     WHERE p1.Type = 'C' AND p.Employee_code = p1.Employee_code) > 
    (SELECT COUNT(*)
     FROM pets p1
     WHERE p1.Type = 'D' AND p.Employee_code = p1.Employee_code)
GROUP BY e.Employee_code;

这可能不是最佳解决方案，但应该完成这项工作。

Answer 4

select min(e.First_Name), min(e.Last_Name)
from emp e left outer join pets p
    on p.Employee_Code = e.Employee_Code
group by e.Employee_Code
having
    count(case when Type = 'C' then 1 end) >
    count(case when Type = 'D' then 1 end)

MYSQL-帮助进行查询

4 个答案: