如何获得每个公司每个项目类型的前3名最高评级用户的表？
contact users
-------------------------- ----------------
id id
Contact_Title Name_Title
Contact_First Name_First
Contact_Middle Name_Middle
Contact_Last Name_Last
Lead_Referral_Source Email
Date_of_Initial_Contact Password
Title User_Roles
Company User_Status
Industry
Address
Address_Street_1
Address_Street_2
Address_City
Address_State
Address_Zip
Address_Country
Phone
Email
Status
Website
LinkedIn_Profile
Background_Info
Sales_Rep
Rating
Project_Type
Project_Description
Proposal_Due_Date
Budget
Deliverables
在SQL方面，我只是一个初学者，在逻辑方面遇到了困难。我在这个问题上还没有取得任何进展，很想看看你们可能会如何尝试并听到一些建议。非常感谢您的帮助：）
答案 0 :(得分：0)
我找不到通过单个查询执行此操作的方法，但这是一个存储过程，它将创建一个表以存储结果，然后使用所需的结果填充该表：
DROP PROCEDURE IF EXISTS Get_Top_3_Rated_Salespeople_Per_Company_and_Project_Type;
DELIMITER //
CREATE PROCEDURE IF NOT EXISTS Get_Top_3_Rated_Salespeople_Per_Company_and_Project_Type()
BEGIN
DECLARE ct int;
DECLARE comboCT int;
DECLARE lt int;
/* Create table where results will be stored */
/* WARNING: Drops existing table first! */
DROP TABLE IF EXISTS `Top_3_Rated_Salespeople_Per_Company_and_Project_Type`;
CREATE TABLE IF NOT EXISTS `Top_3_Rated_Salespeople_Per_Company_and_Project_Type` (
Name_First varchar(30),
Name_Middle varchar(30),
Name_Last varchar(100),
Email varchar(100),
Rating numeric(6,4),
Company varchar(100),
Project_Type varchar(100)
);
/* Create temp table to store distinct Company/Project_Type combinations */
CREATE TEMPORARY TABLE IF NOT EXISTS temp_distinct_combos
SELECT Project_Type, Company FROM contact GROUP BY Project_Type, Company;
SET ct = 0;
SET comboCT = (SELECT COUNT(*) FROM temp_distinct_combos);
/* For each combination in temp table, get top three ratings with associated user data
/* and inset into Top_3_Rated_Salespeople_Per_Company_and_Project_Type */
WHILE ct <= (comboCT + 1) DO
SET lt = 1;
INSERT INTO
`Top_3_Rated_Salespeople_Per_Company_and_Project_Type`
SELECT
u.Name_First,
u.Name_Middle,
u.Name_Last,
u.Email,
c.Rating,
c.Company,
c.Project_Type
FROM
contact as c
LEFT OUTER JOIN
users AS u
ON
u.id = c.Sales_Rep
WHERE
/* Company/Project_Type combination matches current record in temp_distinct_combos */
CONCAT(c.Company, c.Project_Type) = (
SELECT
CONCAT(Company, Project_Type)
FROM
temp_distinct_combos
LIMIT
ct, lt /* this will limit this subquery to returning only one row at a time */
)
ORDER BY
c.Company,
c.Project_Type,
c.Rating DESC
LIMIT 3;
SET ct = ct + 1;
END WHILE;
END //
DELIMITER ;
我用来设置测试的代码（注意：发布问题时，您应该包括这些内容；周围的大多数人都不会这么帮助您）：
DROP TABLE IF EXISTS contact;
CREATE TABLE IF NOT EXISTS contact (
id int,
Contact_Title varchar(40),
Contact_First varchar(40),
Contact_Middle varchar(40),
Contact_Last varchar(40),
Lead_Referral_Source varchar(40),
Date_of_Initial_Contact varchar(40),
Title varchar(40),
Company varchar(40),
Industry varchar(40),
Address varchar(40),
Address_Street_1 varchar(40),
Address_Street_2 varchar(40),
Address_City varchar(40),
Address_State varchar(40),
Address_Zip varchar(40),
Address_Country varchar(40),
Phone varchar(40),
Email varchar(40),
Status varchar(40),
Website varchar(40),
LinkedIn_Profile varchar(40),
Background_Info varchar(40),
Sales_Rep int,
Rating numeric(6,4),
Project_Type varchar(40),
Project_Description varchar(40),
Proposal_Due_Date varchar(40),
Budget varchar(40),
Deliverables varchar(40)
);
DROP TABLE IF EXISTS users;
CREATE TABLE IF NOT EXISTS users (
id int,
Name_Title varchar(40),
Name_First varchar(40),
Name_Middle varchar(40),
Name_Last varchar(40),
Email varchar(40),
Password varchar(40),
User_Roles varchar(40),
User_Status varchar(40)
);
INSERT INTO
users
(id, Name_First, Name_Last, Email)
VALUES
(1, 'Name_First_1', 'Name_Last_1', 'Email_1'),
(2, 'Name_First_2', 'Name_Last_2', 'Email_2'),
(3, 'Name_First_3', 'Name_Last_3', 'Email_3'),
(4, 'Name_First_4', 'Name_Last_4', 'Email_4'),
(5, 'Name_First_5', 'Name_Last_5', 'Email_5'),
(6, 'Name_First_6', 'Name_Last_6', 'Email_6'),
(7, 'Name_First_7', 'Name_Last_7', 'Email_7'),
(8, 'Name_First_8', 'Name_Last_8', 'Email_8'),
(9, 'Name_First_9', 'Name_Last_9', 'Email_9'),
(10, 'Name_First_10', 'Name_Last_10', 'Email_10'),
(11, 'Name_First_11', 'Name_Last_11', 'Email_11');
INSERT INTO
contact
(id, Contact_First, Contact_Last, Sales_Rep, Rating, Project_Type, Company)
VALUES
(1, 'Contact_First_1', 'Contact_Last_1', 1, 10, 'Type_1', 'Company_1'),
(2, 'Contact_First_2', 'Contact_Last_2', 2, 9.9, 'Type_1', 'Company_1'),
(3, 'Contact_First_3', 'Contact_Last_3', 3, 9.8, 'Type_1', 'Company_1'),
(4, 'Contact_First_4', 'Contact_Last_4', 4, 9.7, 'Type_1', 'Company_1'),
(5, 'Contact_First_5', 'Contact_Last_5', 5, 9.6, 'Type_1', 'Company_1'),
(6, 'Contact_First_6', 'Contact_Last_6', 6, 9.5, 'Type_2', 'Company_1'),
(7, 'Contact_First_7', 'Contact_Last_7', 7, 9.4, 'Type_2', 'Company_1'),
(8, 'Contact_First_8', 'Contact_Last_8', 8, 9.3, 'Type_2', 'Company_1'),
(9, 'Contact_First_9', 'Contact_Last_9', 9, 9.2, 'Type_2', 'Company_1'),
(10, 'Contact_First_10', 'Contact_Last_10', 10, 9.1, 'Type_2', 'Company_1'),
(11, 'Contact_First_11', 'Contact_Last_11', 11, 9, 'Type_2', 'Company_1'),
(12, 'Contact_First_12', 'Contact_Last_12', 1, 10, 'Type_1', 'Company_2'),
(13, 'Contact_First_13', 'Contact_Last_13', 2, 9.9, 'Type_1', 'Company_2'),
(14, 'Contact_First_14', 'Contact_Last_14', 3, 9.8, 'Type_1', 'Company_2'),
(15, 'Contact_First_15', 'Contact_Last_15', 4, 9.7, 'Type_1', 'Company_2'),
(16, 'Contact_First_16', 'Contact_Last_16', 5, 9.6, 'Type_2', 'Company_2'),
(17, 'Contact_First_17', 'Contact_Last_17', 6, 9.5, 'Type_2', 'Company_2'),
(18, 'Contact_First_18', 'Contact_Last_18', 7, 9.4, 'Type_2', 'Company_2'),
(19, 'Contact_First_19', 'Contact_Last_19', 8, 9.3, 'Type_2', 'Company_2'),
(20, 'Contact_First_20', 'Contact_Last_20', 9, 9.2, 'Type_2', 'Company_2'),
(21, 'Contact_First_21', 'Contact_Last_21', 10, 9.1, 'Type_2', 'Company_2'),
(22, 'Contact_First_22', 'Contact_Last_22', 11, 9, 'Type_2', 'Company_2'),
(23, 'Contact_First_23', 'Contact_Last_23', 1, 10, 'Type_1', 'Company_3'),
(24, 'Contact_First_24', 'Contact_Last_24', 2, 9.9, 'Type_1', 'Company_3'),
(25, 'Contact_First_25', 'Contact_Last_25', 3, 9.8, 'Type_2', 'Company_3'),
(26, 'Contact_First_26', 'Contact_Last_26', 4, 9.7, 'Type_2', 'Company_3'),
(27, 'Contact_First_27', 'Contact_Last_27', 5, 9.6, 'Type_2', 'Company_3'),
(28, 'Contact_First_28', 'Contact_Last_28', 6, 9.5, 'Type_2', 'Company_3'),
(29, 'Contact_First_29', 'Contact_Last_29', 7, 9.4, 'Type_2', 'Company_3'),
(30, 'Contact_First_30', 'Contact_Last_30', 8, 9.3, 'Type_2', 'Company_3');
如果这可以解决您的问题，请记住标记为已回答，这样每个人都会知道您不再需要帮助（因此，我花了两个小时为您服务：-），我将得到奖励。如果这不能满足您的需求，请尽可能详细地说明它与您期望的结果有何不同，我将尝试进行修改。