Question

我正在玩SQL，所以如果我在求职面试中被问到，我并不是完全无知。我的朋友最近在接受采访时被问到以下问题而且他无法得到它，我问工作中哪些人体面地了解SQL并且他不知道。你能为我解答这个问题，然后解释它是如何工作的吗？请？

* 问题 *

数据库规范化（或缺乏规范化）通常会给开发人员带来挑战。

考虑一个包含三个字段的员工数据库表：

EmployeeID
EmployeeName
EmailAddresses

每位员工（由唯一的EmployeeID标识）可能在EmailAddresses字段中包含一个或多个以逗号分隔的@ rockauto.com电子邮件地址。

数据库表定义如下：

CREATE TABLE Employees
(
  EmployeeID int UNSIGNED NOT NULL PRIMARY KEY,
  EmployeeName varchar(50) NOT NULL,
  EmailAddresses varchar(40) NOT NULL ,
  PRIMARY KEY(EmployeeID)
);

出于测试目的，这里有一些示例数据：

INSERT INTO Employees (EmployeeID, EmployeeName, EmailAddresses) VALUES
('1', 'Bill', 'bill@companyx.com'),
('2', 'Fred', 'fred@companyx.com,freddie@companyx.com'),
('3', 'Fred', 'fredsmith@companyx.com'),
('4', 'Joe', 'joe@companyx.com,joe_smith@companyx.com');

您的任务是编写一个MySQL SELECT查询，该查询将显示以上示例数据的以下输出：

Employee    EmailAddress
Bill    bill@companyx.com
Fred (2)    fred@companyx.com
Fred (2)    freddie@companyx.com
Fred (3)    fredsmith@companyx.com
Joe     joe@companyx.com
Joe     joe_smith@companyx.com

请注意，因为有多个人姓名相同（在本例中为“Fred”），所以EmployeeID包含在括号中。

您的查询需要使用MySQL版本5.1.41兼容语法编写。您应该假设使用标准数据库升序排序完成排序：“ORDER BY EmployeeID ASC”

对于此问题，您需要提交单个SQL SELECT查询。您的查询应该能够在合理的时间内处理1000条记录的表。

Answer 1

只有当您的电子邮件少于10000封时......可以接受吗？

select 
       if(t1.c > 1, concat(e.employeename, ' (', e.employeeid, ')'), e.employeename) as Employee,
       replace(substring(substring_index(e.EmailAddresses, ',', n.row), length(substring_index(e.EmailAddresses, ',', n.row - 1)) + 1), ',', '') EmailAddress 
from 
       (select employeename, count(*) as c from Employees group by employeename) as t1, 
       (select EmployeeID, length(EmailAddresses) - length(replace(EmailAddresses,',','')) + 1 as emails from Employees) as t2,
       (SELECT @row := @row + 1 as row FROM 
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x,
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x2, 
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x3, 
       (select 0 union all select 1 union all select 3 union all select 4 union all select 5 union all select 6 union all select 6 union all select 7 union all select 8 union all select 9) x4, 
       (SELECT @row:=0) as ff) as n,
       Employees e
where 
      e.employeename = t1.employeename and
      e.employeeid = t2.employeeid and
      n.row <= t2.emails
order by e.employeeid;

修改

产生的无用数字越来越少：

select if(t1.c > 1, concat(e.EmployeeName, ' (', e.EmployeeID, ')'), e.EmployeeName) as Employee, replace(substring(substring_index(e.EmailAddresses, ',', n.row), length(substring_index(e.EmailAddresses, ',', n.row - 1)) + 1), ',', '') as EmailAddress from (select EmployeeName, count(*) as c from Employees group by EmployeeName) as t1, (select EmployeeID, length(EmailAddresses) - length(replace(EmailAddresses,',','')) + 1 as emails from Employees) as t2, (select `1` as row from (select 1 union all select 2 union all select 3 union all select 4) x) as n, Employees e where e.EmployeeName = t1.EmployeeName and e.EmployeeID = t2.EmployeeID and n.row <= t2.emails order by e.EmployeeID;

我们学到了什么？糟糕的数据库设计导致糟糕的查询。而且你可以用SQL做一些事情，这可能只是因为人们做不好的数据库设计而得到支持......：）

SELECT将规范化列拆分为单独的记录？

1 个答案: