我使用Wordpress插件创建SQL报告来查询MySQL表。我不熟悉SQL，但我希望查询具有来自相同用户的多个条目的表。我需要由学生和教师分开用户，按用户单独汇总五个类别的总数，并为学生开发一个排行榜，为教师开发一个排行榜。领导者基于高尔夫分数，该高尔夫分数通过在五个类别中的每一个中对用户进行排名并基于组合分数（来自添加排名）以升序排序来计算。
我提到的所有字段（用户，角色 - 学生或教师以及五个类别）都是表格列中名为field_name的值，需要转到标题。与每个字段关联的值位于field_value列下，第三列指定它来自的形式（Numbers）：
form_name field_name field_value
numbers user a
numbers role student
numbers category1 4
numbers category2 9
numbers category3 7
numbers category4 3
numbers category5 2
numbers user b
numbers role student
numbers category1 7
numbers category2 5
numbers category3 8
numbers category4 2
numbers category5 9
numbers user a
numbers role student
numbers category1 2
numbers category2 6
numbers category3 1
numbers category4 8
numbers category5 8
numbers user c
numbers role teacher
numbers category1 2
numbers category2 11
numbers category3 3
numbers category4 8
numbers category5 5
numbers user d
numbers role teacher
numbers category1 1
numbers category2 13
numbers category3 6
numbers category4 8
numbers category5 7
有不确定数量的用户。输出应列出用户名，每个类别的总和，每个类别中的排名以及最终排名。它不需要显示角色（作为学生或教师，因为我将学生排行榜放在一个页面上，而教师放在另一个页面上），表格名称或高尔夫评分（从加起来所有个别类别排名。即：
user c1 c1r c2 c2r c3 c3r c4 c4r c5 c5r golf overall form role
a 6 2 14 1 8 1 11 1 10 1 6 1 numbers student
b 7 1 5 2 8 1 2 2 9 2 8 2 numbers student
第二个但类似的查询将拉出教师编号：
user c1 c1r c2 c2r c3 c3r c4 c4r c5 c5r golf overall form role
d 1 2 13 1 6 1 8 1 7 1 6 1 numbers teacher
c 2 1 11 2 3 2 8 1 5 2 8 2 numbers teacher
c1 = category1， c1r = category1 rank
应隐藏高尔夫球得分，表格和角色列。
任何人都知道如何将其置于查询中？提前谢谢。
答案 0 :(得分：0)
我不得不将数据放入带有标识列的表中，并假设行将按顺序添加。除非你遗漏了一个专栏，否则就没有办法做到这一点。如果您无法在其他任何地方访问SQL Server，则应该可以使用SQL Fiddle。
MS SQL Server 2008架构设置：
create table T (id int not null identity, nm varchar(20), val varchar(20));
insert into T (nm, val) values
('user', 'a'),
('role', 'student'),
('category1', '4'),
('category2', '9'),
('category3', '7'),
('category4', '3'),
('category5', '2'),
('user', 'b'),
('role', 'student'),
('category1', '7'),
('category2', '5'),
('category3', '8'),
('category4', '2'),
('category5', '9'),
('user', 'a'),
('role', 'student'),
('category1', '2'),
('category2', '6'),
('category3', '1'),
('category4', '8'),
('category5', '8'),
('user', 'c'),
('role', 'teacher'),
('category1', '2'),
('category2', '11'),
('category3', '3'),
('category4', '8'),
('category5', '5'),
('user', 'd'),
('role', 'teacher'),
('category1', '1'),
('category2', '13'),
('category3', '6'),
('category4', '8'),
('category5', '7')
查询1 ：
with u0(id, name) as (
select id, val from T where nm = 'user'
),
users(id, [user], [role]) as (
select u.id, u.name, r.val
from u0 as u inner join T as r on r.id = u.id + 1 and r.nm = 'role'
),
scores([user], c1, c2, c3, c4, c5) as (
select
[user],
sum(case when category = 'category1' then score end),
sum(case when category = 'category2' then score end),
sum(case when category = 'category3' then score end),
sum(case when category = 'category4' then score end),
sum(case when category = 'category5' then score end)
from (
select
nm as category, cast(val as int) as score,
(
select [user] from users where id = (
select max(u.id) from u0 as u where u.id < s.id
)
) as [user]
from T as s
where nm like 'category_'
) as s_pivot
group by [user]
)
select
u.[role], u.[user],
c1, rank() over (partition by u.[role] order by c1 desc) as c1r,
c2, rank() over (partition by u.[role] order by c2 desc) as c2r,
c3, rank() over (partition by u.[role] order by c3 desc) as c3r,
c4, rank() over (partition by u.[role] order by c4 desc) as c4r,
c5, rank() over (partition by u.[role] order by c5 desc) as c5r,
rank() over (partition by u.[role] order by c1+c2+c3+c4+c5 desc) as overall
from
(select distinct [user], [role] from users) as u
inner join scores as s on s.[user] = u.[user]
order by
u.[role], u.[user]
<强> Results ：
| ROLE | USER | C1 | C1R | C2 | C2R | C3 | C3R | C4 | C4R | C5 | C5R | OVERALL |
|---------|------|----|-----|----|-----|----|-----|----|-----|----|-----|---------|
| student | a | 6 | 2 | 15 | 1 | 8 | 1 | 11 | 1 | 10 | 1 | 1 |
| student | b | 7 | 1 | 5 | 2 | 8 | 1 | 2 | 2 | 9 | 2 | 2 |
| teacher | c | 2 | 1 | 11 | 2 | 3 | 2 | 8 | 1 | 5 | 2 | 2 |
| teacher | d | 1 | 2 | 13 | 1 | 6 | 1 | 8 | 1 | 7 | 1 | 1 |
这是一个查询，可以让您轻松创建脚本以将数据传输到另一台服务器：
select 'insert into T (nm, val) values (''' + field_name + ''', ''' + field_value + ''')'
from ??