postgresql – count()对应于满足某些条件的不同值的max()

我有以下表格:

USER_GROUP

usergrp_id      bigint     Primary Key
usergrp_name    text  

用户

user_id             bigint     Primary Key
user_name           text
user_usergrp_id     bigint
user_loc_id         bigint

user_usergrp_id具有user_group表中的相应id
user_loc_id具有来自分支表的相应id(branch_id).

branch_id      bigint     Primary Key
branch_name    text
branch_type    smallint

branch_type默认设置为1.虽然它可以包含1到4之间的任何值.

user_projects

proj_id          bigint     Primary Key
proj_name        text
proj_branch_id   smallint

proj_branch_id具有来自分支表的相应id(branch_id).

user_approval

appr_id           bigint     Primary Key
appr_prjt_id      bigint
appr_status       smallint
appr_approval_by  bigint

appr_approval_by具有来自用户表的相应id(user_id)
appr_status可能包含不同的状态值,例如10,20,30 …对于单个appr_prjt_id

USER_GROUP

usergrp_id | usergrp_name
-------------------------
    1      | Admin
    2      | Manager

用户

user_id | user_name | user_usergrp_id |user_loc_id
---------------------------------------------------
    1   | John      |      1          |     1
    2   | Harry     |      2          |     1

branch_id | branch_name | branch_type
-------------------------------------
    1     |  location1  |    2
    2     |  location2  |    1
    3     |  location3  |    4
    4     |  location4  |    2
    5     |  location4  |    2

user_projects

proj_id | proj_name | proj_branch_id
------------------------------------
    1   | test1      |       1
    2   | test2      |       2
    3   | test3      |       1
    4   | test4      |       3
    5   | test5      |       1
    6   | test5      |       4

user_approval

appr_id | appr_prjt_id | appr_status | appr_approval_by
-------------------------------------------------------
    1   |    1         |     10      |     1
    2   |    1         |     20      |     1
    3   |    1         |     30      |     1
    4   |    2         |     10      |     2
    5   |    3         |     10      |     1
    6   |    3         |     20      |     2
    7   |    4         |     10      |     1
    8   |    4         |     20      |     1

条件:输出必须为每个appr_prjt_id获取appr_status的MAX()值并对其进行计数.

即,在上表中,appr_prjt_id = 1具有3种不同的状态:10,20,30.其计数必须仅显示在输出中对应于30的状态(不在状态10和20中),对应于用户组中的特定的branch_name.类似地,对于appr_prjt_id字段中的每个其他id

SQL Fiddle

期望的输出:

                           10   |    20  |  30

         ------> Admin     0    |    1   |   1
         |
location1
         |
         ------> Manager   1    |    1   |   0

我怎样才能做到这一点?

SQL Fiddle

最佳答案
SQL Fiddle

select
    branch_name, usergrp_name,
    sum((appr_status = 10)::integer) "10",
    sum((appr_status = 20)::integer) "20",
    sum((appr_status = 30)::integer) "30"
from
    (
        select distinct on (appr_prjt_id)
            appr_prjt_id, appr_approval_by, appr_status
        from user_approval
        order by 1, 3 desc
    ) ua
    inner join
    users u on ua.appr_approval_by = u.user_id
    inner join
    user_group ug on u.user_usergrp_id = ug.usergrp_id
    inner join
    branch b on u.user_loc_id = b.branch_id
group by branch_name, usergrp_name
order by usergrp_name

适用于大多数DBMS的经典解决方案是使用案例:

select
    branch_name, usergrp_name,
    sum(case appr_status when 10 then 1 else 0 end) "10",

但Postgresql具有布尔类型,并且它具有转换为整数(布尔值::整数),从而产生0或1,这使得代码更简洁.

在这种情况下,也可以进行计数而不是总和:

select
    branch_name, usergrp_name,
    count(appr_status = 10 or null) "10",

我确实更喜欢伯爵,但我的印象是难以理解.诀窍是要知道count计算任何非null值并且a(true或null)为真且a(false或null)为null,因此只要条件为真,它就会计数.

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