Question

Ive a Great selected case in Prof.

(select case tb_usuario.int_id_cargo 
                    when 13
                    then  20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53 
                    when 20
                    then   
                    when 21
                    then  20 
                    when 22
                    then   
                    when 23
                    then  22 
                    when 24
                    then  20,21,22,23 
                    when 25
                    then  20,21,22,23,24 
                    when 26
                    then  20,21,22,23,24,25 
                    when 27
                    then   
                    when 28
                    then  27 
                    when 29
                    then   
                    when 30
                    then  29 
                    when 31
                    then  27,28 
                    when 32
                    then  27,28,31 
                    when 33
                    then  29,30 
                    when 34
                    then  27,28,29,30,31,32,33 
                    when 35
                    then   
                    else   
                    end) as subs

如果你注意的话,你会认识到,有许多案例会恢复空洞。我要检查那些在单一案件中返回空洞的人,即:

  (select case tb_usuario.int_id_cargo 
                        when 20,22,27,29,35
                                        then   
                                         [...]

这是可能的吗?

Answer 1

如何使用表格(临时表格)在不使用<编码>CASE说明的情况下选定这些数值?

migration (table name)
old_id      new_id
---------   --------------
24          20,21,22,23
30          29
.....

您的问候会

select migration.new_id 
from migration inner join tb_usuario
on migration.old_id = tb_usuario.int_id_cargo

EDIT:对于多旧的_指新_和空白str,你必须单独制作条目。

e.g. (假设旧的1 &2点指新_id->空白)

migration (table name)
old_id      new_id
---------   --------------
1          
2

这样做的好处是,将新金从空白改为使用本表的其他东西,而不是改变上文提到的询问。

Answer 2

我将把它留给SE条款。仅指定您希望填写在CASE上。除非你有理由在案件陈述中保留这些身份证。

(select case tb_usuario.int_id_cargo 
                    when 13
                    then  20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53 
                    when 21
                    then  20 
                    when 23
                    then  22 
                    when 24
                    then  20,21,22,23 
                    when 25
                    then  20,21,22,23,24 
                    when 26
                    then  20,21,22,23,24,25 
                    when 28
                    then  27 
                    when 30
                    then  29 
                    when 31
                    then  27,28 
                    when 32
                    then  27,28,31 
                    when 33
                    then  29,30 
                    when 34
                    then  27,28,29,30,31,32,33 
                    else   
                    end) as subs

Answer 3

使用:

(SELECT CASE 
         WHEN tb_usuario.int_id_cargo = 13 THEN
           20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53 
         WHEN tb_usuario.int_id_cargo IN(20, 22, 27, 29, 35) THEN
              
         WHEN tb_usuario.int_id_cargo = 21 THEN
             20 
         WHEN tb_usuario.int_id_cargo = 23 THEN
             22 
         WHEN tb_usuario.int_id_cargo = 24 THEN
             20,21,22,23 
         WHEN tb_usuario.int_id_cargo = 25 THEN
             20,21,22,23,24 
         WHEN tb_usuario.int_id_cargo = 26 THEN
             20,21,22,23,24,25 
         WHEN tb_usuario.int_id_cargo = 28 THEN
             27 
         WHEN tb_usuario.int_id_cargo = 30 THEN
             29 
         WHEN tb_usuario.int_id_cargo = 31 THEN
             27,28 
         WHEN tb_usuario.int_id_cargo = 32 THEN
             27,28,31 
         WHEN tb_usuario.int_id_cargo = 33 THEN
            29,30 
         WHEN tb_usuario.int_id_cargo = 34 THEN
            27,28,29,30,31,32,33 
         ELSE   
        END) as subs

尽管我看不出区分应当恢复零长度的特有价值,而只是让ELSE(回报同样的价值)达到这些价值:

(SELECT CASE tb_usuario.int_id_cargo 
         WHEN 13 THEN
            20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53 
         WHEN 21 THEN
             20 
         WHEN 23 THEN
             22 
         WHEN 24 THEN
             20,21,22,23 
         WHEN 25 THEN
             20,21,22,23,24 
         WHEN 26 THEN
             20,21,22,23,24,25 
         WHEN 28 THEN
             27 
         WHEN 30 THEN
             29 
         WHEN 31 THEN
             27,28 
         WHEN 32 THEN
             27,28,31 
         WHEN 33 THEN
            29,30 
         WHEN 34 THEN
            27,28,29,30,31,32,33 
         ELSE   
        END) as subs

Answer 4

select (case 
       when tbl_usuario.int_id_cargo in (20,22,27,29,35,.....) then    
       else
            case tbl_usuario.int_id_cargo
            when 23 then  22 
            when 25 then  20,21,22,23,24 
            when ....
            end
        end) as subs

replace .... with the rest of the ids and then fix the second case statement so that matches up for all the other values

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