在 Oracle SQL Query 中转换 Text 中的一系列 Number 值

Question

在 Oracle 数据库中，我有字符串值 (VARCHAR2)，例如 1、4、7、8。数字表示为 1=car, 2= bus, 3=BB, 4=SB, 5=Ba, 6=PA, 7=HB, 8=G 并想在我的查询结果中将上述示例转换为“car,SB,HB,G” 我尝试使用“解码”，但它不起作用。请告知如何使其工作。将不胜感激。

谢谢` 最初，我使用了以下查询：

    Select Clientid as C#, vehicletypeExclusions as vehicle  from 
    clients 

结果样本是：

   C#   Vehicle
   20   1,19,20,23,24,7,5
   22   1,19,20,23,24,7,5

我还尝试了以下方法，使我获得了车辆的空值：

  Select Clientid as C#,  Decode (VEHICLETYPEEXCLUSIONS, '1', 'car', 
  '3','bus', '5','ba' ,'7','HB', '8','G'
  , '9','LED1102', '10','LED1104', '13','LED8-2',
  '14','Flip4-12', '17','StAT1003', '19','Taxi-Min', '20','Tax_Sed', 
   '21','Sup-veh'  , '22','T-DATS', '23','T-Mini',
   '24','T-WAM') as vehicle_Ex  from clients >

Answer 1

这是一个选项。在代码中读取 cmets。第 1 - 13 行中的样本数据；查询从第 14 行开始。

SQL> with
  2  expl (id, name) as
  3    (select 1, 'car' from dual union all
  4     select 2, 'bus' from dual union all
  5     select 3, 'BB'  from dual union all
  6     select 4, 'SB'  from dual union all
  7     select 5, 'Ba'  from dual union all
  8     select 6, 'PA'  from dual union all
  9     select 7, 'HB'  from dual union all
 10     select 8, 'G'   from dual
 11    ),
 12  temp (col) as
 13    (select '1,4,7,8' from dual),
 14  -- split COL to rows
 15  spl as
 16    (select regexp_substr(col, '[^,]+', 1, level) val,
 17            level lvl
 18     from temp
 19     connect by level <= regexp_count(col, ',') + 1
 20    )
 21  -- join SPL with EXPL; aggregate the result
 22  select listagg(e.name, ',') within group (order by s.lvl) result
 23  from expl e join spl s on s.val = e.id;

RESULT
--------------------------------------------------------------------------------
car,SB,HB,G

SQL>

Answer 2

使用来自https://stackoverflow.com/a/68537479/429100的函数f_subst：

create or replace
function f_subst(str varchar2, template varchar2, subst sys.odcivarchar2list) return varchar2
as
    res varchar2(32767):=str;
begin
    for i in 1..subst.count loop
        res:=replace(res, replace(template,'%d',i), subst(i));
    end loop;
    return res;
end;
/

我已将ora_name_list_t（嵌套表）替换为sys.odcivarchar2list（varray）以简化此示例，但我建议您创建自己的集合，例如create type varchar2_table as table of varchar2(4000);

例子：

select
   f_subst(
      '1,4,7,8'
     ,'%d'
     ,sys.odcivarchar2list('car','bus','BB','SB','Ba','PA','HB','G')
   ) s 
from dual; 

S
----------------------------------------
car,SB,HB,G

Answer 3

假设您有一个查找表（将数字代码与描述相关联）和一个输入字符串表，我在测试中将其称为sample_inputs，如下所示：

create table lookup (code, descr) as
    select 1, 'car' from dual union all
    select 2, 'bus' from dual union all
    select 3, 'BB'  from dual union all
    select 4, 'SB'  from dual union all
    select 5, 'Ba'  from dual union all
    select 6, 'PA'  from dual union all
    select 7, 'HB'  from dual union all
    select 8, 'G'   from dual
;

create table sample_inputs (str) as
    select '1,4,7,8' from dual union all
    select null      from dual union all
    select '3'       from dual union all
    select '5,5,5'   from dual union all
    select '6,2,8'   from dual
;

解决问题的一种策略是拆分输入 - 稍作修改以使其成为 JSON 数组，以便我们可以使用 json_table 拆分它 - 然后加入查找表并重新聚合。

select s.str, l.descr_list
from   sample_inputs s cross join lateral
       ( select listagg(descr, ',') within group (order by ord) as descr_list
         from   json_table( '[' || str || ']', '$[*]'
                            columns code number path '$', ord for ordinality)
                join lookup l using (code)
       ) l
;

STR     DESCR_LIST                    
------- ------------------------------
1,4,7,8 car,SB,HB,G                   
                                      
3       BB                            
5,5,5   Ba,Ba,Ba                      
6,2,8   PA,bus,G