sql - ORACLE SQL 查询表使用来自其他表的条件

Question

TABLEA 包含数据，而 TABLEB 包含搜索条件

这是一个包含数据的 SQL Fiddle

1. 表

   TABLEA
   visited_states_time
   AL= Alabama,2, AK=Alaska,5
   AR=Arkansas,6
   AZ=Arizona,10
   CA=California, 10,CT=Connecticut,20
   
   TABLEB
   CRITERIA
   AL
   HI
   CA
   CT
   AK
   

2. 期望的结果

   visited_states ...................................     total_time_spent
   AL= Alabama, AK=Alaska ............................     7
   CA=California, CT=Connecticut...................        30
   

Answer 1

这是一个糟糕的数据模型。你也没有说条件tableb。如果有任何状态匹配，或者如果全部匹配？

因为我们需要将行拆分（到 sum()）然后重新组合它们，您可以使用：

SQL> with v as (select rownum r,
  2                    ','||visited_states_time||',' visited_states_time,
  3                    length(
  4                      regexp_replace(visited_states_time, '[^,]', '')
  5                    )+1 fields
  6               from tablea)
  7  select trim(both ',' from visited_states_time) visited_states_time,
  8         sum(total_time_spent) total_time_spent
  9    from (select *
 10            from v
 11                 model
 12                 partition by (r)
 13                 dimension by (0 as f)
 14                 measures (visited_states_time, cast('' as varchar2(2)) state,
 15                           0 as total_time_spent, fields)
 16                 rules (
 17                   state[for f from 0 to fields[0]-1  increment 2]
 18                     = trim(
 19                        substr(visited_states_time[0],
 20                               instr(visited_states_time[0], ',', 1, cv(f)+1)+1,
 21                               instr(visited_states_time[0], '=', 1, (cv(f)/2)+1)
 22                               - instr(visited_states_time[0], ',', 1,  cv(f)+1)-1
 23                              )),
 24                   visited_states_time[any]= visited_states_time[0],
 25                   total_time_spent[any]
 26                      = substr(visited_states_time[0],
 27                               instr(visited_states_time[0], ',', 1, (cv(f)+2))+1,
 28                               instr(visited_states_time[0], ',', 1,  (cv(f)+3))
 29                               - instr(visited_states_time[0], ',', 1,  (cv(f)+2))-1
 30                             )
 31                 ))
 32   where state in (select criteria from tableb)
 33   group by visited_states_time;

VISITED_STATES_TIME                   TOTAL_TIME_SPENT
------------------------------------- ----------------
CA=California, 10,CT=Connecticut,20                 30
AL=Alabama,2, AK=Alaska,5                            7

但说真的，重写该数据模型以将它们分开存储。