mysql - 在连接表时，我们可以在连接条件中使用 Case When 创建列吗

Question

如果我们在连接条件中创建列时使用案例...代码将运行。但这是正确的吗？如果是的话怎么执行？

select *,
case when position('/' in pax_name)>0 
         then SUBSTR(pax_name, 1, position('/' in pax_name)- 1) 
          end as **lastname**, 
    CASE WHEN position('/' in pax_name)>0 
         THEN SUBSTR(pax_name, position('/' in pax_name) + 1, LENGTH(pax_name))  
         END as **firstname**
from o
inner join m
on o.record=m.record
and o.pax_first_name = **firstname**
and o.pax_last_name = **lastname**

Answer 1

select 中定义的列别名在 select 同一级别的大多数查询中不可用。特别是，它们不可用于 where 或 from 子句。

您可以使用having来完成此操作:

select *,
       (case when position('/' in pax_name) > 0 
             then SUBSTR(pax_name, 1, position('/' in pax_name)- 1) 
        end) as lastname, 
       (case when position('/' in pax_name)  >0 
             then substr(pax_name, position('/' in pax_name) + 1, length(pax_name))  
        end) as firstname
from o inner join
     m
     on o.record = m.record
having o.pax_first_name = firstname and
       o.pax_last_name = lastname;

您可以简化逻辑。我想你只是想要:

select *,
       (case when pax_name like '%'
             then substring_index(pax_name, '/', 1)
        end) as firstname,
       (case when pax_name like '%'
             then substring_index(pax_name, '/', -1)
        end) as lastname
from o inner join
     m
     on o.record = m.record
having o.pax_first_name = firstname and
       o.pax_last_name = lastname;

我还建议放弃拥有，所以:

select *,
       (case when pax_name like '%'
             then substring_index(pax_name, '/', 1)
        end) as firstname,
       (case when pax_name like '%'
             then substring_index(pax_name, '/', -1)
        end) as lastname
from o inner join
     m
     on o.record = m.record
        m.pax_name = concat_ws('/', o.pax_first_name, o.pax_last_name);