从两个不同的条件在一个查询中获取不同的值

Question

我需要有像结果如下

section_one | section_two 
56788    678 

，这是我写有上述结果的查询分别

Select count(*) as section_one from tableNew 
WHERE columnone LIKE %hjk% 

而且

Select count(*) as section_two from tableNew 
WHERE columnone NOT LIKE %hjk%; 

如何在一个查询中组合这两个并获得如上所示的结果？

我想下面的东西，但似乎并没有正常工作

select *, 
CASE 
WHEN columnone LIKE %hjk% THEN count(1) as section_one 
ELSE count(1) as section_two 
END 
from tableNew 
Group by tableid; 

是否有任何其他的方式，以及中都有一个代替的条件？ 任何帮助，这将是伟大的！

PS：也就是有任何其他方式以及检查boolean条件等代替columnone LIKE %hjk%目标columnone IS NOT NULL

Answer 1

试试这个使用SUM

select 
SUM(columnone LIKE %hjk%) as section_one, 
SUM(columnone NOT LIKE %hjk%) as section_two 
from tableNew 
Group by tableid; 

或者你可以使用CASE

select 
SUM(CASE WHEN columnone LIKE %hjk% THEN 1 ELSE 0 END) as section_one, 
SUM(CASE WHEN columnone NOT LIKE %hjk% THEN 1 ELSE 0 END) as section_two 
from tableNew 
Group by tableid;