Question

我有一个名为TEST的表。某些字段具有NULL值

Country City   Street   House
US      NULL   Avenue   14
UK      London NULL     15
NULL    NULL   NULL     NULL
NULL    NULL   Central  16

我需要显示只有一个NULL的项目。

所以输出应该是：

Country City   Street House
US      NULL   Avenue 14
UK      London NULL   15

我有愚蠢的解决方案，只需比较所有可能的列对并检查NULL。

就像这样：

 SELECT * FROM TEST
 WHERE NOT (
 (country='NULL' AND city='NULL')
 OR (country='NULL' AND street='NULL')
 OR (coutry='NULL' AND house='NULL')
 OR (city='NULL' AND street='NULL')
 OR (city='NULL' AND house='NULL')
 OR (street='NULL' AND house='NULL')
 )

它有效，但要明白这是非常低效的。

您能推荐更优雅的解决方案吗？

Answer 1

select * from TEST
WHERE  
  CASE WHEN country = 'NULL' THEN 1 ELSE 0 END
+ CASE WHEN city= 'NULL' THEN 1 ELSE 0 END
+ CASE WHEN Street = 'NULL' THEN 1 ELSE 0 END
+ CASE WHEN House = 'NULL' THEN 1 ELSE 0 END = 1

Answer 2

select *
from test1 
where
length(nvl(country,'@')||nvl(City,'@')||nvl(street,'@')||nvl(house,'@'))-
length(replace(nvl(country,'@')||nvl(City,'@')||nvl(street,'@')||nvl(house,'@'),'@',null))=1

连续查找重复项

2 个答案: