Question

我正在使用Oracle，我希望将结果从选择计数转换为＆＃34;二进制＆＃34; 0/1值... 0 = 0 ...非零= 1.从我在线阅读的内容中，在MS SQL中，您可以将其转换为＆＃34; bit＆＃34;但甲骨文似乎并不支持这一点。

这是我的简单示例查询（真正的查询要复杂得多）。我希望MATCH_EXISTS始终为0或1.这可能吗？

select count(*) as MATCH_EXISTS 
from MY_TABLE 
where MY_COLUMN is not null;

Answer 1

这应该是最快的...最多获得一行。

SELECT COUNT(*) AS MATCH_EXISTS
FROM MY_TABLE 
WHERE MY_COLUMN IS NOT NULL
  AND rownum <= 1;

Answer 2

如果使用exists子句，对于大型表来说这应该更快，因为Oracle不需要扫描整个表。只要有一行，就可以停止检索它：

select count(*) as match_exists
from dual
where exists (select * 
              from my_table 
              where my_column is not null);

Answer 3

您可以使用CASE WHEN：

SELECT CASE WHEN MATCH_EXISTS = 0 THEN 0
            ELSE 1
       END AS MATCH_EXISTS 
FROM (SELECT COUNT(*) AS MATCH_EXISTS 
      FROM MY_TABLE 
      WHERE MY_COLUMN IS NOT NULL) AS t;

修改

SELECT CASE WHEN COUNT(*) = 0 THEN 0 ELSE 1 END as MATCH_EXISTS FROM MY_TABLE WHERE MY_COLUMN IS NOT NULL;

Answer 4

不要做count()。为了提高性能，请使用exists：

select (case when exists (select 1 as MATCH_EXISTS 
                          from MY_TABLE 
                          where MY_COLUMN is not null
                         )
              then 1 else 0
        end)
from dual;

这可以明显加快。

Oracle SQL - 将选择计数（*）转换为零或一个

4 个答案: