oracle:计算一行中的空字段数

我有一个有5个可选字段的表。我想知道有多少行的5个字段全部为空，有多少行的1个字段为非空，等等

我试过几件事，比如：

select 
 count(*),
 ( (if field1 is null then 1 else 0) +
   (if field2 is null then 1 else 0) +
 etc.

但这当然行不通

理想情况下，我要寻找的输出是

Nulls   Cnt
0        200
1        345
...
5        40

有优雅的解决方案吗？

关键字不是if，而是case，您必须使用端到端case语句

以下是一个适合您的查询：

WITH subQuery AS
(
  SELECT My_Table.*, (CASE WHEN My_Table.field1 IS NULL THEN 1 ELSE 0 END +
                      CASE WHEN My_Table.field2 IS NULL THEN 1 ELSE 0 END +
                      CASE WHEN My_Table.field3 IS NULL THEN 1 ELSE 0 END +
                      CASE WHEN My_Table.field4 IS NULL THEN 1 ELSE 0 END +
                      CASE WHEN My_Table.field5 IS NULL THEN 1 ELSE 0 END ) NumberOfNullFields
    FROM My_Table
)
SELECT NumberOfNullFields, COUNT(*)
  FROM subQuery
 GROUP BY NumberOfNullFields;

---

虽然在计算时，这个案例没有什么问题，但我只是想看看是否还有其他方法

WITH SAMPLEDATA AS(--just generate some data with nulls for 5 cols
    select  level ,
            (case when mod(level,2) = 0 then 1 else null end) colA,
            (case when mod(level,3) = 0 then 1 else null end) colB,
            (case when mod(level,5) = 0 then 1 else null end) colC,
            (case when mod(level,7) = 0 then 1 else null end) colD,
            (case when mod(level,11) = 0 then 1 else null end) colE

      from dual
      connect by level < 1000
    ), --utilize the count(Aggregate)'s avoidance of nulls to our summation advantage
    nullCols as(
    SELECT COUNT(COLA) aNotNull
           ,cOUNT(*)-COUNT(COLA) aNull
           ,count(colB) bNotNull
           ,cOUNT(*)-count(colB) bNull
           ,count(colc) cNotNull
           ,cOUNT(*)-count(colc) cNull
           ,count(cold) dNotNull
           ,cOUNT(*)-count(cold) dNull
           ,count(cole) eNotNull
           ,cOUNT(*)-count(cole) eNull
           , cOUNT(*) TotalCountOfRows
     from SAMPLEDATA  )
    SELECT (select count(*) from sampledata where cola is null and colb is null and colc is null and cold is null and cole is null) allIsNull     
           ,nullCols.*
    FROM nullCols;

ALLISNULL              ANOTNULL               ANULL                  BNOTNULL               BNULL                  CNOTNULL               CNULL                  DNOTNULL               DNULL                  ENOTNULL               ENULL                  TOTALCOUNTOFROWS       
---------------------- ---------------------- ---------------------- ---------------------- ---------------------- ---------------------- ---------------------- ---------------------- ---------------------- ---------------------- ---------------------- ---------------------- 
207                    499                    500                    333                    666                    199                    800                    142                    857                    90                     909                    999                    

这利用了

countexpression中的If表达式 计算结果为null，不计算该值： 如从

如上所述，这种方法不能很好地“雄辩地”求和所有空列。只是想看看如果没有案例逻辑，这是否可行