Question

我有一个列名检查，数据类型为bit（true或False），我想要做的是计算列中有多少false和true。

SELECT 
COUNT(FeatureState) AS tot_true, 
COUNT(*)-COUNT(FeatureState) AS tot_false
FROM productDetail
 WHERE FeatureState= 1

这有效

Answer 1

怎么样;

SELECT 
  COUNT(CASE fld WHEN 0 THEN 1 ELSE NULL END) AS ZEROS,
  COUNT(CASE fld WHEN 1 THEN 1 ELSE NULL END) AS ONES
FROM
  T

--alternative

SELECT 
  SUM(CASE fld WHEN 0 THEN 1 ELSE 0 END) AS ZEROS,
  SUM(CASE fld WHEN 1 THEN 1 ELSE 0 END) AS ONES
FROM
  T

Answer 2

试试这个，应该有效：

SELECT 
    COUNT(field) AS tot_true, 
    COUNT(*)-COUNT(field) AS tot_false
FROM table
WHERE field=1

Answer 3

我使用SIGN将位更改为int

SELECT 
    SUM(SIGN(field)) AS tot_true, 
    SUM(1-SIGN(field)) AS tot_false
FROM table

或者使用COUNT忽略NULL的事实

SELECT 
    COUNT(NULLIF(field, 0)) AS tot_true, 
    COUNT(NULLIF(field, 1)) AS tot_false
FROM table

如果您想要其他值MAX（SomeOtherField），其中最高值来自在其他解决方案中过滤掉的“false”行，则此方法有效

Answer 4

SELECT 
SUM(CASE WHEN FeatureState = 1     THEN 1 ELSE 0 END) AS TrueCount,
SUM(CASE WHEN FeatureState = 0     THEN 1 ELSE 0 END) AS FalseCount,
SUM(CASE WHEN FeatureState IS NULL THEN 1 ELSE 0 END) AS NullCount,
COUNT(1)                                              AS TotalCount
FROM ProductDetail

使用数据类型位查询计数列

4 个答案: