Question

下面的查询是计算缺陷＆＃39;在我的数据中，意味着任何时候EncounterID匹配但其他列不匹配。如何修改查询以不包含任何“缺陷”。由NULL值引起？

SELECT FacilityCode,
       AdmitDate,
       SUM(CASE
             WHEN NumFirstName <> 1 THEN 1
             ELSE 0
           END) AS DifferentFirstNames,
       SUM(CASE
             WHEN NumLastName <> 1 THEN 1
             ELSE 0
           END) AS DifferentLastNames,
       SUM(CASE
             WHEN NumPhone <> 1 THEN 1
             ELSE 0
           END) AS DifferentPhone,
       'Defect' Label
FROM   (SELECT EncounterId,
               FacilityCode,
               AdmitDate,
               COUNT(*)                           AS Num,
               COUNT(DISTINCT LEFT(FirstName, 4)) AS NumFirstName,
               COUNT(DISTINCT LastName)           AS NumLastName,
               COUNT(DISTINCT Phone)              AS NumPhone,
        FROM   [BINextGen].[dbo].[tbFCCDefectReport]
        GROUP  BY EncounterId,
                  facilitycode,
                  AdmitDate) e
GROUP  BY FacilityCode,
          AdmitDate;

以下是目前的情况：

示例数据：

FacilityCode     AdmitDate     FirstName              LastName
BOHG             7-1-15        Justin                 Kelley
BOHG             7-1-15        Justin                 NULL
AARD             7-1-15        James                  Smith
AARD             7-1-15        James                  Smith

示例输出：

FacilityCode     AdmitDate     DifferentFirstNames    DifferentLastNames
BOHG             7-1-15        0                      1
AARD             7-1-15        0                      0

我想忽略第一个姓氏更改，因为它是NULL并且在输出中全部为0（我只关心是否有更改，例如Kelley更改为Kelly）：

期望的输出：

FacilityCode     AdmitDate     DifferentFirstNames    DifferentLastNames
BOHG             7-1-15        0                      0
AARD             7-1-15        0                      0

Answer 1

此SQL fiddle为您提供了所要求的结果......但我不确定它会一直有效。我不确定为什么你的方法不起作用我所做的就是消除代码中的错误并删除不需要的字段...计数按预期工作。

SELECT FacilityCode,
       AdmitDate,
       SUM(CASE
             WHEN NumFirstName <> 1 THEN 1
             ELSE 0
           END) AS DifferentFirstNames,
       SUM(CASE
             WHEN NumLastName <> 1 THEN 1
             ELSE 0
           END) AS DifferentLastNames
FROM   (SELECT 
               FacilityCode,
               AdmitDate,
               COUNT(*)                           AS Num,
               COUNT(DISTINCT LEFT(FirstName, 4)) AS NumFirstName,
               COUNT(DISTINCT LastName)           AS NumLastName
        FROM  [tbFCCDefectReport]
        GROUP  BY facilitycode,
                  AdmitDate) e
GROUP  BY FacilityCode,
          AdmitDate;

可能还有其他两种方法可以提供更好的缩放功能。

一种方法是生成两个不同的数据集，一个＆＃34;之前＆＃34;一个＆＃34;之后＆＃34;但我不知道如何在没有桌子钥匙的情况下关联数据。

类似

SELECT A.Key, 
sum(case when (coalesce(Before.Field1,'') = coalesce(After.Field1,'') OR After.Field1 is null) then 0 else 1 end) as countField1Difference,
sum(case when coalesce(Before.Field2,'') = coalesce(After.Field2,'') OR After.Field2 is null) then 0 else 1 end) as countField2Difference
FROM TableName Before
LEFT JOIN TableName After
 on A.Key = B.Key
and B.Bal = 'Something which denotes it as the After set'
and A.Val = 'something which denotes it as the Before set'
GROUP BY A.Key

coalesce处理空值，允许字符串compairison报告

或者你可以使用

Select * from tableName where B.Val = 'something denoting it's the before value'
EXCEPT
Select * from tableName where B.Val = 'something denoting it's the after value'

然而，这显示出任何差异，因为整个差异不是逐列

修改Case Query以不计算NULL值

1 个答案: