将不匹配的记录作为 NULL 获取结果

Question

我正在尝试在 wonderware 中查询 table 以获取某些数据。我知道查询的 'a' 部分中有一些数据，TagName 等于 'Weightdata2.uiID'。但是 'b' 部分没有匹配的数据，因此查询 returns 空数据集。但是我想获取 'a' 部分和 'b' 的数据，如果那里没有匹配的数据，则 uiWater 列中的值为 NULL 或零。

这是我的查询：

DECLARE @StartDate DateTime
DECLARE @EndDate DateTime
set @StartDate = '2018-09-18 08:00:00.000'
set @EndDate = '2018-09-18 09:00:00.000'

SELECT a.Value as uiID, b.value as uiWater, cast(a.datetime as datetime2(0)) 
as dtCreated, 2 as Weightdata
FROM [INSQL].[Runtime].[dbo].[History] a
JOIN [INSQL].[Runtime].[dbo].[History] b ON a.datetime=b.datetime
WHERE a.TagName IN ('Weightdata2.uiID') and a.datetime>=@StartDate and 
a.datetime<=@EndDate and a.Value!=0
and b.TagName IN ('Weightdata2.uiWater') and b.datetime>=@StartDate and 
b.datetime<=@EndDate

Answer 1

您只需将 JOIN 替换为 LEFT JOIN。如果值为 null

，则可以将 isnull 用于 return 0

DECLARE @StartDate DateTime
DECLARE @EndDate DateTime
set @StartDate = '2018-09-18 08:00:00.000'
set @EndDate = '2018-09-18 09:00:00.000'

SELECT a.Value as uiID, ISNULL(b.value, 0) as uiWater, cast(a.datetime as datetime2(0)) 
as dtCreated, 2 as Weightdata
FROM [INSQL].[Runtime].[dbo].[History] a
LEFT JOIN [INSQL].[Runtime].[dbo].[History] b ON a.datetime=b.datetime
WHERE a.TagName IN ('Weightdata2.uiID') and a.datetime>=@StartDate and 
a.datetime<=@EndDate and a.Value!=0
and ((b.TagName IN ('Weightdata2.uiWater') and b.datetime>=@StartDate and 
b.datetime<=@EndDate) OR b.datetime is null)

Answer 2

这更像是 PIVOT 的工作：

;with cteData as (
    SELECT t.datetime, t.TagName, t.value
    FROM [INSQL].[Runtime].[dbo].[History] t
    WHERE t.datetime>=@StartDate and t.datetime<=@EndDate
    AND t.TagName IN ('Weightdata2.uiID', 'Weightdata2.uiWater')
)
SELECT 
  d.dtCreated,
  NULLIF(p.[Weightdata2.uiID], 0) as uiID,
  p.[Weightdata2.uiWater] as uiWater
FROM (
  SELECT 
    cast(d.datetime as datetime2(0)) as dtCreated,
    d.TagName, 
    d.value
  FROM cteData d
) d
PIVOT (
  MAX(d.value) for d.TagName in ([Weightdata2.uiID], [Weightdata2.uiWater])
) p

在所有情况下都将 return 数据：当有 uiID 行但没有 uiWater 时，当两者都存在时，当没有 uiID 但 uiWater 存在。

并且很容易针对更长的标签列表进行调整。

Answer 3

可能只是这样：

with Perimeter as (
    SELECT t.datetime, t.TagName, t.value
    FROM [INSQL].[Runtime].[dbo].[History] t
    WHERE t.datetime between @StartDate and @EndDate
    AND t.TagName IN ('Weightdata2.uiID', 'Weightdata2.uiWater')
)
select f1.Value as uiID, ISNULL(f2.value, 0) as uiWater, 
cast(f1.datetime as datetime2(0)) as dtCreated, 2 as Weightdata
from Perimeter f1 
left outer join Perimeter f2 on f1.datetime=f2.datetime and f2.TagName='Weightdata2.uiWater'
where f1.TagName='Weightdata2.uiID'

将不匹配的记录作为 NULL 获取结果

Getting result with unmatched records as NULL

tsql

sql-server

wonderware