我被投入了一个新项目，它显然不仅仅是过时的。该应用程序在数据库中以一种非常奇怪的模式保存开放时间，这让我疯狂了一个多星期。
请看这张图片:
如您所见，营业时间的保存格式如下:
dayFrom | dayTo | timeFrom | timeTo
=======================================
monday | friday | 07:00 | 17:00
saturday | | 08:00 | 12:00
为了防止误会:
Open MO - FR from 07:00 to 17:00
Open SA from 08:00 to 12:00
Closed on Sunday
现在，这似乎已经有点不对劲了，但坚持下去，表格可能看起来像这样:
dayFrom | dayTo | timeFrom | timeTo
=======================================
monday | tuesday | 07:00 | 14:00
wednesday | | 08:00 | 12:00
thursday | friday | 07:30 | 13:00
saturday | | 08:00 | 12:00
那么，现在我的问题是:我需要创建一个循环(或类似的东西)来创建一个有效的 json 字符串，其中包含所有这些开放时间。
现在，我有这个:
jsonAppendix = "{""openingHours"":["
for i = 1 To cint(hoechsterTag)
jsonAppendix = jsonAppendix & "{""dayOfWeek"":" & i & ", ""from1"":""" & rs("ZeitVon") & """, ""to1"":""" & rs("ZeitBis") & """},"
next
'Remove last comma
jsonAppendix = LEFT(jsonAppendix, (LEN(jsonAppendix)-1))
jsonAppendix = jsonAppendix & "]}"
如果我只有一个“星期一至星期五”，它已经可以工作了，但不会考虑第二个(或下一个条目)。
输出看起来像这样，这显然是正确的:
{
"openingHours":[
{
"dayOfWeek":1,
"from1":"07:00",
"to1":"17:00"
},
{
"dayOfWeek":2,
"from1":"07:00",
"to1":"17:00"
},
{
"dayOfWeek":3,
"from1":"07:00",
"to1":"17:00"
},
{
"dayOfWeek":4,
"from1":"07:00",
"to1":"17:00"
},
{
"dayOfWeek":5,
"from1":"07:00",
"to1":"17:00"
}
]
}
但“星期六”未被识别。
我的函数如下所示:
SQL = "SELECT * FROM StandortOpen WHERE S_ID = " & iStandortId & " AND OpenArt = '" & sArt & "' ORDER BY Sort,OpenArt DESC"
call openRS(SQL)
'day-mapping
tageV(0) = replace(rs("TagVon"),"Mo", 1)
tageV(1) = replace(rs("TagVon"),"Di", 2)
tageV(2) = replace(rs("TagVon"),"Mi", 3)
tageV(3) = replace(rs("TagVon"),"Do", 4)
tageV(4) = replace(rs("TagVon"),"Fr", 5)
tageV(5) = replace(rs("TagVon"),"Sa", 6)
tageV(6) = 7
tageB(0) = replace(rs("TagBis"),"Mo", 1)
tageB(1) = replace(rs("TagBis"),"Di", 2)
tageB(2) = replace(rs("TagBis"),"Mi", 3)
tageB(3) = replace(rs("TagBis"),"Do", 4)
tageB(4) = replace(rs("TagBis"),"Fr", 5)
tageB(5) = replace(rs("TagBis"),"Sa", 6)
'for example: mo - fr
for each item in tageV
'save smallest weekday
if(isNumeric(item) AND item > "") then
if(cint(item) <= cint(niedrigsterTag)) then
niedrigsterTag = cint(item)
end if
end if
next
for each item in tageB
'save highest weekday
if(isNumeric(item) AND item > "") then
if(cint(item) >= cint(hoechsterTag)) then
hoechsterTag = cint(item)
end if
end if
next
还有 openRS() 函数:
sub openRS(str_sql)
'Response.write "SQL: " & str_sql & "<br>"
set rs = CreateObject("ADODB.Recordset")
rs.open str_sql,conn,1,3
end sub
基本上:将数字映射到天数，遍历(或比较它们以获得时间跨度)。
我也在使用 RecordSet。也许我需要使用数组或类似的东西？非常感谢任何帮助。
I can't alter the table nor the design of that, I have to stick with that gargabe
最佳答案
如果要在 SQL Server 中创建数据集，请考虑以下事项
示例
Declare @YourTable table (dayFrom varchar(25),dayTo varchar(25),timeFrom varchar(25),timeTo varchar(25))
Insert Into @YourTable values
('monday' ,'tuesday','07:00','14:00'),
('wednesday','' ,'08:00','12:00'),
('thursday' ,'friday' ,'07:30','13:00'),
('saturday' ,'' ,'08:00','12:00')
;with cteD as (Select * From (Values(1,'Monday'),(2,'Tuesday'),(3,'Wednesday'),(4,'Thursday'),(5,'Friday'),(6,'Saturday'),(7,'Sunday')) DDD(DD,DDD) ),
cteR as (
Select A.*
,R1 = B.DD
,R2 = IsNull(C.DD,B.DD)
From @YourTable A
Left Join cteD B on dayFrom = B.DDD
Left Join cteD C on dayTo = C.DDD
Where 1=1 -- Your WHERE STATEMENT HERE
)
Select daySeq = A.DD
,dayOfWeek = A.DDD
,from1 = IsNull(B.TimeFrom,'Closed')
,from2 = IsNull(B.TimeTo,'Closed')
From cteD A
Left Join cteR B on A.DD between B.R1 and B.R2
Order By 1
返回
注意: Closed 是可选的。去掉最后查询中的“LEFT”Join
现在，如果您想在 SQL Server 中创建 JSON 字符串，并且您不是在 2016 年，我们可以调整最终查询并添加一个 UDF。
Select JSON=[dbo].[udf-Str-JSON](0,0,(
Select daySeq = A.DD
,dayOfWeek = A.DDD
,from1 = IsNull(B.TimeFrom,'Closed')
,from2 = IsNull(B.TimeTo,'Closed')
From cteD A
Left Join cteR B on A.DD between B.R1 and B.R2
Order By 1
For XML RAW
))
返回的 JSON 字符串
[{
"daySeq": "1",
"dayOfWeek": "Monday",
"from1": "07:00",
"from2": "14:00"
}, {
"daySeq": "2",
"dayOfWeek": "Tuesday",
"from1": "07:00",
"from2": "14:00"
}, {
"daySeq": "3",
"dayOfWeek": "Wednesday",
"from1": "08:00",
"from2": "12:00"
}, {
"daySeq": "4",
"dayOfWeek": "Thursday",
"from1": "07:30",
"from2": "13:00"
}, {
"daySeq": "5",
"dayOfWeek": "Friday",
"from1": "07:30",
"from2": "13:00"
}, {
"daySeq": "6",
"dayOfWeek": "Saturday",
"from1": "08:00",
"from2": "12:00"
}, {
"daySeq": "7",
"dayOfWeek": "Sunday",
"from1": "Closed",
"from2": "Closed"
}]
UDF(如果有兴趣)
CREATE FUNCTION [dbo].[udf-Str-JSON] (@IncludeHead int,@ToLowerCase int,@XML xml)
Returns varchar(max)
AS
Begin
Declare @Head varchar(max) = '',@JSON varchar(max) = ''
; with cteEAV as (Select RowNr =Row_Number() over (Order By (Select NULL))
,Entity = xRow.value('@*[1]','varchar(100)')
,Attribute = xAtt.value('local-name(.)','varchar(100)')
,Value = xAtt.value('.','varchar(max)')
From @XML.nodes('/row') As R(xRow)
Cross Apply R.xRow.nodes('./@*') As A(xAtt) )
,cteSum as (Select Records=count(Distinct Entity)
,Head = IIF(@IncludeHead=0,IIF(count(Distinct Entity)<=1,'[getResults]','[[getResults]]'),Concat('{"status":{"successful":"true","timestamp":"',Format(GetUTCDate(),'yyyy-MM-dd hh:mm:ss '),'GMT','","rows":"',count(Distinct Entity),'"},"retults":[[getResults]]}') )
From cteEAV)
,cteBld as (Select *
,NewRow=IIF(Lag(Entity,1) over (Partition By Entity Order By (Select NULL))=Entity,'',',{')
,EndRow=IIF(Lead(Entity,1) over (Partition By Entity Order By (Select NULL))=Entity,',','}')
,JSON=Concat('"',IIF(@ToLowerCase=1,Lower(Attribute),Attribute),'":','"',Value,'"')
From cteEAV )
Select @JSON = @JSON+NewRow+JSON+EndRow,@Head = Head From cteBld, cteSum
Return Replace(@Head,'[getResults]',Stuff(@JSON,1,1,''))
End
-- Parameter 1: @IncludeHead 1/0
-- Parameter 2: @ToLowerCase 1/0 (converts field name to lowercase
-- Parameter 3: (Select * From ... for XML RAW)
-- Syntax : Select [dbo].[udf-Str-JSON](0,1,(Select Top 2 RN=Row_Number() over (Order By (Select NULL)),* from [Chinrus-Shared].[dbo].[ZipCodes] Where StateCode in ('RI') for XML RAW))
/*
Declare @User table (ID int,Active bit,First_Name varchar(50),Last_Name varchar(50),EMail varchar(50))
Insert into @User values
(1,1,'John','Smith','john.smith@email.com'),(2,0,'Jane','Doe' ,'jane.doe@email.com')
Declare @XML xml = (Select * from @User for XML RAW)
Select A.ID
,B.JSON
From @User A
Cross Apply (Select JSON=[dbo].[udf-Str-JSON](0,0,(Select A.* For XML Raw)) ) B
*/
关于sql - 从 asp 中错误的 mysql 格式日期生成 json，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43140800/