从asp中错误的mysql格式日期生成json

Question

我被投入了一个新项目，这显然不仅仅是过时的。该应用程序以一种非常奇怪的模式在数据库中保存了开放时间，这让我疯狂了一个多星期。

请看这张图片：

如您所见，营业时间以如下模式保存：

dayFrom  | dayTo  | timeFrom | timeTo 
=======================================
monday   | friday | 07:00    | 17:00
saturday |        | 08:00    | 12:00

只是为了防止任何误解：

周一至周五 07:00 至 17:00 开放

从 08:00 到 12:00 开放 SA

周日休息

现在，这似乎已经有点不对了，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

所以，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“周一至周五”，它已经可以使用，但不考虑第二个（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"
        }
    ]
}

但是“星期六”没有被识别。

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：将数字映射到日期，迭代（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？任何帮助将不胜感激。

我不能改变桌子和它的设计，我必须坚持那个gargabe

Answer 1

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

退货

注意：关闭是可选的。删除最终查询中的“LEFT”联接

现在，如果您想在 SQL Server 中创建 JSON 字符串，并且您不在 2016 年，我们可以调整最终查询并添加 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"
}]

---

有兴趣的 UDF

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
*/