3

私はこのようなテーブル（データテーブル）を持っています

 Hotelid    Room#   Description visitor Name    amount
    1          2    of             5    sam     10 
    1          2    of             5    sam      5
    1          2    of             5    sam     50
    1          2    of             8    james   50
    1          2    of             8    james   50
    1          2    of             6    justin  50
    2          3    sm             4    john     5
    2          4    al             3    jose     8
    3          5    ms             2    tim     10
    3          5    ms             7    tom     20

そこから XML を作成したいと考えています。私はこれにLINQを使用していますが、完全に混乱して疲れています。私はそれを取得していません

<Hotels>
    <Hotel id="1" room="2" description="of">
    <Room="2" descr="of" visitor="5" name="sam"/>
        <fine amount="10"/>
        <fine amount="5"/>
        <fine amount="50"/>
        <Room="2" descr="of" visitor="8" name="james"/>
         <fine amount="50"/>
        <fine amount="50"/>
        <Room="2" descr="of" visitor="6" name="justin"/>
        <fine amount="50"/>
        </hotel>
        <Hotel id="2" room="3" description="sm">
        <Room="3" descr="sm" visitor="4" name="john"/>
        <fine amount="5"/>
        </hotel>
        <Hotel id="2" room="4" description="al">
        <Room="4" descr="al" visitor="3" name="jose"/>
        <fine amount="8"/>
        </hotel>

        <Hotel id="3" room="5" description="ms">
        <Room="5" descr="ms" visitor="2" name="tim"/>
        <fine amount="10"/>
        <Room="5" descr="ms" visitor="7" name="tom"/>
        <fine amount="20"/>
        </hotel>
    </Hotels>

これは私のコードがこのように見えるものです

   var query =
    from row in Hotels.AsEnumerable()
    group row by new
    {
        Hotelid = row.Field<string>("Hotelid"),
        room = row.Field<string>("room"),
        descr = row.Field<string>("descr"),
    }
        into g
        select new XElement("Hotel",
                new XAttribute("Hotelid", g.Key.Hotelid),
                new XAttribute("room", g.Key.room),
                new XAttribute("desc", g.Key.desc),
                from row in g
                    select new XElement(
                    "Room",
                    new XAttribute("room", row.Field<string>("room#")),
                    new XAttribute("desc", row.Field<string>("desc")),
                    new XAttribute("visitor", row.Field<string>("visitor")),
                    new XAttribute("name", row.Field<string>("name")),
                from row in g 
                    select new XElement(
                    "fine",
                    new XAttribute("amount", row.Field<string>("amount"))));

                    var document = new XDocument(new XElement("Hotels", query));

しかし、同じ値を持つ「部屋」の複数のノードを取得しています。何か助けて???? :(

4

1 に答える 1

0

ハハハ、私はこの悪い男の子を思いついたことで少し正気を失ったと思います. それがあなたを助けるかどうか私に知らせてください。

void Main()
{
    var query =
    from row in Hotels.AsEnumerable()
    group row by new
    {
        row.Hotelid, row.Room, row.Description
    }
        into g
        select new XElement("Hotel",
                new XAttribute("Hotelid", g.Key.Hotelid),
                new XAttribute("room", g.Key.Room),
                new XAttribute("desc", g.Key.Description),
                from row in g
                    group row by new {row.Room, row.Description, row.Visitor, row.Name} into r
                    select new XElement(
                    "Room",
                    new XAttribute("room", r.Key.Room),
                    new XAttribute("desc", r.Key.Description),
                    new XAttribute("visitor", r.Key.Visitor),
                    new XAttribute("name", r.Key.Name),
                    new XAttribute("fineSum", r.Sum (x => x.Amount)),
                    from row in r                   
                    group row by new {row.Amount} into a
                    select new XElement("fine", a.Key.Amount )
              ));

    var document = new XDocument(new XElement("Hotels", query));                    

}

申し訳ありませんが、DataTable を使用していることを忘れていました...

void Main()
{

    var query =
    from row in Hotels.AsEnumerable()
    group row by new
    {
        Hotelid = row.Field<string>("Hotelid"),
        room = row.Field<string>("room"),
        descr = row.Field<string>("Description")        
    }
        into g
        select new XElement("Hotel",
                new XAttribute("Hotelid", g.Key.Hotelid),
                new XAttribute("room", g.Key.room),
                new XAttribute("desc", g.Key.descr),
                from row in g
                    group row by new {
                        room = row.Field<string>("Room"),
                        descr = row.Field<string>("Description"), 
                        visitor = row.Field<string>("Visitor"), 
                        name = row.Field<string>("Name")
                    } into r
                    select new XElement(
                    "Room",
                    new XAttribute("room", r.Key.room),
                    new XAttribute("desc", r.Key.descr),
                    new XAttribute("visitor", r.Key.visitor),
                    new XAttribute("name", r.Key.name),
                    new XAttribute("fineSum", r.Sum (x => x.Field<int>("Amount"))),
                    from row in r                   
                    group row by new {fine = row.Field<int>("Amount")} into a
                    select new XElement("fine", a.Key.fine)
              ));

    var document = new XDocument(new XElement("Hotels", query));                    

}

// Define other methods and classes here
于 2013-09-26T04:00:04.960 に答える