[英]Create XML using LINQ
我有一個表(數據表),看起來像這樣
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
我想用它創建一個XML。 我正在使用LINQ,我感到很困惑和累。 我沒有得到它
<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>
這就是我的代碼看起來像這樣
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));
但是我得到了具有相同值的“房間”的多個節點。 任何幫助???? :(
哈哈哈,我想我已經失去了這個壞男孩的一點理智。 如果它對您有幫助,請告訴我。
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));
}
對不起,忘了你正在使用DataTable ......
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
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