I posted another post earlier on on how to transform an XmlDocument with XSLT but maybe it is possible to do with LINQ instead.
I have an XmlDocument that looks something like:
<DocumentElement>
<Customer>
<CustomerId>2315</CustomerId>
<VersionNumber>1</VersionNumber>
<GUID>2E05DE20-02A0-425D-944D-65E5E744FF8A</GUID>
</Customer>
<Customer>
<CustomerId>2316</CustomerId>
<VersionNumber>2</VersionNumber>
<GUID>2E05DE20-02A0-425D-944D-65E5E744FF8A</GUID>
</Customer>
<Customer>
<CustomerId>2317</CustomerId>
<VersionNumber>1</VersionNumber>
<GUID>9995DE20-02A0-425D-944D-65E5E744FF8A</GUID>
</Customer>
</DocumentElement>
Can I, with LINQ, extract one customer element for each unique GUID and get the customer with the highest version number?
Ie the new/transformed document will look like:
<DocumentElement>
<Customer>
<CustomerId>2316</CustomerId>
<VersionNumber>2</VersionNumber>
<GUID>2E05DE20-02A0-425D-944D-65E5E744FF8A</GUID>
</Customer>
<Customer>
<CustomerId>2317</CustomerId>
<VersionNumber>1</VersionNumber>
<GUID>9995DE20-02A0-425D-944D-65E5E744FF8A</GUID>
</Customer>
</DocumentElement>
Thanks in advance.
Update:
So I should do (if I understand you correctly?):
XDocument xdoc = new XDocument();
using (StringWriter sw = new StringWriter())
{
// Converts the datatable to XML
dt.WriteXml(sw);
xdoc = XDocument.Parse(sw.ToString());
}
var query = xdoc.Root
.Elements("Customer")
.GroupBy(x => x.Element("GUID").Value)
.Select(g => g.OrderByNumberDescending(x =>(int)x.Element("VersionNumber"))
.First());
I still get the error on GroupBy though, or am I doing something else wrong here?
Absolutely:
var query = parent
.Elements("Customer")
.GroupBy(x => x.Element("GUID").Value)
.Select(g => g.OrderByNumberDescending(x => (int) x.Element("VersionNumber"))
.First());
Or using MoreLINQ :
var query = parent
.Elements("Customer")
.GroupBy(x => x.Element("GUID").Value)
.Select(g => g.MaxBy(x => (int) x.Element("VersionNumber")));
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.