c# - linq to XML: unique attribute value count per group -

i have got xml nodes below.

... <parentnode>     <node id="2343" name="some name" mode="some mode">          //some child nodes here     </node>     <node id="2344" name="some other name" mode="some mode">          //some child nodes here     </node>     ... </parentnode> <parentnode>     <node id="2343" name="some name" mode="some other mode">          //some child nodes here     </node>     <node id="2344" name="some other name" mode="some mode">          //some child nodes here     </node> </parentnode> .... 

what need

id      name             distinct-mode-count -------------------------------------------- 2343    name        2 2344    other name  1 

i have tried below this.

xelement myxml = xelement.load(filepath);  ienumberable<xelement> parentnodes = myxml.descendants("parentnode");  var nodeattributes = parentnodes.select(le => le.descendants("node")     .groupby(x => new {                            id = x.attribute("id").value,                            name = x.attribute("name").value                        }).select(g => new {                            id = g.key.id,                            name = g.key.name,                            distinct_mode_count = // stuck                       })); 

i not sure how distinct_mode_count in above query.

edit

i need distinct attribute value count attribute "mode", regardless of parentnode in.

assuming want count of distinct "mode" attribute values within nodes same id/name, need project each element in group mode, take distinct sequence of modes, count it:

you need take count of group, , use selectmany "flatten" parent nodes. (or use myxml.descendants("node") start with.)

short complete example gives desired results:

using system; using system.collections.generic; using system.linq; using system.xml.linq;  class test {     static void main()     {         xelement myxml = xelement.load("test.xml");          ienumerable<xelement> parentnodes = myxml.descendants("parentnode");         var nodeattributes = parentnodes             .selectmany(le => le.descendants("node"))             .groupby(x => new {                  id = x.attribute("id").value,                  name = x.attribute("name").value              })             .select(g => new {                 g.key.id,                 g.key.name,                 distinctmodecount = g.select(x => x.attribute("mode").value)                                      .distinct()                                      .count()             });         foreach (var item in nodeattributes)         {             console.writeline(item);         }     } } 

alternatively:

xelement myxml = xelement.load("test.xml");  var nodeattributes = myxml     .descendants("node")     .groupby(...)     // remaining code before