c# - Creating XML from Datatable -

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• how convert datatable xml? 1 answer

i hava datatable following entries:

**companyid   deptid    location    employeeid employeename    employeeage**        001           d001      ca           0001      jason bourne     57 001           d001      ca           0002      smith       45 001           d001      nv           0003      kurt rusell      47 002           d002      ca           0008      panda            57 002           d002      ca           0009      fox              45 002           d002      nv           0010      wolf             35 

i want xml created in c# using linq. combination of comnpanyid,deptid , location treated unique. 1 of change want new company elemnt created , employee under company element. xml should like

<companies> <company companyid="001" deptid="d001" location="ca">   <employee id="0001" employeename="jason bourne" employeeage=57/>   <employee id="0002" employeename="will smith"   employeeage=45/> </company> <company companyid="001" deptid="d001" location="nv">   <employee id="0003" employeename="kurt rusell" employeeage=47/>  </company> <company companyid="002" deptid="d002" location="ca">   <employee id="0008" employeename="panda" employeeage=57/>   <employee id="0009" employeename="fox"   employeeage=45/> </company> <company companyid="002" deptid="d002" location="nv"> <employee id="0010" employeename="wolf"  employeeage=35/> </company> </companies> 

any appreciated. datatable sorted on companyid,deptid,location

var query =     row in table.asenumerable()     group row new     {         companyid = row.field<string>("companyid"),         deptid = row.field<string>("deptid"),         location = row.field<string>("location")     }     g     select new xelement("company",             new xattribute("companyid", g.key.companyid),             new xattribute("deptid", g.key.deptid),             new xattribute("location", g.key.location),             row in g             select new xelement(                 "employee",                 new xattribute("id", row.field<string>("id")),                 new xattribute("employeename", row.field<string>("employeename")),                 new xattribute("employeeage", row.field<string>("employeeage"))));  var document = new xdocument(new xelement("companies", query));