我怎么能在linq做:
IEnumerable<DataRow> query =
from rec in dt.AsEnumerable()
where rec.Field<decimal>("column2") == 1 && foo(rec.Field<decimal>("column1"))
select new {
column1 = rec.Field<decimal>("column1"),
column2 = rec.Field<decimal>("column2"),
column3 = rec.Field<decimal>("column3")} ;
这不起作用。我试图选择一些列作为新数据表,然后将其与其他数据表连接起来。
答案 0 :(得分:10)
您可以使用http://code.msdn.microsoft.com/LinqEntityDataReader 然后选择你喜欢的值:
DataTable query =
(from rec in dt.AsEnumerable()
where rec.Field<decimal>("column2") == 1 && foo(rec.Field<decimal>("column1"))
select new {
column1 = rec.Field<decimal>("column1"),
column2 = rec.Field<decimal>("column2"),
column3 = rec.Field<decimal>("column3")}).ToDataTable();
一个更简单的解决方案,不需要外部库,就是创建一个DataTable(如果你还没有),并按如下方式填充它:
// declaring a data table.. Replace it with whatever code you want
var table = new DataTable();
table.Columns.Add("column1", typeof(Decimal));
table.Columns.Add("column2", typeof(Decimal));
table.Columns.Add("column3", typeof(Decimal));
// Populate the table
(from rec in dt.AsEnumerable()
where rec.Field<decimal>("column2") == 1 && foo(rec.Field<decimal>("column1"))
select new {
column1 = rec.Field<decimal>("column1"),
column2 = rec.Field<decimal>("column2"),
column3 = rec.Field<decimal>("column3")})
.Aggregate(table, (dt, r) => { dt.Rows.Add(r.column1, r.column2, r.column3); return dt; });
// at this point your table variable is populated