var result = from row1 in raptorDS.Tables[RaptorTable.DurationByCurrency].AsEnumerable()
join row2 in fundDT.AsEnumerable()
on row1.Field<string>("FundCode") equals row2.Field<string>("FundCode")
where row1.Field<string>("value")
equals "2" select row1;
result.ToList().ForEach(row => row.Delete());
为linqpad样品测试代码:
void Main()
{
//sample data for test
DataSet ds = new DataSet();
ds.Tables.Add(GetTable1());
ds.Tables.Add(GetTable2());
var result = (from rec1 in ds.Tables[0].AsEnumerable()
join rec2 in ds.Tables[1].AsEnumerable()
on rec1.Field<string>("FC") equals rec2.Field<string>("FC")
where rec2.Field<int>("Value") == 2 select rec1);
result.ToList().ForEach(row => row.Delete());
//now you have only "ABCD" and "AZY" in table 1
//ds.Tables[0].Dump(); linqpad display result
}
DataTable GetTable1()
{
DataTable table = new DataTable();
table.Columns.Add("FC", typeof(string));
table.Rows.Add("ABCD");
table.Rows.Add("XYZ");
table.Rows.Add("AZY");
return table;
}
DataTable GetTable2()
{
DataTable table = new DataTable();
table.Columns.Add("FC", typeof(string));
table.Columns.Add("Value", typeof(int));
table.Rows.Add("ABCD", 1);
table.Rows.Add("XYZ", 2);
table.Rows.Add("AZY",3);
return table;
}
在这一行
其中rec1.Field(“类型”),我得到的错误声明之前,“不能使用局部变量REC1做我。需要申报rec1? –
@ManivannanKG检查我的更新代码 – Damith