未找到列 [Convert(LateTime,'System.Double')]是什么意思？

Question

我从数据库读取数据到datatable中，然后进行统计：table.Compute("Sum([Convert(LateTime,'System.Double')])", "Numbers='2012'")，原来的LateTime字段是字符型，我现在要统计这一列中的所有不为空的数值，现在报错“未找到列 [Convert(LateTime,'System.Double')]”是怎么回事呢，请教各位高手指点一下。

Answer 1

table.Compute("Sum(Convert(LateTime,'System.Double'))", "Numbers='2012'")

Answer 2

table.Compute("Sum(Convert(LateTime,'System.Double'))", "Numbers='2012'")

感谢朋友的回复，你这个是错的，会弹出：聚合参数中的语法错误: 需要具有可能的“Child”限定符的单个列参数。

Answer 3

 DataTable dt = new DataTable();
            dt.Columns.Add("Id", typeof(string));
            dt.Columns.Add("Name", typeof(string));
            dt.Rows.Add("11.00");
            dt.Rows.Add("2.00");
            dt.Rows.Add("11.00");
            var value = (dt.AsEnumerable().Where(row => Convert.ToDouble(row["Id"]) > 2.0)).Sum(row => Convert.ToDouble(row["Id"]));
            Console.WriteLine(value);

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Answer 4

 DataTable dt = new DataTable();
            dt.Columns.Add("Id", typeof(string));
            dt.Columns.Add("Name", typeof(string));
            dt.Rows.Add("11.00");
            dt.Rows.Add("2.00");
            dt.Rows.Add("11.00");
            var value = (dt.AsEnumerable().Where(row => Convert.ToDouble(row["Id"]) > 2.0)).Sum(row => Convert.ToDouble(row["Id"]));
            Console.WriteLine(value);

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感谢朋友的回复，我对LINQ和Lambda表达式不是很熟，如果列值有可能有空值的情况下怎么办呢？

Answer 5

DataTable dt = new DataTable();
            dt.Columns.Add("Id", typeof(string));
            dt.Columns.Add("Name", typeof(string));
            dt.Rows.Add("11.00");
            dt.Rows.Add("2.00");
            dt.Rows.Add("11.00");
            var value = (dt.AsEnumerable().Where(row => row["Id"]!=DBNull.Value && Convert.ToDouble(row["Id"]) > 2.0)).Sum(row => Convert.ToDouble(row["Id"]));
            Console.WriteLine(value);

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Answer 6

DataTable dt = new DataTable();
            dt.Columns.Add("Id", typeof(string));
            dt.Columns.Add("Name", typeof(string));
            dt.Rows.Add("11.00");
            dt.Rows.Add("2.00");
            dt.Rows.Add("11.00");
            var value = (dt.AsEnumerable().Where(row => row["Id"]!=DBNull.Value && Convert.ToDouble(row["Id"]) > 2.0)).Sum(row => Convert.ToDouble(row["Id"]));
            Console.WriteLine(value);

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谢谢朋友，你的方法非常好，我成功决定了三个条件下的筛选统计。