2011-05-30 547 views
3

我有这样的代码来新列添加到数据表:C#的DataTable小数精度

 
DataColumn col = new DataColumn("column", typeof(decimal));  
col.Caption = "Column"; 
mytable.Columns.Add(col); 

如何指定此列小数精度所以价值总是会在我希望它是格式?

+1

的格式应该由你的任何UI处理使用时,DataTable不应该关心格式化,只是数据。 – 2011-05-30 13:12:22

回答

1

你不能。但是,您可以格式化值,当您使用String.Format函数从表中检索:

String.Format("{0:0.##}", (Decimal) myTable.Rows[rowIndex].Columns[columnIndex]); 
+0

好的。这意味着我有2个选项:1)要么使用数据定义来限制小数数据类型的精度,我想在创建表时使用sql 2)使用Math.Round()或String.Format()进行显示。 – Blablablaster 2011-05-30 13:17:12

+0

@Blablablaster:你将如何使用数据定义来精确限制小数数据类型? – 2011-05-30 13:21:57

+0

create table mytab(somefield decimal(18,2)) – Blablablaster 2011-05-30 13:43:09

1

我有同样的问题我自己,我通过加载应用程序启动时的整个架构,然后引用列固定它根据需要从架构获取信息。

我只有一个Visual Basic例子,但希望它是很容易转换成C#

设置

' in whatever class you do your database communication: 
Private _database As SqlDatabase 
Private Shared _schema As DataTable 

Sub New() 
    ' or however you handle the connection string/database creation 
    Dim connectionString as String = GetConnectionString() 
    _database = New SqlDatabase(connectionString) 

    RetrieveSchema() 
End Sub 


Private Function RetrieveSchema() as DataTable 
    If _schema Is Nothing Then 
    Using connection As SqlConnection = _database.CreateConnection() 
     connection.Open() 
     _schema = connection.GetSchema("Columns") 
    End Using 
    End If 

    return _schema 
End Function 


Public Function GetColumnInformation(tableName As String, columnName As String) as DataRow 
    Dim firstMatchingRow as DataRow = (
    From row In _schema.Rows _ 
    Where (
     row("TABLE_NAME") = tableName AndAlso row("COLUMN_NAME") = columnName) 
    )).FirstOrDefault() 

    Return firstMatchingRow 
End Function 

使用

Dim columnInformation As DataRow = Dal.GetColumnInformation(tableName, columnName) 

' find the precision 
Dim precision = columnInformation("NUMERIC_PRECISION") 
Dim scale = columnInformation("NUMERIC_SCALE") 

' convert the decimal to the column's format 
' e.g.: 2.345 with a scale of 2 would result in 
'  2.35 
value = Decimal.Round(value, scale)