我正在寻找将ADO.NET Datatable转换为csv文件的c#代码,但我想保存/恢复
- 列名称,
- 列数据类型和
- 列值
在csv中。我找到的大部分解决方案都是在字符串列类型中从CSV恢复数据表。我还希望可以为null的值恢复为DBNull.Value。应保存DateTime列并将其还原为DateTime Type。这个概念是使用来自Oracle / Sqlserver数据库的DataAdapter填充数据表,然后将该表保存为CSV文件,然后从CSV恢复。
我已使用下面的代码链接将DataTable保存到CSV文件,使用DataTableExtensions类c# datatable to csv
为了将CSV文件读回DataTable,我使用了以下链接 http://www.codeproject.com/Articles/11698/A-Portable-and-Efficient-Generic-Parser-for-Flat-F
问题是当我将CSV文件恢复到datatable时,我必须从DataTable行创建实体。但是他们在InvalidCast上抛出异常。
答案 0 :(得分:1)
假设您要将列名存储在第一行中,而将类型存储在第二行中,并且数据从第三行开始,则可以使用以下代码。样本数据:
DataTable tblExport = new DataTable();
tblExport.Columns.Add("ID", typeof(int));
tblExport.Columns.Add("Name", typeof(string));
tblExport.Columns.Add("DateofBirth", typeof(DateTime)).AllowDBNull = false;
tblExport.Columns.Add("DateofDeath", typeof(DateTime)).AllowDBNull = true;
tblExport.Rows.Add(1, "Tim", new DateTime(1973, 7, 9), DBNull.Value);
tblExport.Rows.Add(2, "Jim", new DateTime(1953, 3, 19), new DateTime(2011, 1, 2));
tblExport.Rows.Add(3, "Toby", new DateTime(1983, 4, 23), DBNull.Value);
由于您需要将所有值转换为value.ToString
字符串,我将文化更改为InvariantCulture
以强制使用特定的DateTime格式,然后存储旧文件以便您可以再次启用它在末尾。我希望代码能够自我解释:
var oldCulture = CultureInfo.CurrentCulture;
System.Threading.Thread.CurrentThread.CurrentCulture = CultureInfo.InvariantCulture;
string delimiter = "\t"; // tab separated
StringBuilder sb = new StringBuilder();
// first line column-names
IEnumerable<string> columnNames = tblExport.Columns.Cast<DataColumn>()
.Select(column => column.ColumnName);
sb.AppendLine(string.Join(delimiter, columnNames));
// second line column-types
IEnumerable<string> columnTypes = tblExport.Columns.Cast<DataColumn>()
.Select(column => column.DataType.ToString());
sb.AppendLine(string.Join(delimiter, columnTypes));
// rest: table data
foreach (DataRow row in tblExport.Rows)
{
IEnumerable<string> fields = row.ItemArray.Select(field => field.ToString());
sb.AppendLine(string.Join(delimiter, fields));
}
string path = @"C:\Temp\Testfile.csv";
File.WriteAllText(path, sb.ToString());
string[] lines = File.ReadAllLines(path);
string[] columns = lines[0].Split(new[] { delimiter }, StringSplitOptions.None);
string[] types = lines[1].Split(new[] { delimiter }, StringSplitOptions.None);
DataTable tblImport = new DataTable();
for (int i = 0; i < columns.Length; i++)
{
string colName = columns[i];
string typeName = types[i];
tblImport.Columns.Add(colName, Type.GetType(typeName));
}
// import data
// use a typeValueConverter dictionary to convert values:
var typeValueConverter = new Dictionary<Type, Func<string, object>> {
{ typeof(DateTime), value => value.TryGetDateTime(null, null) },
{ typeof(Decimal), value => value.TryGetDecimal(null) },
{ typeof(int), value => value.TryGetInt32(null) },
};
foreach (string line in lines.Skip(2))
{
string[] fields = line.Split(new[]{ delimiter }, StringSplitOptions.None);
DataRow r = tblImport.Rows.Add(); // already added at this point
for (int i = 0; i < tblImport.Columns.Count; i++)
{
DataColumn col = tblImport.Columns[i];
string rawValue = fields[i];
object val = rawValue;
if (typeValueConverter.ContainsKey(col.DataType))
val = typeValueConverter[col.DataType](rawValue);
else if (col.DataType != typeof(string) && string.IsNullOrEmpty(rawValue))
val = DBNull.Value;
r.SetField(col, val);
}
}
System.Threading.Thread.CurrentThread.CurrentCulture = oldCulture;
当然,您应该将两种方法分开,一种用于导出,另一种用于导入。
我使用了我的扩展方法TryGetDateTime
,TryGetDecimal
和TryGetInt32
,它们将字符串解析为DateTime?
,Decimal?
和int?
(null如果无法解析)。它们在LINQ查询中特别方便:
public static DateTime? TryGetDateTime(this string item, DateTimeFormatInfo dfi, params string[] allowedFormats)
{
if (dfi == null) dfi = DateTimeFormatInfo.InvariantInfo;
DateTime dt;
bool success;
if(allowedFormats == null)
success = DateTime.TryParse(item, dfi, DateTimeStyles.None, out dt);
else
success = DateTime.TryParseExact(item, allowedFormats, dfi, DateTimeStyles.None, out dt);
if (success) return dt;
return null;
}
public static decimal? TryGetDecimal(this string item, IFormatProvider formatProvider = null, NumberStyles nStyles = NumberStyles.Any)
{
if (formatProvider == null) formatProvider = NumberFormatInfo.InvariantInfo;
decimal d = 0m;
bool success = decimal.TryParse(item, nStyles, formatProvider, out d);
if (success)
return d;
else
return null;
}
public static int? TryGetInt32(this string item, IFormatProvider formatProvider = null, NumberStyles nStyles = NumberStyles.Any)
{
if (formatProvider == null) formatProvider = NumberFormatInfo.InvariantInfo;
int i = 0;
bool success = int.TryParse(item, nStyles, formatProvider, out i);
if (success)
return i;
else
return null;
}