我想使用C#代码获取上传的excel文件的工作表名称。该文件可能是.xls或.xlsx格式。我使用的代码如下:
protected void btnGenerateCSV_Click(object sender, EventArgs e)
{
string sourceFile = ExcelFileUpload.PostedFile.FileName;
string worksheetName = ??? (How to get the first sheetname of the uploaded file)
string strConn = "Provider=Microsoft.Jet.OLEDB.4.0;Data Source=" + sourceFile + ";Extended Properties=\"Excel 8.0;HDR=Yes;IMEX=1\"";
if (sourceFile.Contains(".xlsx"))
strConn = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" + sourceFile + ";Extended Properties=\"Excel 12.0;HDR=Yes;IMEX=1\"";
try
{
conn = new OleDbConnection(strConn);
conn.Open();
cmd = new OleDbCommand("SELECT * FROM [" + worksheetName + "$]", conn);
cmd.CommandType = CommandType.Text;
wrtr = new StreamWriter(targetFile);
da = new OleDbDataAdapter(cmd);
DataTable dt = new DataTable();
da.Fill(dt);
for (int x = 0; x < dt.Rows.Count; x++)
{
string rowString = "";
for (int y = 0; y < dt.Columns.Count; y++)
{
rowString += "\"" + dt.Rows[x][y].ToString() + "\",";
}
wrtr.WriteLine(rowString);
}
}
catch (Exception exp)
{
}
finally
{
if (conn.State == ConnectionState.Open)
conn.Close();
conn.Dispose();
cmd.Dispose();
da.Dispose();
wrtr.Close();
wrtr.Dispose();
}
}
string worksheetName = ???
(如何获取上传文件的第一个表名)
任何人请帮忙......
答案 0 :(得分:5)
我使用它从.xlsx文件中获取工作表名称并循环遍历所有名称以逐页阅读工作表
OleDbConnection connection = new OleDbConnection(@"Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" + filename + ";Extended Properties='Excel 12.0 xml;HDR=YES;'");
connection.Open();
DataTable Sheets = connection.GetOleDbSchemaTable(OleDbSchemaGuid.Tables, null);
foreach (DataRow dr in Sheets.Rows)
{
string sht = dr[2].ToString().Replace("'", "");
OleDbDataAdapter dataAdapter = new OleDbDataAdapter("select * from [" + sht + "]", connection);
}
答案 1 :(得分:1)
DataTable Sheets = oleConnection.GetOleDbSchemaTable(OleDbSchemaGuid.Tables, null);
for(int i=0;i<Sheets.Rows.Count;i++)
{
string worksheets= Sheets.Rows[i]["TABLE_NAME"].ToString();
string sqlQuery = String.Format("SELECT * FROM [{0}]", worksheets);
}
答案 2 :(得分:0)
如果Excel太大,此代码将浪费大量时间(conn.open())。使用Openxml会更好(使用更少的时间),但如果Excel是Open - 使用openxml读取将有异常,但oldbhelper wile也没有例外。我的英语是游泳池,对不起.-----中国男孩
答案 3 :(得分:0)
我使用Microsoft excel库Microsoft.Office.Interop.Excel。然后,您可以使用index来获取工作表名称。
string path = @"C\Desktop\MyExcel.xlsx" //Path for excel
using Excel = Microsoft.Office.Interop.Excel;
xlAPP = new Excel.Application();
xlAPP.Visible = false;
xlWbk = xlAPP.Workbooks.Open(path);
string worksheetName = xlWbk.Worksheets.get_Item(1).Name //pass Index here. Reemember that index starts from 1.
xlAPP.Quit();
releaseObject(xlWbk);
releaseObject(xlAPP);
//Always handle unmanaged code.
private void releaseObject(object obj)
{
try
{
System.Runtime.InteropServices.Marshal.ReleaseComObject(obj);
obj = null;
}
catch (Exception ex)
{
obj = null;
MessageBox.Show("Unable to release the Object " + ex.ToString());
}
finally
{
GC.Collect();
}
}