如何使用C＃獲取上傳的Excel文件的sheetname？

Question

我想使用C＃代碼獲取上傳的Excel文件的表名。該文件可能是.xls或.xlsx格式。我用的代碼如下：

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 = ???（如何獲得上傳文件的第一SHEETNAME）

有人請大家幫幫忙...

Answer 1

我使用它來獲取片從.xlsx文件，並遍歷所有名稱的名稱由一個

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); 
    } 

Answer 2

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); 
} 

Answer 3

讀表一個如果Excel太大，這段代碼會浪費很多時間（conn.open（））。使用Openxml會更好（使用更少的時間），但是如果Excel是打開的---使用openxml讀取將會有異常，但oldbhelper不會有任何異常。我的英語是游泳池，對不起.-----中國男孩

Answer 4

我使用微軟的excel庫Microsoft.Office.Interop.Excel。然後，您可以使用索引來獲取工作表名稱，如下所示。

 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(); 
     } 
    }