2016-03-03 97 views
0

你好,我想上傳一個excel文件和讀取上傳的excel表到另一個數據的Excel工作表,但我得到的文件不可用error.can有人來幫它無法讀取上傳的Excel文件

string filetest1; 


protected void UploadButton_Click(object sender, EventArgs e) 
    { 
     if (FileUpload1.HasFile) 
      SaveFile(FileUpload1.PostedFile); 
     else 
      UploadStatusLabel1.Text = "You did not specify a file to upload."; 
    } 
     void SaveFile(HttpPostedFile file) 
     { 
      string savePath = "c:\\Users\\*****\\Desktop\\"; 

     // Get the name of the file to upload. 
     string fileName = file.FileName; 

     // Create the path and file name to check for duplicates. 
     string pathToCheck = savePath + fileName; 

     // Create a temporary file name to use for checking duplicates. 
     string tempfileName = ""; 

     // Check to see if a file already exists with the 
     // same name as the file to upload.   
     if (System.IO.File.Exists(pathToCheck)) 
     { 
      int counter = 2; 
      while (System.IO.File.Exists(pathToCheck)) 
      { 
       // if a file with this name already exists, 
       // prefix the filename with a number. 
       tempfileName = counter.ToString() + fileName; 
       pathToCheck = savePath + tempfileName; 
       counter++; 
      } 

      fileName = tempfileName; 

      // Notify the user that the file name was changed. 
      UploadStatusLabel1.Text = "A file with the same name already exists." + 
       "<br />Your file was saved as " + fileName; 
     } 
     else 
     { 
      // Notify the user that the file was saved successfully. 
      UploadStatusLabel1.Text = "Your file was uploaded successfully."; 
     } 

     // Append the name of the file to upload to the path. 
     savePath += fileName; 
     filetest1 = fileName; 
     MessageBox.Show(filetest1); 
     // Call the SaveAs method to save the uploaded 
     // file to the specified directory. 
     file.SaveAs(savePath); 
    } 

    protected void Compare_Click(object sender, EventArgs e) 
    { 

     string filetest1 = FileUpload1.PostedFile.FileName; 
     **MessageBox.Show(filetest1);** 
    ------------- 

    obook = oAPP.Workbooks.Open(filetest1); 
} 

有2個消息框和第一個消息框返回上傳的文件,但第二個消息框中的名稱返回空消息框

回答

-1

存儲的文件名在會話變量並使用它不斷,其中u希望