Update gridview with fileupload

Question

how Update gridview with fileupload

hothorasman panjaitan · Answer

thaks Mr prancis this my code if I update fileupload1 then the fileupload2 changed / deleted protected void Updatedata( object sender GridViewUpdateEventArgs e) { MySqlConnection con = new MySqlConnection (strConnString); string id_peraturan = GridView1.DataKeys[e.RowIndex].Value.ToString(); FileUpload FileUpload1 = ( FileUpload )GridView1.Rows[e.RowIndex].FindControl( FileUpload1 ) as FileUpload ; FileUpload FileUpload2 = ( FileUpload )GridView1.Rows[e.RowIndex].FindControl( FileUpload2 ) as FileUpload ; FileUpload FileUpload3 = ( FileUpload )GridView1.Rows[e.RowIndex].FindControl( FileUpload3 ) as FileUpload ; string fileName1 = string .Empty; string fileName2 = string .Empty; string fileName3 = string .Empty; string perda1 = ( Pembentukan/ ); //fle Pembentukan string perda2 = ( Pengelolaan/ ); //file pengelolaan string perda3 = ( Retribusi/ ); //file retribusi if (FileUpload1.HasFile) fileName1 = Path .GetFileName(FileUpload1.FileName); if (FileUpload2.HasFile) fileName2 = Path .GetFileName(FileUpload2.FileName); if (FileUpload3.HasFile) fileName3 = Path .GetFileName(FileUpload3.FileName); if (FileUpload1.HasFile) { perda1 += FileUpload1.FileName; FileUpload1.SaveAs(Server.MapPath( Pembentukan/ ) + fileName1); } if (FileUpload2.HasFile) { perda2 += FileUpload2.FileName; FileUpload2.PostedFile.SaveAs(Server.MapPath(perda2)); } if (FileUpload3.HasFile) { perda3 += FileUpload3.FileName; FileUpload3.PostedFile.SaveAs(Server.MapPath(perda3)); } con.Open(); MySqlCommand update = new MySqlCommand ( UPDATE tbl_peraturan SET namafile1=' + fileName1 + 'perda_pembentukan=' + perda1 + 'fileName1=' + fileName2 + 'perda_pengelolaan=' + perda2 + 'fileName3=' +fileName3+ 'perda_retribusi=' + perda3 + ' WHERE ID_PERATURAN= + id_peraturan +    con); update.ExecuteNonQuery(); con.Close(); GridView1.EditIndex = -1; tampilgird(); }

hothorasman panjaitan · Answer

thaks Mr prancis this my code if I update fileupload1 then the fileupload2 changed / deleted protected void Updatedata( object sender GridViewUpdateEventArgs e) { MySqlConnection con = new MySqlConnection (strConnString); string id_peraturan = GridView1.DataKeys[e.RowIndex].Value.ToString(); FileUpload FileUpload1 = ( FileUpload )GridView1.Rows[e.RowIndex].FindControl( FileUpload1 ) as FileUpload ; FileUpload FileUpload2 = ( FileUpload )GridView1.Rows[e.RowIndex].FindControl( FileUpload2 ) as FileUpload ; FileUpload FileUpload3 = ( FileUpload )GridView1.Rows[e.RowIndex].FindControl( FileUpload3 ) as FileUpload ; string fileName1 = string .Empty; string fileName2 = string .Empty; string fileName3 = string .Empty; string perda1 = ( Pembentukan/ ); //fle Pembentukan string perda2 = ( Pengelolaan/ ); //file pengelolaan string perda3 = ( Retribusi/ ); //file retribusi if (FileUpload1.HasFile) fileName1 = Path .GetFileName(FileUpload1.FileName); if (FileUpload2.HasFile) fileName2 = Path .GetFileName(FileUpload2.FileName); if (FileUpload3.HasFile) fileName3 = Path .GetFileName(FileUpload3.FileName); if (FileUpload1.HasFile) { perda1 += FileUpload1.FileName; FileUpload1.SaveAs(Server.MapPath( Pembentukan/ ) + fileName1); } if (FileUpload2.HasFile) { perda2 += FileUpload2.FileName; FileUpload2.PostedFile.SaveAs(Server.MapPath(perda2)); } if (FileUpload3.HasFile) { perda3 += FileUpload3.FileName; FileUpload3.PostedFile.SaveAs(Server.MapPath(perda3)); } con.Open(); MySqlCommand update = new MySqlCommand ( UPDATE tbl_peraturan SET namafile1=' + fileName1 + 'perda_pembentukan=' + perda1 + 'fileName1=' + fileName2 + 'perda_pengelolaan=' + perda2 + 'fileName3=' +fileName3+ 'perda_retribusi=' + perda3 + ' WHERE ID_PERATURAN= + id_peraturan +    con); update.ExecuteNonQuery(); con.Close(); GridView1.EditIndex = -1; tampilgird(); }

Munesh Sharma · Answer

http://stackoverflow.com/questions/5975360/file-upload-in-gridview http://www.dotnetfox.com/articles/file-upload-in-gridview-Asp-Net-1008.aspx

Francis · Answer

What kind of errors that you have received while implement?