以下は、Excelファイルを読み取るための私のコードです。
コード。
FileInfo newFile = new FileInfo("C:\\Excel\\SampleStockTakeExceptionReport.xls");
ExcelPackage pck = new ExcelPackage(newFile);
var ws = pck.Workbook.Worksheets.Add("Content");
ws.View.ShowGridLines = false;
ws.Cells["J12"].Value = "Test Write";
pck.Save();
System.Diagnostics.Process.Start("C:\\Excel\\SampleStockTakeExceptionReport.xls");
コードを実行すると、ランタイム エラーがスローされます。
エラー
System.Exception: Can not open the package. Package is an OLE compound document. If this is an encrypted package, please supply the password ---> System.IO.FileFormatException: File contains corrupted data.
at MS.Internal.IO.Zip.ZipIOEndOfCentralDirectoryBlock.FindPosition(Stream archiveStream)
at MS.Internal.IO.Zip.ZipIOEndOfCentralDirectoryBlock.SeekableLoad(ZipIOBlockManager blockManager)
at MS.Internal.IO.Zip.ZipArchive..ctor(Stream archiveStream, FileMode mode, FileAccess access, Boolean streaming, Boolean ownStream)
at MS.Internal.IO.Zip.ZipArchive.OpenOnStream(Stream stream, FileMode mode, FileAccess access, Boolean streaming)
at System.IO.Packaging.ZipPackage..ctor(Stream s, FileMode mode, FileAccess access, Boolean streaming)
at System.IO.Packaging.Package.Open(Stream stream, FileMode packageMode, FileAccess packageAccess, Boolean streaming)
at System.IO.Packaging.Package.Open(Stream stream, FileMode packageMode, FileAccess packageAccess)
at OfficeOpenXml.ExcelPackage.ConstructNewFile(Stream stream, String password)
--- End of inner exception stack trace ---
at OfficeOpenXml.ExcelPackage.ConstructNewFile(Stream stream, String password)
at OfficeOpenXml.ExcelPackage..ctor(FileInfo newFile)
at Report.Form1.ExportToExcel1(DataTable Tbl, String ExcelFilePath) in C:\SMARTAG_PROJECT\SUREREACH\EXCEL\Report\Report\Form1.cs:line 39
誰かがこれについてアドバイス/助けてくれれば感謝します。ありがとう。