I already have the path of the Excel file, but whenever I call save method, the program writes new content to UTC.xlsx file.
string pathString = "C:\Users\ADMIN-PC\Documents\SUMMER 2018\PRN192\CreateFile\SubFolder\UTC.xlsx";
ef.Save(pathString);
var path = #"C:\Users\ADMIN-PC\Documents\SUMMER 2018\PRN192\CreateFile\SubFolder\UTC.xlsx";
// Load the workbook:
var workbook = ExcelFile.Load(path);
// Manipulate the workbook …
// … and save it over the original
workbook.Save(path);
Related
I have a program which exports a System.Data.DataTable to an XLSX / OpenXml Spreadsheet. Finally have it mostly working. However when opening the Spreadsheet in Excel, Excel complains about the file being invalid, and needing repair, giving this message...
We found a problem with some content in . Do you want us to
try to recover as much as we can? If you trust the source of the
workbook, clik Yes.
If I click Yes, it comes back with this message...
Clicking the log file and opening that, just shows this...
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<recoveryLog xmlns="http://schemas.openxmlformats.org/spreadsheetml/2006/main">
<logFileName>error268360_01.xml</logFileName>
<summary>Errors were detected in file 'C:\Users\aabdi\AppData\Local\Temp\data.20190814.152538.xlsx'</summary>
<repairedRecords>
<repairedRecord>Repaired Records: Cell information from /xl/worksheets/sheet1.xml part</repairedRecord>
</repairedRecords>
</recoveryLog>
Obviously, we don't want to deploy this into a production environment like this. So I've been trying to figure out how to fix this. I threw together a quick little sample to validate the XML and show the errors, based on this link from MSDN. But when I run the program and load the exact same XLSX document that Excel complains about, the Validator comes back saying that the file is perfectly Valid. So I'm not sure where else to go from there.
Any better tools for trying to validate my XLSX XML? Following is the complete code I'm using to generate the XLSX file. (Yes, it's in VB.NET, it's a legacy app.)
If I comment out the line in the For Each dr As DataRow loop, then the XLSX file opens fine in Excel, (just without any data). So it's something with the individual cells, but I'm not really DOING much with them. Setting a value and data type, and that's it.
I also tried replacing the For Each loop in ConstructDataRow with the following, but it still outputs the same "bad" XML...
rv.Append(
(From dc In dr.Table.Columns
Select ConstructCell(
NVL(dr(dc.Ordinal), String.Empty),
MapSystemTypeToCellType(dc.DataType)
)
).ToArray()
)
Also tried replacing the call to Append with AppendChild for each cell too, but that didn't help either.
The zipped up XLSX file (erroring, with dummy data) is available here:
https://drive.google.com/open?id=1KVVWEqH7VHMxwbRA-Pn807SXHZ32oJWR
Full DataTable to Excel XLSX Code
#Region " ToExcel "
<Extension>
Public Function ToExcel(ByVal target As DataTable) As Attachment
Dim filename = Path.GetTempFileName()
Using doc As SpreadsheetDocument = SpreadsheetDocument.Create(filename, DocumentFormat.OpenXml.SpreadsheetDocumentType.Workbook)
Dim data = New SheetData()
Dim wbp = doc.AddWorkbookPart()
wbp.Workbook = New Workbook()
Dim wsp = wbp.AddNewPart(Of WorksheetPart)()
wsp.Worksheet = New Worksheet(data)
Dim sheets = wbp.Workbook.AppendChild(New Sheets())
Dim sheet = New Sheet() With {.Id = wbp.GetIdOfPart(wsp), .SheetId = 1, .Name = "Data"}
sheets.Append(sheet)
data.AppendChild(ConstructHeaderRow(target))
For Each dr As DataRow In target.Rows
data.AppendChild(ConstructDataRow(dr)) '// THIS LINE YIELDS THE BAD PARTS
Next
wbp.Workbook.Save()
End Using
Dim attachmentname As String = Path.Combine(Path.GetDirectoryName(filename), $"data.{Now.ToString("yyyyMMdd.HHmmss")}.xlsx")
File.Move(filename, attachmentname)
Return New Attachment(attachmentname, "application/vnd.openxmlformats-officedocument.spreadsheetml.sheet")
End Function
Private Function ConstructHeaderRow(dt As DataTable) As Row
Dim rv = New Row()
For Each dc As DataColumn In dt.Columns
rv.Append(ConstructCell(dc.ColumnName, CellValues.String))
Next
Return rv
End Function
Private Function ConstructDataRow(dr As DataRow) As Row
Dim rv = New Row()
For Each dc As DataColumn In dr.Table.Columns
rv.Append(ConstructCell(NVL(dr(dc.Ordinal), String.Empty), MapSystemTypeToCellType(dc.DataType)))
Next
Return rv
End Function
Private Function ConstructCell(value As String, datatype As CellValues) As Cell
Return New Cell() With {
.CellValue = New CellValue(value),
.DataType = datatype
}
End Function
Private Function MapSystemTypeToCellType(t As System.Type) As CellValues
Dim rv As CellValues
Select Case True
Case t Is GetType(String)
rv = CellValues.String
Case t Is GetType(Date)
rv = CellValues.Date
Case t Is GetType(Boolean)
rv = CellValues.Boolean
Case IsNumericType(t)
rv = CellValues.Number
Case Else
rv = CellValues.String
End Select
Return rv
End Function
#End Region
For anyone else coming in and finding this, I finally tracked this down to the Cell.DataType
Setting a value of CellValues.Date will cause Excel to want to "fix" the document.
(apparently for dates, the DataType should be NULL, and Date was only used in Office 2010).
Also, if you specify a DataType of CellValues.Boolean, then the CellValue needs to be either 0 or 1. "true" / "false" will also cause Excel to want to "fix" your spreadsheet.
Also, Microsoft has a better validator tool already built for download here:
https://www.microsoft.com/en-us/download/details.aspx?id=30425
Is there a way to configure the Apache Tikka so that it only extracts the metadata properties from the file and does not access the content of the file. ? We need a way to do this so as to avoid reading the entire content in larger files.
The code to extract we are using is as follows:
var tikaConfig = TikaConfig.getDefaultConfig();
var metadata = new Metadata();
AutoDetectParser parser = new AutoDetectParser(tikaConfig);
BodyContentHandler handler = new BodyContentHandler();
using (TikaInputStream stream = TikaInputStream.get(new File(filename), metadata))
{
parser.parse(stream, handler, metadata, new ParseContext());
Array metadataKeys = metadata.names();
Array.Sort(metadataKeys);
}
With the above code sample, when we try to extract the metadata even the content is being read. We would need a way to avoid the same.
I have a custom template in which I'd like to control (as best I can) the types of content that can exist in a document. To that end, I disable controls, and I also intercept pastes to remove some of those content types, e.g. charts. I am aware that this content can also be drag-and-dropped, so I also check for it later, but I'd prefer to stop or warn the user as soon as possible.
I have tried a few strategies:
RTF manipulation
Open XML manipulation
RTF manipulation is so far working fairly well, but I'd really prefer to use Open XML as I expect it to be more useful in the future. I just can't get it working.
Open XML Manipulation
The wonderfully-undocumented (as far as I can tell) "Embed Source" appears to contain a compound document object, which I can use to modify the copied content using the Open XML SDK. But I have been unable to put the modified content back into an object that lets it be pasted correctly.
The modification part seems to work fine. I can see, if I save the modified content to a temporary .docx file, that the changes are being made correctly. It's the return to the clipboard that seems to be giving me trouble.
I have tried assigning just the Embed Source object back to the clipboard (so that the other types such as RTF get wiped out), and in this case nothing at all gets pasted. I've also tried re-assigning the Embed Source object back to the clipboard's data object, so that the remaining data types are still there (but with mismatched content, probably), which results in an empty embedded document getting pasted.
Here's a sample of what I'm doing with Open XML:
using OpenMcdf;
using DocumentFormat.OpenXml;
using DocumentFormat.OpenXml.Packaging;
using DocumentFormat.OpenXml.Wordprocessing;
...
object dataObj = Forms.Clipboard.GetDataObject();
object embedSrcObj = dateObj.GetData("Embed Source");
if (embedSrcObj is Stream)
{
// read it with OpenMCDF
Stream stream = embedSrcObj as Stream;
CompoundFile cf = new CompoundFile(stream);
CFStream cfs = cf.RootStorage.GetStream("package");
byte[] bytes = cfs.GetData();
string savedDoc = Path.GetTempFileName() + ".docx";
File.WriteAllBytes(savedDoc, bytes);
// And then use the OpenXML SDK to read/edit the document:
using (WordprocessingDocument openDoc = WordprocessingDocument.Open(savedDoc, true))
{
OpenXmlElement body = openDoc.MainDocumentPart.RootElement.ChildElements[0];
foreach (OpenXmlElement ele in body.ChildElements)
{
if (ele is Paragraph)
{
Paragraph para = (Paragraph)ele;
if (para.ParagraphProperties != null && para.ParagraphProperties.ParagraphStyleId != null)
{
string styleName = para.ParagraphProperties.ParagraphStyleId.Val;
Run run = para.LastChild as Run; // I know I'm assuming things here but it's sufficient for a test case
run.RunProperties = new RunProperties();
run.RunProperties.AppendChild(new DocumentFormat.OpenXml.Wordprocessing.Text("test"));
}
}
// etc.
}
openDoc.MainDocumentPart.Document.Save(); // I think this is redundant in later versions than what I'm using
}
// repackage the document
bytes = File.ReadAllBytes(savedDoc);
cf.RootStorage.Delete("Package");
cfs = cf.RootStorage.AddStream("Package");
cfs.Append(bytes);
MemoryStream ms = new MemoryStream();
cf.Save(ms);
ms.Position = 0;
dataObj.SetData("Embed Source", ms);
// or,
// Clipboard.SetData("Embed Source", ms);
}
Question
What am I doing wrong? Is this just a bad/unworkable approach?
I have an application that uploads a file. I need to pass this file into another program but for that I need the file name only. Is there any simple code for that, using only Java or a servlet procedure?
while (files.hasMoreElements())
{
name = (String)files.nextElement();
type = multipartRequest.getContentType(name);
filename = multipartRequest.getFilesystemName(name);
originalFilename = multipartRequest.getOriginalFileName(name);
//extract the file extension - this can be use to reject a
//undesired file extension
extension1 = filename.substring
(filename.length() - 4, filename.length());
extension2 = originalFilename.substring
(originalFilename.length() - 4, originalFilename.length());
//return a File object for the specified uploaded file
File currentFile = multipartRequest.getFile(name);
//InputStream inputStream = new BufferedInputStream
(new FileInputStream(currentFile));
if(currentFile == null) {
out.println("There is no file selected!");
return;
}
There's a method in apache commons-io to get the file's extension. There's also guava Files class, with its getFileExtension method.
I have built a Server that you can upload files to and download, using Eclipse, servlet and jsp, it's all very new to me. (more info).
Currently the upload system works with the file's name. I want to programmatically assign each file a random key. And with that key the user can download the file. That means saving the data in a config file or something like : test.txt(file) fdjrke432(filekey). And when the user inputs the filekey the servlet will pass the file for download.
I have tried using a random string generator and renameTo(), for this. But it doesn't work the first time, only when I upload the same file again does it work. And this system is flawed, the user will receive the file "fdjrke432" instead of test.txt, their content is the same but you can see the problem.
Any thoughts, suggestions or solutions for my problem?
Well Sebek, I'm glad you asked!! This is quite an interesting one, there is no MAGIC way to do this. The answer is indeed to rename the file you uploaded. But I suggest adding the random string before the name of the file; like : fdjrke432test.txt.
Try this:
filekey= RenameRandom();
File renamedUploadFile = new File(uploadFolder + File.separator+ filekey+ fileName);
item.write(renamedUploadFile);
//remember to give the user the filekey
with
public String RenameRandom()
{
final int LENGTH = 8;
StringBuffer sb = new StringBuffer();
for (int x = 0; x < LENGTH; x++)
{
sb.append((char)((int)(Math.random()*26)+97));
}
System.out.println(sb.toString());
return sb.toString();
}
To delete or download the file from the server you will need to locate it, the user will input the key, you just need to search the upload folder for a file that begins with that key:
filekey= request.getParameter("filekey");
File f = new File(getServletContext().getRealPath("") + File.separator+"data");
File[] matchingFiles = f.listFiles(new FilenameFilter() {
public boolean accept(File dir, String name) {
return name.startsWith(filekey);
}
});
String newfilename = matchingFiles[0].getName();
// now delete or download newfilename