How can I convert a string to RTF format read from the database?
When read, in the richedit appears the string with tag : /par {ansistring.......
I tried using this code but the result is the same.
rtfString:= set1.fieldbyname('corpo_rtf').asansistring;
stream := TMemoryStream.Create;
stream.Clear;
stream.Write(PAnsiChar(rtfString)^, Length(rtfString));
stream.Position := 0;
corpo.PlainText := False;
corpo.Lines.LoadFromStream(stream);
stream.Free;
Your code works fine if the content of the memory stream is valid RTF. Ergo, that cannot be the case.
You need to dig deeper into the actual content of the memory stream. Write it out to a text file with .rtf extension. Try to load it with Wordpad. See what happens. You should see the same as your Delphi application displays.
So, where could this be going wrong? Some possible causes include:
The data in the database is not valid RTF.
There are some undesired text conversions being performed. We assume that rtfString is of type AnsiString. Is it?
Related
I am new to Oracle Apex 5.1, and I have been asked to implement a button that when clicked, the user gets (downloads) a .doc file of an Interactive report.
I have noticed that the Interactive Report gives you the option to download it as .pdf, .xls, and so, but I need it to be a Word (.doc) file.
In addition, the file must be in a specific format (with heading, indentation, font, etc.) that I was given (as a template) in a Word file.
Any help would be appreciated.
Additional Information: I was able to open the template (.doc) file in NotePad++ and get the <html> version of it, so I could edit it in both NotePad++ and Word.
One of the best actually to do that is APEX OFFICE PRINT(AOP) but isn't free licence.
otherwise you can check this solution
How do we export a ms-word (or rtf) document (from a web browser) to generated by pl/sql?
I end up finding information in this page: http://davidsgale.com/apex-how-to-download-a-file/ and I wrote this code:
declare
l_clob clob;
begin
l_clob := null;
sys.htp.init;
sys.owa_util.mime_header('application/vnd.ms-word', FALSE,'utf-8');
sys.htp.p('Content-length: ' || sys.dbms_lob.getlength( l_clob ));
sys.htp.p('Content-Disposition: inline; filename="test_file.doc"' );
sys.owa_util.http_header_close;
sys.htp.p(SET_DOC_HEADER);
sys.htp.p(SET_TABLE_HEADER);
sys.htp.p(ADD_TABLE_ENTRY("arguments"));
sys.htp.p(SET_TABLE_FOOTER);
sys.wpg_docload.download_file(l_clob);
apex_application.stop_apex_engine;
exception when others then
--sys.htp.prn('error: '||sqlerrm);
apex_application.stop_apex_engine;
end;
It works, but I had to create functions in the SQL Workshop because writting a table in html is really long.
I am building a code that lets the user open some files.
reference = warndlg('Choose the files for analysis.');
uiwait(reference);
filenames2 = uigetfile('./*.txt','MultiSelect', 'on');
if ~iscell(filenames2)
filenames2 = {filenames2}; % force it to be a cell array of strings
end
numberOfFiles = numel(filenames2);
data = importdata(filenames2{i},delimiterIn,headerlinesIn);
When I run the code, the prompts show up, I press OK, and then nothing happens. The code just stops, telling me :
Error using importdata (line 137)
Unable to open file.
Error in FreqVSChampB_no_spec (line 119)
data=importdata(filenames2{1},delimiterIn,headerlinesIn);
I just don't have the opportunity to select a file. The cellarray stays empty as showed in the following image.
MATLAB can't find the file that you have selected. Your variable filenames2 contains only the name of the file, not its full path. If you don't provide the full path to importdata, it will search for whatever file name you provide on the MATLAB path, and if it can't find it it will error as you see.
Try something like this - I'm just doing it with single selection for ease of description, but you can do something similar with multiple selection.
[fileName, pathName] = uigetfile('*.txt');
fullNameWithPath = fullfile(pathName, fileName);
importdata(fullNameWithPath)
fullfile is useful, as it inserts the correct character between pathName and fileName (\ on Windows, / on Unix).
You can try to add
pause(0.1);
just after uiwait(reference);
For me it works. In fact I've noticed the active windows changes when we use uiwait and uigetfile.
I downloaded a data set which is supposed to be in RDF format http://iw.rpi.edu/wiki/Dataset_1329, using Notepad++ I opened it but can't read it. Any suggestions?
The file, uncompressed, is about 140MB. Notepad++ is probably failing due to the size of the file. The RDF format used in this dataset is Ntriples, one triple per line with three components (subject, predicate, object), very human readable. Sample data from the file:
<http://data-gov.tw.rpi.edu/raw/1329/data-1329-00017.rdf#entry8389> <http://data-gov.tw.rpi.edu/vocab/p/1329/race_other_multi_racial> "0" .
<http://data-gov.tw.rpi.edu/raw/1329/data-1329-00017.rdf#entry8389> <http://data-gov.tw.rpi.edu/vocab/p/1329/race_black_and_white> "0" .
<http://data-gov.tw.rpi.edu/raw/1329/data-1329-00017.rdf#entry8389> <http://data-gov.tw.rpi.edu/vocab/p/1329/national_origin_hispanic> "0" .
<http://data-gov.tw.rpi.edu/raw/1329/data-1329-00017.rdf#entry8389> <http://data-gov.tw.rpi.edu/vocab/p/1329/filed_cases> "1" .
If you want to have a look at the data then try to open it with a tool that streams the file rather than loading it all at once, for instance less or head.
If you want to use the data you might want to look into loading it in a triple store (4store, Virtuoso, Jena TDB, ...) and use SPARQL to query it.
Try Google Refine (possibly with RDF extension: http://lab.linkeddata.deri.ie/2010/grefine-rdf-extension/ )
I'm currently trying to attach image files to a model directly from a zip file (i.e. without first saving them on a disk). It seems like there should be a clearer way of converting a ZipEntry to a Tempfile or File that can be stored in memory to be passed to another method or object that knows what to do with it.
Here's my code:
def extract (file = nil)
Zip::ZipFile.open(file) { |zip_file|
zip_file.each { |image|
photo = self.photos.build
# photo.image = image # this doesn't work
# photo.image = File.open image # also doesn't work
# photo.image = File.new image.filename
photo.save
}
}
end
But the problem is that photo.image is an attachment (via paperclip) to the model, and assigning something as an attachment requires that something to be a File object. However, I cannot for the life of me figure out how to convert a ZipEntry to a File. The only way I've seen of opening or creating a File is to use a string to its path - meaning I have to extract the file to a location. Really, that just seems silly. Why can't I just extract the ZipEntry file to the output stream and convert it to a File there?
So the ultimate question: Can I extract a ZipEntry from a Zip file and turn it directly into a File object (or attach it directly as a Paperclip object)? Or am I stuck actually storing it on the hard drive before I can attach it, even though that version will be deleted in the end?
UPDATE
Thanks to blueberry fields, I think I'm a little closer to my solution. Here's the line of code that I added, and it gives me the Tempfile/File that I need:
photo.image = zip_file.get_output_stream image
However, my Photo object won't accept the file that's getting passed, since it's not an image/jpeg. In fact, checking the content_type of the file shows application/x-empty. I think this may be because getting the output stream seems to append a timestamp to the end of the file, so that it ends up looking like imagename.jpg20110203-20203-hukq0n. Edit: Also, the tempfile that it creates doesn't contain any data and is of size 0. So it's looking like this might not be the answer.
So, next question: does anyone know how to get this to give me an image/jpeg file?
UPDATE:
I've been playing around with this some more. It seems output stream is not the way to go, but rather an input stream (which is which has always kind of confused me). Using get_input_stream on the ZipEntry, I get the binary data in the file. I think now I just need to figure out how to get this into a Paperclip attachment (as a File object). I've tried pushing the ZipInputStream directly to the attachment, but of course, that doesn't work. I really find it hard to believe that no one has tried to cast an extracted ZipEntry as a File. Is there some reason that this would be considered bad programming practice? It seems to me like skipping the disk write for a temp file would be perfectly acceptable and supported in something like Zip archive management.
Anyway, the question still stands:
Is there a way of converting an Input Stream to a File object (or Tempfile)? Preferably without having to write to a disk.
Try this
Zip::ZipFile.open(params[:avatar].path) do |zipfile|
zipfile.each do |entry|
filename = entry.name
basename = File.basename(filename)
tempfile = Tempfile.new(basename)
tempfile.binmode
tempfile.write entry.get_input_stream.read
user = User.new
user.avatar = {
:tempfile => tempfile,
:filename => filename
}
user.save
end
end
Check out the get_input_stream and get_output_stream messages on ZipFile.
I have an old application developed in Delphi 2006 that I now needed to make some changes in.
In this app, I'm loading images from a MsSQL "image" field
but when i compile with delphi 2010 I get the error :
"raised exception class EJPEG with message 'JPEG error #51'."
The code that get's the image from the database:
aStream := TMemoryStream.Create;
Try
If LoadFromBlob(FieldByName('Picture'), aStream) then
begin
Pic:=TJpegImage.Create;
try
Try
Pic.LoadFromStream(aStream);
Picture.Assign(Pic); // <------ JPEG Error #51 here w D2010
Except
//something went wrong loading
HandleImageError();
End;
finally
Pic.Free;
end;
end;
Finally
aStream.Free;
End;
// ............
function LoadFromBlob(const AField: TField; const Stream: TStream): boolean;
var
ResultStr: string;
PResultStr: PChar;
begin
Result := false;
if (Assigned(AField)) and (Assigned(Stream)) then begin
try
ResultStr := AField.Value;
If ResultStr <> '' then
begin
PResultStr := PChar(ResultStr);
Stream.Write(PResultStr^, Length(ResultStr));
Stream.Seek(0,0);
Result := true;
end;
except
end;
end;
end;
Googled around a bit and found out that error #51 means: JERR_NO_QUANT_TABLE, whatever that means.
When I compile with Delphi 2006, the same code works fine with no errors, so what's going wrong with D2010?
In D2010 Char changed from Ansi to Unicode, which means it occupies two bytes instead of one.
Change your variables in LoadFromBlob to AnsiString and PAnsiChar and it should work.
Update: Instead of Field.Value (Variant) you should better access Field.AsAnsiString; The variant access might include an implicit Unicode->Ansi conversion which introduces the error.
You are using a rather awkward way to save a blob field to a stream. Why are you not using TBlobField.SaveToStream ?
Save your aStream out to a file and examine with a hex editor. Compare the D2006 and D2010 results.
What you find there should convince you that it's nothing to do with the JPEG header, and it's likely a Unicode/Widechar issue.
Why going the way over the strings anyway? What about directly writing to the stream? The image field probably is a TBlobField or TGraphicField. Those have a SaveToStream method that would perfectly suit your needs.