I'm using JasperReports to generate PDFs, and it gives me the PDF as a byte array, byte[].
I want to pass the raw bytes to another function that needs the file in terms of a FileItem object. In particular, the FileItem is from the Apache Commons library org.apache.commons.fileupload.FileItem.
// the function I want to pass it into
public DocumentDO toDocumentDO(FileItem fileItem);
Is there any way to do that or is it not possible (ie. the byte[] doesn't contain the metadata needed for it to be a FileItem like filename, mime type, etc)?
Your byte array is just what it is - bunch ow raw bytes, all meta data that you mentioned needs to be provided separately unless you read it into the file and then parse the file for the embedded meta information
Related
I am trying to find an example, or documentation, on creating a picture (floating shape object) inside the Excel sheet. The source is supposed to be a numeric bitmap data, stored in a VBA array acquired using external I/O libraries. Using Excel cells as an intermediary storage is possible, but not desired, since the RGB bitmap data is expected to be huge.
The task itself seems to be extremely simple in matlab-like environments, or python. But I just have no Idea how to make it in Excel and VBA without importing an independent image file from the file system.
In terms of storing the file, how huge is 'huge'? If you convert the image into Base64, it'll be a fairly trivial task to split it up amongst the cells and then reconstitute it when converting it into an image.
Alternatively, you can store the Base64 string in a standard module - I'm currently doing much the same thing, but my image only clocks in at 100kb (better to save it as a PNG rather than BMP).
In terms of converting the Base64 string to an image, the Windows Image Acquisition COM object will convert a byte array into a stdPicture image type (and further to my point above, it will also accept PNG files...]. The following function accepts a Base64 string, converts it into a byte array, and returns an stdPicture object:
Function Base64toStdPicture(ByVal Base64Code As String) As StdPicture
Dim ImgVector As Object
Dim Node As Object
Set ImgVector = CreateObject("WIA.Vector")
Set Node = CreateObject("Msxml2.DOMDocument.3.0").createElement("base64")
Node.DataType = "bin.base64"
Node.Text = Base64Code
ImgVector.BinaryData = Node.nodeTypedValue
Set Base64toStdPicture = ImgVector.BinaryData.Picture
Set Node = Nothing
Set ImgVector = Nothing
End Function
From that point, you can out it in an image control, or copy it to / from the clipboard, etc.
I am developing a LightSwitch application that generates barcodes (QR images) for tickets. I am calling an encode function that converts text to bitmap.
I just need to save this in an LightSwitch Image field.
I have this:
QRCodeEncoder qrCodeEncoder = new QRCodeEncoder();
EditableImage image = qrCodeEncoder.Encode(data);
I want this:
ticket.QRImage = .....???
I am using this library for the QR
http://www.jeff.wilcox.name/2009/09/quick-read-silverlight-barcodes/
http://www.codeproject.com/Articles/20574/Open-Source-QRCode-Library
You can get the bytes by calling image.GetStream(), and then using one of the standard methods to get the bytes out of the stream (see How to convert an Stream into a byte[] in C#?)
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 am trying to append 2 images (as byte[] ) in GoogleAppEngine Java and then ask HttpResponseServlet to display it.
However, it does not seem like the second image is being appended.
Is there anything wrong with the snippet below?
...
resp.setContentType("image/jpeg");
byte[] allimages = new byte[1000000]; //1000kB in size
int destPos = 0;
for(Blob savedChart : savedCharts) {
byte[] imageData = savedChart.getBytes(); //imageData is 150k in size
System.arraycopy(imageData, 0, allimages, destPos, imageData.length);
destPos += imageData.length;
}
resp.getOutputStream().write(allimages);
return;
Regards
I would expect the browser/client to issue 2 separate requests for these images, and the servlet would supply each in turn.
You can't just concatenate images together (like most other data structures). What about headers etc.? At the moment you're providing 2 jpegs butted aainst one another and a browser won't handle that at all.
If you really require 2 images together, you're going to need some image processing library to do this for you (or perhaps, as noted, AWT). Check out the ImageIO library.
Seem that you have completely wrong concept about image file format and how they works in HTML.
In short, the arrays are copied very well without problem. But it is not the way how image works.
You will need to do AWT to combine images in Java
I'm using GWT and GAE for my project. I'm using data transfer objects and dozer to move data between client and server. Dozer had been working great, but I have some classes that need to store text that is over 500 characters, so I must use com.google.appengine.api.datastore.Text datatype in my server side object, but a regular String in my client side object. How do I map these two types using dozer? I know somehow I can specify an XML file, but how do I write that XML file?
specify a mapping between both the datatypes as below. Dozer will use it at run-time to convert.
<mapping>
<class-a>com.google.appengine.api.datastore.Text</class-a>
<class-b>java.lang.String</class-b>
</mapping>
In case you don't know how to use config file ,
In your code,
DozerMapper beanMapper = new DozerMapper();
beanMapper.mapping(new ArrayList<String>(){
{
add("name Of the dozer mapping file");
}
});