Save WordprocessingMLPackage to ByteArrayInputStream

问题

How can I save a org.docx4j.openpackaging.packages.WordprocessingMLPackage instance into ByteArrayInputStream, then It can be downloaded from server.

Thanks.

回答1:

You cannot save to a ByteArrayInputStream ... ever. A ByteArrayInputStream is an InputStream and you don't / can't write to an InputStream.

However you can write something to a ByteArrayOutputStream, get the byte array, and create a ByteArrayInputStream wrapper for the array.

(I'm assuming that there is a way to save one of those instances to an OutputStream or Writer ...)

Well, my assumption was wrong, and WordprocessingMLPackage's only save method saves to a File. (I guess someone didn't get the memo on how to design flexible I/O apis ...)

But the source code ( here ) offers some clues on how you could implement it yourself. The method is as follows:

public void save(java.io.File docxFile) throws Docx4JException {

    if (docxFile.getName().endsWith(".xml")) {

        // Create a org.docx4j.wml.Package object
        FlatOpcXmlCreator worker = new FlatOpcXmlCreator(this);
        org.docx4j.xmlPackage.Package pkg = worker.get();

        // Now marshall it
        JAXBContext jc = Context.jcXmlPackage;
        try {
            Marshaller marshaller=jc.createMarshaller();

            marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT,
                                                   Boolean.TRUE);
            NamespacePrefixMapperUtils.setProperty(marshaller, 
                    NamespacePrefixMapperUtils.getPrefixMapper());          

            marshaller.marshal(pkg, new FileOutputStream(docxFile));
        } catch (Exception e) {
            throw new Docx4JException("Error saving Flat OPC XML", e);
        }   
        return;
    }

    SaveToZipFile saver = new SaveToZipFile(this); 
    saver.save(docxFile);
}

It looks like you should be able to copy this code in a helper class, and tweak it to save to a OutputStream rather than (specifically) a FileOutputStream. Note that the SaveToZipFile class has alternative save methods that write to an OutputStream.

回答2:

I had the same issue and found an easier way to do it without having to change the save() function. Source here and I made a few edits:

For a WordMLPackage p and HttpServletResponse response:

response.setContentType("application/vnd.openxmlformats-officedocument.wordprocessingml.document");
String fileName = "MyDocument.docx";
response.setHeader("Content-disposition", "attachment;filename=${fileName}");
SaveToZipFile saver = new SaveToZipFile(p);
saver.save( response.getOutputStream() );

import statement:

import org.docx4j.openpackaging.io.*

回答3:

Since 3.1.0 version you can use save(OutputStream outStream):

/**
     *  Save this pkg to an OutputStream in the usual zipped up format
     *  (Docx4J.FLAG_SAVE_ZIP_FILE)
     *  
     *  @since 3.1.0
     */ 
    public void save(OutputStream outStream) throws Docx4JException {
        save(outStream, Docx4J.FLAG_SAVE_ZIP_FILE);                     
    }

来源：https://stackoverflow.com/questions/13360867/save-wordprocessingmlpackage-to-bytearrayinputstream

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