如何将WordOpenXML属性转换成System.IO.Packaging.Package？

问题描述

我一直在尝试用C＃构建的Outlook 2010插件，使用Visual Studio 2008中，我能够获得 Microsoft.Office.Interop.Word.DocumentClass 从 Microsoft.Office.Interop.Outlook.Inspector 对象，它代表电子邮件对象当前正编辑。我明白，这的DocumentClass 其他各种环境中使用过多（一般为Microsoft Word扩展，毫不奇怪）。

I've been experimenting with building an Outlook 2010 addin in C#, using Visual Studio 2008. I am able to obtain a Microsoft.Office.Interop.Word.DocumentClass object from a Microsoft.Office.Interop.Outlook.Inspector object, which represents the e-mail being currently edited. I understand that this DocumentClass is used in various other contexts too (usually for Microsoft Word extensions, unsurprisingly).

那类有益有一个叫做物业 WordOpenXML ，这似乎是构成所有文件的XML表示的 .DOCX 套餐，保存此Word文档另存为 .DOCX 。

That class helpfully has a property called WordOpenXML, which seems to be an XML representation of all the files constituting the the .docx package which would be saved to disk when saving this Word document as a .docx.

一个该属性的方式时，可以保存到磁盘将是非常有用的，如果它可以转换成内存中的 System.IO.Packaging.Package 对象; IE浏览器。反序列化。有谁知道是否有一个简单的方法来做到这一点，或者我需要写一些XML解析代码做手工？

One way this property would be very useful is if it could be converted into a System.IO.Packaging.Package object in memory; ie. deserialize it. Does anyone know whether there is an easy way to do this, or do I need to write some XML parsing code to do it manually?

推荐答案

我最终修改一些代码，我发现在网上创建一个 WordOpenXML 转换成一个包，这是保存到磁盘的方法：

I ended up modifying some code I found online to create a method which converts WordOpenXML to a Package, which is saved to disk:

using System;
using System.Xml;
using System.Xml.XPath;
using System.Text;
using System.IO;
using System.IO.Packaging;

// [...]
// Call like this: createPackageFromWordOpenXML(wordEditor.WordOpenXML, @"C:\outputFiles\testOut.docx");

/// <summary>
/// Creates a ZIP package (ie. Word's .docx format) from a WordOpenXML string, which is saved to the file at the path specified.
/// </summary>
/// <param name="wordOpenXML">The WordOpenXML string to get the ZIP package data from.</param>
/// <param name="filePath">The path of the file to save the ZIP package to.</param>
private void createPackageFromWordOpenXML(string wordOpenXML, string filePath)
{
    string packageXmlns = "http://schemas.microsoft.com/office/2006/xmlPackage";
    Package newPkg = System.IO.Packaging.ZipPackage.Open(filePath, FileMode.Create);

    try
    {
        XPathDocument xpDocument = new XPathDocument(new StringReader(wordOpenXML));
        XPathNavigator xpNavigator = xpDocument.CreateNavigator();

        XmlNamespaceManager nsManager = new XmlNamespaceManager(xpNavigator.NameTable);
        nsManager.AddNamespace("pkg", packageXmlns);
        XPathNodeIterator xpIterator = xpNavigator.Select("//pkg:part", nsManager);

        while (xpIterator.MoveNext())
        {
            Uri partUri = new Uri(xpIterator.Current.GetAttribute("name", packageXmlns), UriKind.Relative);

            PackagePart pkgPart = newPkg.CreatePart(partUri, xpIterator.Current.GetAttribute("contentType", packageXmlns));

            // Set this package part's contents to this XML node's inner XML, sans its surrounding xmlData element.
            string strInnerXml = xpIterator.Current.InnerXml
                .Replace("<pkg:xmlData xmlns:pkg=\"" + packageXmlns + "\">", "")
                .Replace("</pkg:xmlData>", "");
            byte[] buffer = Encoding.UTF8.GetBytes(strInnerXml);
            pkgPart.GetStream().Write(buffer, 0, buffer.Length);
        }

        newPkg.Flush();
    }
    finally
    {
        newPkg.Close();
    }
}

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