【发布时间】:2014-02-10 22:06:22
【问题描述】:
我对来自 OpenXML 的 MemoryStream 有疑问。如果我在一个方法中完成所有步骤,我可以成功打开一个 Word 文件、更改它并通过 HttpResponse 下载它。
但是,如果我尝试通过返回 MemoryStream 在两个不同的类(或方法)中执行此操作,我会得到一个损坏的 word 文件。我考虑过冲洗或缓冲问题,但没有找到解决方案。
这是工作代码:
public void FillTemplateOpenXmlWord(HttpResponse response)
{
string filePath = @"c:\template.docx";
byte[] filebytes = File.ReadAllBytes(filePath);
using (MemoryStream stream = new MemoryStream(filebytes))
{
using (WordprocessingDocument myDoc = WordprocessingDocument.Open(stream, true))
{
// do some changes
...
myDoc.MainDocumentPart.Document.Save();
}
string docx = "docx";
response.Clear();
response.ClearHeaders();
response.ClearContent();
response.AddHeader("content-disposition", "attachment; filename=\"" + docx + ".docx\"");
response.ContentType = "application/vnd.openxmlformats-officedocument.wordprocessingml.document";
response.ContentEncoding = Encoding.GetEncoding("ISO-8859-1");
stream.Position = 0;
stream.CopyTo(response.OutputStream);
response.End();
}
}
这是无效的代码:
public void OpenFile(HttpResponse response)
{
MemoryStream stream = this.FillTemplateOpenXmlWord();
string docx = "docx";
response.Clear();
response.ClearHeaders();
response.ClearContent();
response.AddHeader("content-disposition", "attachment; filename=\"" + docx + ".docx\"");
response.ContentType = "application/vnd.openxmlformats-officedocument.wordprocessingml.document";
response.ContentEncoding = Encoding.GetEncoding("ISO-8859-1");
stream.Position = 0;
stream.CopyTo(response.OutputStream);
response.End();
}
public MemoryStream FillTemplateOpenXmlWord()
{
string filePath = @"c:\template.docx";
byte[] filebytes = File.ReadAllBytes(filePath);
using (MemoryStream stream = new MemoryStream(filebytes))
{
using (WordprocessingDocument myDoc = WordprocessingDocument.Open(stream, true))
{
// do some changes
...
myDoc.MainDocumentPart.Document.Save();
}
return stream;
}
}
有什么想法吗?
谢谢
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标签: c# .net ms-word openxml memorystream