blocks|key|3021213|text|不是的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3021214|Zip包不是(普通)+Zip文件。它们必须遵循强制结构，并在根目录中包含该Content_Types.xml文件。ZipPackage+=+ZipArchive+%2B+Structure。|offset|length|style|CODE|3021215|如果你想创建(尤其是如果你想要阅读)普通的Zip压缩文件，你将需要一个额外的库，在BCL中没有支持。|3021216|请参阅SharpZipLib+(GPL)和DotNetZip|3021217|entityMap|0|LINK|mutability|MUTABLE|url|http://www.icsharpcode.net/opensource/sharpziplib/|1|http://dotnetzip.codeplex.com/^0|0|11|H|1L|Z|0|0|3|B|0|L|9|1|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@$D|X|E|Y|F|G]|$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|12|8|@]|9|@$D|13|E|14|1|15]|$D|16|E|17|1|18]]|A|$]]|$1|L|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]|T|$5|O|P|Q|A|$R|U]]]]

No.

Zip Packages are not (normal) Zip files. They must follow an imposed structure and contain that <code>Content_Types.xml</code> file in the root. <code>ZipPackage = ZipArchive + Structure</code>.

If you want to create (and especially if you want to read) normal Zip archives you will need an extra library, there is no support in the BCL.

See <a href="http://www.icsharpcode.net/opensource/sharpziplib/" rel="nofollow noreferrer">SharpZipLib</a> (GPL) and <a href="http://dotnetzip.codeplex.com/" rel="nofollow noreferrer">DotNetZip</a>

blocks|key|294060|text|如果不调用.Flush()方法，就不会有这样的文件|type|unstyled|depth|inlineStyleRanges|entityRanges|data|294061|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

If you don't call the .Flush() method, there will be no such file

blocks|key|294065|text|是，您可以在不添加额外XML内容文件的情况下创建压缩包|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|294066|灵感来自这个链接：Using+System.IO.Packaging+to+generate+a+ZIP+file|294067|使用Yiping提到的上述发现，您可以避免将额外的xml文件添加到包中。在关闭zip存档之前，将内存流中的zip流保存到物理zip文件中，如下所示：|294068|public+static+void+AddFilesToZip(string+zipFilename,+List<String>+filesToAdd)
{
++++using+(var+memStream+=+new+MemoryStream())
++++{
++++++++using+(Package+zip+=+System.IO.Packaging.Package.Open(memStream,+FileMode.Create))
++++++++{

++++++++++++foreach+(var+fileToAdd+in+filesToAdd)
++++++++++++{
++++++++++++++++string+destFilename+=+".\\"+%2B+Path.GetFileName(fileToAdd);
++++++++++++++++Uri+uri+=+PackUriHelper.CreatePartUri(new+Uri(destFilename,+UriKind.Relative));

++++++++++++++++//Existing+parts+not+likely+in+fresh+memory+stream
++++++++++++++++if+(zip.PartExists(uri))
++++++++++++++++{
++++++++++++++++++++zip.DeletePart(uri);
++++++++++++++++}

++++++++++++++++PackagePart+part+=+zip.CreatePart(uri,+"",+CompressionOption.Normal);

++++++++++++++++using+(FileStream+fileStream+=+new+FileStream(fileToAdd,+FileMode.Open,+FileAccess.Read))
++++++++++++++++{

++++++++++++++++++++using+(Stream+dest+=+part.GetStream())
++++++++++++++++++++{
++++++++++++++++++++++++CopyStream(fileStream,+dest);
++++++++++++++++++++}
++++++++++++++++}
++++++++++++}

++++++++++++//The+zip+Package+will+add+an+XML+content+type+file+to+memeory+stream+when+it+closes
++++++++++++//so+before+it+closes+we+save+the+memorystream+to+physical+zip+file.
++++++++++++using+(FileStream+zipfilestream+=+new+FileStream(zipFilename,+FileMode.Create,+FileAccess.Write))
++++++++++++{
++++++++++++++++memStream.Position+=+0;
++++++++++++++++CopyStream(memStream,+zipfilestream);
++++++++++++}

++++++++++++//+That's+it.+Zip+file+saved+to+file.+Things+added+by+package+after+this+point+will+be+to+memory+stream+finally+disposed.
++++++++}
++++}
}|code-block|syntax|javascript|294069|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/6386113/using-system-io-packaging-to-generate-a-zip-file|1|https://stackoverflow.com/users/3559256/yiping^0|0|R|0|9|1C|0|0|2|6|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@$9|12|A|13|1|14]]|E|$]]|$1|H|3|I|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|J|3|K|5|L|7|19|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

Yes you can create zip packages without the extra XML content file added 

Inspired by this link: <a href="https://stackoverflow.com/questions/6386113/using-system-io-packaging-to-generate-a-zip-file">Using System.IO.Packaging to generate a ZIP file</a>

Using above discovery mentioned by <a href="https://stackoverflow.com/users/3559256/yiping">Yiping</a> you can avoid the extra xml file added into the package. Save zip stream from memory stream to a physical zip file before zip archive is closed like this:

<pre><code>public static void AddFilesToZip(string zipFilename, List&lt;String&gt; filesToAdd)
{
 using (var memStream = new MemoryStream())
 {
 using (Package zip = System.IO.Packaging.Package.Open(memStream, FileMode.Create))
 {

 foreach (var fileToAdd in filesToAdd)
 {
 string destFilename = ".\\" + Path.GetFileName(fileToAdd);
 Uri uri = PackUriHelper.CreatePartUri(new Uri(destFilename, UriKind.Relative));

 //Existing parts not likely in fresh memory stream
 if (zip.PartExists(uri))
 {
 zip.DeletePart(uri);
 }

 PackagePart part = zip.CreatePart(uri, "", CompressionOption.Normal);

 using (FileStream fileStream = new FileStream(fileToAdd, FileMode.Open, FileAccess.Read))
 {

 using (Stream dest = part.GetStream())
 {
 CopyStream(fileStream, dest);
 }
 }
 }

 //The zip Package will add an XML content type file to memeory stream when it closes
 //so before it closes we save the memorystream to physical zip file.
 using (FileStream zipfilestream = new FileStream(zipFilename, FileMode.Create, FileAccess.Write))
 {
 memStream.Position = 0;
 CopyStream(memStream, zipfilestream);
 }

 // That's it. Zip file saved to file. Things added by package after this point will be to memory stream finally disposed.
 }
 }
}
</code></pre>

blocks|key|297163|text|天哪，这对我帮助很大！你太棒了！我不得不使用它，因为我的旧框架(3.5)。仅作为补充，请参见下面CopyStream函数的实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|297164|++++private+void+CopyStream(Stream+source,+Stream+target)
++++{
++++++++const+int+bufSize+=+0x1000;
++++++++byte[]+buf+=+new+byte[bufSize];
++++++++int+bytesRead+=+0;
++++++++while+((bytesRead+=+source.Read(buf,+0,+bufSize))+>+0)
++++++++++++target.Write(buf,+0,+bytesRead);
++++}|code-block|syntax|javascript|297165|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Man, this helped me a lot! You rock! I had to use this because my old framework (3.5). Only to complement, see below the implementation of the CopyStream function:

<pre><code> private void CopyStream(Stream source, Stream target)
 {
 const int bufSize = 0x1000;
 byte[] buf = new byte[bufSize];
 int bytesRead = 0;
 while ((bytesRead = source.Read(buf, 0, bufSize)) &gt; 0)
 target.Write(buf, 0, bytesRead);
 }
</code></pre>

blocks|key|297167|text|您也可以将System.IO.Packaging.Package的私有字段_contentTypeHelper设置为null+(使用DynamicObject+)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|297168|entityMap^0|5|R|11|I|1M|4|1U|D|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

You can also set private field <code>_contentTypeHelper</code> of <code>System.IO.Packaging.Package</code> to <code>null</code> ( using <code>DynamicObject</code> ).

blocks|key|3021311|text|在完成zip文件的打包后，运行此函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3021312|public+static+void+Remove_Content_Types_FromZip(string+zipFileName)
++++{
++++++++string+contents;
++++++++using+(ZipFile+zipFile+=+new+ZipFile(File.Open(zipFileName,+FileMode.Open)))
++++++++{
++++++++++++/*
++++++++++++ZipEntry+startPartEntry+=+zipFile.GetEntry("[Content_Types].xml");
++++++++++++using+(StreamReader+reader+=+new+StreamReader(zipFile.GetInputStream(startPartEntry)))
++++++++++++{
++++++++++++++++contents+=+reader.ReadToEnd();
++++++++++++}
++++++++++++XElement+contentTypes+=+XElement.Parse(contents);
++++++++++++XNamespace+xs+=+contentTypes.GetDefaultNamespace();
++++++++++++XElement+newDefExt+=+new+XElement(xs+%2B+"Default",+new+XAttribute("Extension",+"sab"),+new+XAttribute("ContentType",+@"application/binary;+modeler=Acis;+version=18.0.2application/binary;+modeler=Acis;+version=18.0.2"));
++++++++++++contentTypes.Add(newDefExt);
++++++++++++contentTypes.Save("[Content_Types].xml");
++++++++++++zipFile.BeginUpdate();
++++++++++++zipFile.Add("[Content_Types].xml");
++++++++++++zipFile.CommitUpdate();
++++++++++++File.Delete("[Content_Types].xml");
++++++++++++*/
++++++++++++zipFile.BeginUpdate();
++++++++++++try
++++++++++++{
++++++++++++++++zipFile.Delete("[Content_Types].xml");
++++++++++++++++zipFile.CommitUpdate();
++++++++++++}
++++++++++++catch{}
++++++++}
++++}|code-block|syntax|javascript|3021313|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Run this function when you finish packing your zip file:

<pre><code>public static void Remove_Content_Types_FromZip(string zipFileName)
 {
 string contents;
 using (ZipFile zipFile = new ZipFile(File.Open(zipFileName, FileMode.Open)))
 {
 /*
 ZipEntry startPartEntry = zipFile.GetEntry("[Content_Types].xml");
 using (StreamReader reader = new StreamReader(zipFile.GetInputStream(startPartEntry)))
 {
 contents = reader.ReadToEnd();
 }
 XElement contentTypes = XElement.Parse(contents);
 XNamespace xs = contentTypes.GetDefaultNamespace();
 XElement newDefExt = new XElement(xs + "Default", new XAttribute("Extension", "sab"), new XAttribute("ContentType", @"application/binary; modeler=Acis; version=18.0.2application/binary; modeler=Acis; version=18.0.2"));
 contentTypes.Add(newDefExt);
 contentTypes.Save("[Content_Types].xml");
 zipFile.BeginUpdate();
 zipFile.Add("[Content_Types].xml");
 zipFile.CommitUpdate();
 File.Delete("[Content_Types].xml");
 */
 zipFile.BeginUpdate();
 try
 {
 zipFile.Delete("[Content_Types].xml");
 zipFile.CommitUpdate();
 }
 catch{}
 }
 }
</code></pre>

I wish to know if there is any way to avoid to have a [Content_Types].xml file inside the zip file while using .net's ZipPackage class.

how to avoid [Content_Types].xml in .net's ZipPackage class

我想知道在使用.Net的ZipPackage类时，是否有任何方法可以避免在压缩文件中包含Content_Types.xml文件。

问如何在.Net的ZipPackage类中避免[Content_Types].xml
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