创建一个从字节zip文件[] [英] Create zip file from byte[]

问题描述

我想创建一个Zip从一系列字节数组的文件中.NET 4.5（System.IO.Com pression）。作为一个例子，从我使用一个API我结束了一个列表＆LT;附件＆GT; ，每个附件有物业名为机身这是一个字节[] 。我怎么可以遍历该列表，并创建一个包含每个附件的zip文件？

现在我的IM pression，我将不得不写每个附件到磁盘，从创建zip文件下。

  //这是伟大的，如果我对磁盘上的文件
ZipFile.CreateFromDirectory（startPath，zipPath）;
//如何从一系列的字节数组的创造呢？
 

解决方案

在多一点玩耍和阅读我能想出解决办法。这里是你可以创建多个文件的zip文件（存档），而无需编写任何临时数据到磁盘：

 使用（VAR COM pressedFileStream =新的MemoryStream（））{
    //创建一个归档和数据流存储在内存中。
    使用（VAR zipArchive =新的ZipArchive（COM pressedFileStream，ZipArchiveMode.Update，FALSE））{
        的foreach（在caseAttachmentModels VAR caseAttachmentModel）{
            //创建为每个附件的ZIP条目
            VAR的ZipEntry = zipArchive.CreateEntry（caseAttachmentModel.Name）;            //获取附件的流
            使用（VAR originalFileStream =新的MemoryStream（caseAttachmentModel.Body））{
                使用（VAR zipEntryStream = zipEntry.Open（））{
                    //附件流复制到ZIP入口流
                    originalFileStream.CopyTo（zipEntryStream）;
                }
            }
        }    }    返回新FileContentResult（COM pressedFileStream.ToArray（），应用程序/压缩）{FileDownloadName =Filename.zip};
}
 

I am trying to create a Zip file in .NET 4.5 (System.IO.Compression) from a series of byte arrays. As an example, from an API I am using I end up with a List<Attachment> and each Attachment has a property called Body which is a byte[]. How can I iterate over that list and create a zip file that contains each attachment?

Right now I am under the impression that I would have to write each attachment to disk and create the zip file from that.

//This is great if I had the files on disk
ZipFile.CreateFromDirectory(startPath, zipPath);
//How can I create it from a series of byte arrays?

解决方案

After a little more playing around and reading I was able to figure this out. Here is how you can create a zip file (archive) with multiple files without writing any temporary data to disk:

using (var compressedFileStream = new MemoryStream()) {
    //Create an archive and store the stream in memory.
    using (var zipArchive = new ZipArchive(compressedFileStream, ZipArchiveMode.Update, false)) {
        foreach (var caseAttachmentModel in caseAttachmentModels) {
            //Create a zip entry for each attachment
            var zipEntry = zipArchive.CreateEntry(caseAttachmentModel.Name);

            //Get the stream of the attachment
            using (var originalFileStream = new MemoryStream(caseAttachmentModel.Body)) {
                using (var zipEntryStream = zipEntry.Open()) {
                    //Copy the attachment stream to the zip entry stream
                    originalFileStream.CopyTo(zipEntryStream);
                }
            }
        }

    }

    return new FileContentResult(compressedFileStream.ToArray(), "application/zip") { FileDownloadName = "Filename.zip" };
}

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