Csharp 无效的Zip文件Zip存档

发布于11月23日

我正在try 从byte[]创建一个Zip文件，并将Zip文件作为byte[]返回，但是当我将字节上传到BLOB存储时，当我下载该Zip文件时，它是无效的.

代码:

byte[] zipBytes;
using (var zipMemoryStream = new MemoryStream())
{
    using (var archive = new ZipArchive(zipMemoryStream, ZipArchiveMode.Create, true))
    {
        var zipExcelEntry = archive.CreateEntry(
           $"SAR-{subjectAccessRequest.Basic.FirstName}-{subjectAccessRequest.Basic.LastName}_{DateTime.Now:yyyyMMddHHmmss}.xlsx",
           CompressionLevel.Fastest);

        using (var excelEntryFile = zipExcelEntry.Open()) excelEntryFile.Write(stream.ToArray(), 0, (int)stream.Length);

        zipBytes = zipMemoryStream.ToArray();

    };
};

return zipBytes;

在创建zip文件时，我在这里做了什么明显的错误吗？

上传到BLOB存储的代码已经过测试，所以问题应该出在这里

推荐答案

您需要在ToArray流之前关闭存档.

您还应该在内部文件流上使用CopyTo.

using var zipMemoryStream = new MemoryStream();

using (var archive = new ZipArchive(zipMemoryStream, ZipArchiveMode.Create, true))
{
    var zipExcelEntry = archive.CreateEntry(
           $"SAR-{subjectAccessRequest.Basic.FirstName}-{subjectAccessRequest.Basic.LastName}_{DateTime.Now:yyyyMMddHHmmss}.xlsx",
           CompressionLevel.Fastest);

    using (var excelEntryFile = zipExcelEntry.Open())
        stream.CopyTo(excelEntryFile);
};
var zipBytes = zipMemoryStream.ToArray();
return zipBytes;

理想情况下，您只需返回整个压缩流，而不是创建新的字节数组，因为这样效率更高.

var zipMemoryStream = new MemoryStream();

using (var archive = new ZipArchive(zipMemoryStream, ZipArchiveMode.Create, true))
{
    var zipExcelEntry = archive.CreateEntry(
           $"SAR-{subjectAccessRequest.Basic.FirstName}-{subjectAccessRequest.Basic.LastName}_{DateTime.Now:yyyyMMddHHmmss}.xlsx",
           CompressionLevel.Fastest);

    using (var excelEntryFile = zipExcelEntry.Open())
        stream.CopyTo(excelEntryFile);
};

zipMemoryStream.Position = 0;  // reset position to beginning
return zipMemoryStream;