如何在不将它们放入内存的情况下返回大型文件的集合?

时间:2018-03-14 19:00:45

标签: c# .net-core

我需要编写一个API来检索文件流的集合,然后返回文件。这些文件可能会变大(~1GB)。我需要尽可能快,但这项服务不会消耗太多内存。

过去当我不得不做这样的事情时,文件并不大,所以我只是在内存中创建了一个ZIP并返回了。由于内存限制,我此次无法做到这一点。据我所知,多部分回复不存在,所以我也做不到。我有什么选择?有什么方法可以将拉链回流作为回复吗?

public async Task GetFiles(string someId)
{
    List<Stream> streamList = GetStreams(someId);
    using (ZipArchive archive = new ZipArchive(responseStream /* ?? */, ZipArchiveMode.Create, true))
    {
        ...
    }
}

1 个答案:

答案 0 :(得分:0)

您可以尝试使用gzipsteam,这可以避免将文件加载到内存中。

https://msdn.microsoft.com/en-us/library/system.io.compression.gzipstream(v=vs.110).aspx

以下是该页面的示例:

using System;
using System.IO;
using System.IO.Compression;

namespace zip
{
    public class Program
    {
        private static string directoryPath = @"c:\temp";
        public static void Main()
        {
            DirectoryInfo directorySelected = new DirectoryInfo(directoryPath);
            Compress(directorySelected);

            foreach (FileInfo fileToDecompress in directorySelected.GetFiles("*.gz"))
            {
                Decompress(fileToDecompress);
            }
        }

        public static void Compress(DirectoryInfo directorySelected)
        {
            foreach (FileInfo fileToCompress in directorySelected.GetFiles())
            {
                using (FileStream originalFileStream = fileToCompress.OpenRead())
                {
                    if ((File.GetAttributes(fileToCompress.FullName) & 
                    FileAttributes.Hidden) != FileAttributes.Hidden & fileToCompress.Extension != ".gz")
                    {
                        using (FileStream compressedFileStream = File.Create(fileToCompress.FullName + ".gz"))
                        {
                            using (GZipStream compressionStream = new GZipStream(compressedFileStream, 
                            CompressionMode.Compress))
                            {
                                originalFileStream.CopyTo(compressionStream);

                            }
                        }
                        FileInfo info = new FileInfo(directoryPath + "\\" + fileToCompress.Name + ".gz");
                        Console.WriteLine("Compressed {0} from {1} to {2} bytes.",
                        fileToCompress.Name, fileToCompress.Length.ToString(), info.Length.ToString());
                    }

                }
            }
        }

        public static void Decompress(FileInfo fileToDecompress)
        {
            using (FileStream originalFileStream = fileToDecompress.OpenRead())
            {
                string currentFileName = fileToDecompress.FullName;
                string newFileName = currentFileName.Remove(currentFileName.Length - fileToDecompress.Extension.Length);

                using (FileStream decompressedFileStream = File.Create(newFileName))
                {
                    using (GZipStream decompressionStream = new GZipStream(originalFileStream, CompressionMode.Decompress))
                    {
                        decompressionStream.CopyTo(decompressedFileStream);
                        Console.WriteLine("Decompressed: {0}", fileToDecompress.Name);
                    }
                }
            }
        }
    }
}