C#似乎无法正确添加图像到ziparchive
问题描述:
我试图将一堆文件添加到C#中的zipfile,但它似乎不能正常工作。C#似乎无法正确添加图像到ziparchive
using (var memoryStream = new MemoryStream())
{
using (var zip = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
for (int i = 0; i < kaart_data.GetLength(0); i++)
{
Image img = array[i];
var file = zip.CreateEntry(i + ".bmp");
using (var stream = new MemoryStream())
{
img.Save(stream, ImageFormat.Bmp);
using (var entryStream = file.Open())
{
stream.CopyTo(entryStream);
}
}
}
}
//saves the archive to disk
using (var fileStream = new FileStream(@"C:\Temp\test.zip", FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
事情是它创建并保存压缩文件到磁盘与预期的文件大小。
但是,当我尝试在Windows照片查看器中打开它们时,它们似乎已损坏。
任何帮助表示赞赏。
答
几乎花了3个小时才能找出问题所在。如果你看一下原始图像的大小并提取一个,则会有一点点差异。
using (var memoryStream = new MemoryStream())
{
using (var zip = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
for (var i = 0; i < images.Length; i++)
{
var img = images[i];
var file = zip.CreateEntry(i + ".bmp");
using (var stream = new MemoryStream())
{
img.Save(stream, ImageFormat.Bmp);
using (var entryStream = file.Open())
{
var bytes = stream.ToArray(); -- to keep it as image better to have it as bytes
entryStream.Write(bytes, 0, bytes.Length);
}
}
}
}
using (var fileStream = new FileStream(@"test.zip", FileMode.Create))
{
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
我已经尝试过它像魅力一样工作!
哇,不知道为什么把流转换成bytearray有帮助,但它完全有效。谢啦! –
@john_vanderholt https://stackoverflow.com/questions/46880345/adding-correctly-images-to-ziparchive显然没人知道。如果你找到答案,你可以高兴或回答 –