[英]How to Use DotNetZip to extract XML file from zip
我正在使用最新版本的DotNetZip,我有一个包含5个XML的zip文件。
我想打开zip,读取XML文件并使用XML的值设置String。
我怎样才能做到这一点?
码:
//thats my old way of doing it.But I needed the path, now I want to read from the memory
string xfile = System.IO.File.ReadAllText(strNewFilePath, System.Text.Encoding.Default);
using (ZipFile zip = ZipFile.Read(this.uplZip.PostedFile.InputStream))
{
foreach (ZipEntry theEntry in zip)
{
//What should I use here, Extract ?
}
}
谢谢
ZipEntry
有一个Extract()
重载,它提取到一个流。 (1)
在这个答案中混合如何从MemoryStream中获取字符串? ,你会得到这样的东西(完全未经测试):
string xfile = System.IO.File.ReadAllText(strNewFilePath, System.Text.Encoding.Default);
List<string> xmlContents;
using (ZipFile zip = ZipFile.Read(this.uplZip.PostedFile.InputStream))
{
foreach (ZipEntry theEntry in zip)
{
using (var ms = new MemoryStream())
{
theEntry.Extract(ms);
// The StreamReader will read from the current
// position of the MemoryStream which is currently
// set at the end of the string we just wrote to it.
// We need to set the position to 0 in order to read
// from the beginning.
ms.Position = 0;
var sr = new StreamReader(ms);
var myStr = sr.ReadToEnd();
xmlContents.Add(myStr);
}
}
}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.