[英]How to Use DotNetZip to extract XML file from zip
我正在使用最新版本的DotNetZip,我有一個包含5個XML的zip文件。
我想打開zip,讀取XML文件並使用XML的值設置String。
我怎樣才能做到這一點?
碼:
//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 ?
}
}
謝謝
ZipEntry
有一個Extract()
重載,它提取到一個流。 (1)
在這個答案中混合如何從MemoryStream中獲取字符串? ,你會得到這樣的東西(完全未經測試):
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);
}
}
}
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