如何使用DotNetZip從zip中提取XML文件

Question

我使用的是最新版本的DotNetZip，並且我有一個帶有5個XML的zip文件。
我想打開zip文件，讀取XML文件，並使用XML的值設置String。
我該怎麼做？

代碼：

//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 ? 
    } 
} 

由於

Answer 1

ZipEntry具有Extract()過載，其提取到流。 (1)

這個答案How do you get a string from a MemoryStream?混合，你會得到這樣的事情（經過充分測試）：

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); 
     } 
    } 
}