Question

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文件并设置一个带有XML值的String。
我怎么能这样做？

代码：

//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 ?
    }
}

由于

Answer 1

ZipEntry有一个Extract()重载，它会提取到一个流。 (1)

将这个答案混合到How do you get a string from a MemoryStream?，你会得到类似的东西（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);
        }
    }
}

如何使用DotNetZip从zip中提取XML文件

1 个答案: