zip extractFile(String file, Stream Stream)方法流参数.c#

本文关键字：Stream 方法 参数 extractFile String file zip | 更新日期: 2023-09-27 18:19:12

我正在尝试访问。7z文件中的文件。我知道zip文件夹中的文件名，也知道它存在于。7z文件中。以前我使用了ExtractArchive(模板)，它只是将所有文件转储到一个临时位置。现在我希望能够抓取一个特定的文件，而不需要提取整个。7z文件。

7Zip有一个叫做SevenZipExtractor的类，它有一个ExtractFile方法。我想这就是我要找的，但我找不到任何像样的文件。

我需要澄清的是如何正确地获得传入的流参数。我正在使用这样的代码;

//this grabs the zip file and creates a FileInfo array that hold the .7z file (assume there is only one) 
DirectoryInfo directoryInfo = new DirectoryInfo(ApplicationPath);
FileInfo[] zipFile = directoryInfo.GetFiles("*.7z");
//This creates the zipextractor on the zip file I just placed in the zipFile FileInfo array
using (SevenZip.SevenZipExtractor zipExtractor = new SevenZip.SevenZipExtractor(zipFile[0].FullName)) 
//Here I should be able to use the ExtractFile method, however I don't understand the stream parameter, and I can't find any good documentation on the method itself. What is this method looking for?
{
    zipExtractor.ExtractFile("ConfigurationStore.xml", Stream stream);
}

zip extractFile(String file, Stream Stream)方法流参数.c#

设置一个FileStream，使SevenZip可以写入:

DirectoryInfo directoryInfo = new DirectoryInfo(ApplicationPath);
FileInfo[] zipFile = directoryInfo.GetFiles("*.7z");
using (SevenZip.SevenZipExtractor zipExtractor = new SevenZip.SevenZipExtractor(zipFile[0].FullName)) 
{
    using (FileStream fs = new FileStream("", FileMode.Create))  //replace empty string with desired destination
    {
        zipExtractor.ExtractFile("ConfigurationStore.xml", fs);
    }
}
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