OpenInputFileStream() public method

public OpenInputFileStream ( string path, FileMode mode, FileAccess access, FileShare share, int bufferSize ) : Stream
path string
mode FileMode
access FileAccess
share FileShare
bufferSize int
return Stream
ZipArchiveDirectory Class Documentation
Ejemplo n.º 1
0
 public void OpenInputFileStreamShouldProvideStreamForRequestedFileInRoot()
 {
     //given
     var dir = new ZipArchiveDirectory("Fresh.zip");
     //when
     string result = null;
     using(var reader = new StreamReader(dir.OpenInputFileStream("file1.txt",FileMode.Open,FileAccess.Read,FileShare.Read,0)))
     {
         result = reader.ReadToEnd();
     }
     //then
     Assert.AreEqual("Well hello there little fellah!", result);
 }