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2017-10-04 2017-10-04
using variable expansion in strings in powershell (martin) code wrapping for new design (martin)
Line 51: Line 51:
                UserName = "user",                 UserName = "user",
                Password = "mypassword",                 Password = "mypassword",
-                SshHostKeyFingerprint = "ssh-rsa 2048 xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx"+                SshHostKeyFingerprint = "ssh-rsa 2048 xx:xx:xx:xx:xx:xx:xx:xx..."
            };             };
Line 63: Line 63:
                foreach (RemoteFileInfo fileInfo in directory.Files)                 foreach (RemoteFileInfo fileInfo in directory.Files)
                {                 {
-                    Console.WriteLine("{0} with size {1}, permissions {2} and last modification at {3}", +                    Console.WriteLine( 
-                        fileInfo.Name, fileInfo.Length, fileInfo.FilePermissions, fileInfo.LastWriteTime);+························"{0} with size {1}, permissions {2} and last modification at {3}", 
 +                        fileInfo.Name, fileInfo.Length, fileInfo.FilePermissions, 
 + ·······················fileInfo.LastWriteTime);
                }                 }
            }             }
Line 95: Line 97:
                .UserName = "user"                 .UserName = "user"
                .Password = "mypassword"                 .Password = "mypassword"
-                .SshHostKeyFingerprint = "ssh-rsa 2048 xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx"+                .SshHostKeyFingerprint = "ssh-rsa 2048 xx:xx:xx:xx:xx:xx:xx:xx..."
            End With             End With
Line 102: Line 104:
                session.Open(sessionOptions)                 session.Open(sessionOptions)
-                Dim directory As RemoteDirectoryInfo = session.ListDirectory("/home/martin/public_html")+                Dim directory As RemoteDirectoryInfo = 
 + ···················session.ListDirectory("/home/martin/public_html")
                Dim fileInfo As RemoteFileInfo                 Dim fileInfo As RemoteFileInfo
                For Each fileInfo In directory.Files                 For Each fileInfo In directory.Files
-                    Console.WriteLine("{0} with size {1}, permissions {2} and last modification at {3}", _ +                    Console.WriteLine( 
-                        fileInfo.Name, fileInfo.Length, fileInfo.FilePermissions, fileInfo.LastWriteTime)+························"{0} with size {1}, permissions {2} and last modification at {3}", 
 +                        fileInfo.Name, fileInfo.Length, fileInfo.FilePermissions, 
 + ·······················fileInfo.LastWriteTime)
                Next                 Next
            End Using             End Using
Line 188: Line 193:
    sessionOptions.UserName = "user";     sessionOptions.UserName = "user";
    sessionOptions.Password = "mypassword";     sessionOptions.Password = "mypassword";
-    sessionOptions.SshHostKeyFingerprint = "ssh-rsa 2048 xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx:xx";+    sessionOptions.SshHostKeyFingerprint = "ssh-rsa 2048 xx:xx:xx:xx:xx:xx:xx:xx...";
       
    var session = WScript.CreateObject("WinSCP.Session");     var session = WScript.CreateObject("WinSCP.Session");
Line 199: Line 204:
        var directoryInfo = session.ListDirectory("/home/martin/public_html")         var directoryInfo = session.ListDirectory("/home/martin/public_html")
                       
-        for (var enumerator = new Enumerator(directoryInfo.Files); !enumerator.atEnd(); enumerator.moveNext())+        for (var enumerator = new Enumerator(directoryInfo.Files); 
 + ··················!enumerator.atEnd(); enumerator.moveNext())
        {         {
            var fileInfo = enumerator.item();             var fileInfo = enumerator.item();

Last modified: by martin