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Newbie here

Scheduled Pinned Locked Moved Visual Basic
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  • T thealca

    Dim dlg As New OpenFileDialog() dlg.ShowDialog()

    N Offline
    N Offline
    Nguyen Dinh Quy
    wrote on last edited by
    #3

    So how do i code it to copy the chosen file to another destination?

    M D 2 Replies Last reply
    0
    • N Nguyen Dinh Quy

      So how do i code it to copy the chosen file to another destination?

      M Offline
      M Offline
      Marc 0
      wrote on last edited by
      #4

      IO.File.Copy(sourcefile, desinationfile) so i guess i would be something like:

      Dim dlg As New OpenFileDialog()
      If dlg.ShowDialog() = DialogResult.OK
      IO.File.Copy(dlg.FileName, desinationfile)
      End If


      1 Reply Last reply
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      • N Nguyen Dinh Quy

        So how do i code it to copy the chosen file to another destination?

        D Offline
        D Offline
        Daniel1324
        wrote on last edited by
        #5

        Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\") EndIf ' Will copy the chosen file to C:\

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        • D Daniel1324

          Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\") EndIf ' Will copy the chosen file to C:\

          N Offline
          N Offline
          Nguyen Dinh Quy
          wrote on last edited by
          #6

          Thanks guys..another question.. Can i change the filename while copying the file to the new location? So that it will be renamed when it is copied to the new location?

          T 1 Reply Last reply
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          • N Nguyen Dinh Quy

            Thanks guys..another question.. Can i change the filename while copying the file to the new location? So that it will be renamed when it is copied to the new location?

            T Offline
            T Offline
            toxcct
            wrote on last edited by
            #7

            see the MSDN[^]...


            TOXCCT >>> GEII power
            [toxcct][VisualCalc]

            N 1 Reply Last reply
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            • T toxcct

              see the MSDN[^]...


              TOXCCT >>> GEII power
              [toxcct][VisualCalc]

              N Offline
              N Offline
              Nguyen Dinh Quy
              wrote on last edited by
              #8

              Hmm..but it doesn't teach us how to rename a file? Sorry but i'm really weak in Vb.Net >.<

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              • N Nguyen Dinh Quy

                Hmm..but it doesn't teach us how to rename a file? Sorry but i'm really weak in Vb.Net >.<

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                Yulianto
                wrote on last edited by
                #9

                Copy the code and run it.


                Work hard, Work effectively.

                N 1 Reply Last reply
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                • Y Yulianto

                  Copy the code and run it.


                  Work hard, Work effectively.

                  N Offline
                  N Offline
                  Nguyen Dinh Quy
                  wrote on last edited by
                  #10

                  Private Sub MenuItem16_Click(ByVal sender as System.Object,ByVal e as EventArgs)Handles MenuItem16.Click Dim FileDialog As OpenFileDialog Dim Test As New Test 'The Test class from MSDN if FileDialog.ShowDialog() = DialogResults.Ok then Test.Main() End If There's no build error, but after I chose a file from the dialog box, no file was copied to the new location :( What's my problem here? Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\") EndIf ' Will copy the chosen file to C:\ I tried this as well, but after clicking OK in the dialog box, it gives an error indicating "Directory Exists", so i couldn't copy my chosen file to C:\ as well...why?

                  Y 1 Reply Last reply
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                  • N Nguyen Dinh Quy

                    Private Sub MenuItem16_Click(ByVal sender as System.Object,ByVal e as EventArgs)Handles MenuItem16.Click Dim FileDialog As OpenFileDialog Dim Test As New Test 'The Test class from MSDN if FileDialog.ShowDialog() = DialogResults.Ok then Test.Main() End If There's no build error, but after I chose a file from the dialog box, no file was copied to the new location :( What's my problem here? Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\") EndIf ' Will copy the chosen file to C:\ I tried this as well, but after clicking OK in the dialog box, it gives an error indicating "Directory Exists", so i couldn't copy my chosen file to C:\ as well...why?

                    Y Offline
                    Y Offline
                    Yulianto
                    wrote on last edited by
                    #11

                    Nguyen Dinh Quy wrote: Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\") EndIf ' Will copy the chosen file to C:\ Try Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\a.txt") EndIf


                    Work hard, Work effectively.

                    N 1 Reply Last reply
                    0
                    • Y Yulianto

                      Nguyen Dinh Quy wrote: Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\") EndIf ' Will copy the chosen file to C:\ Try Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\a.txt") EndIf


                      Work hard, Work effectively.

                      N Offline
                      N Offline
                      Nguyen Dinh Quy
                      wrote on last edited by
                      #12

                      Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\a.txt") EndIf Yah, this will create a text file in C:\...but what if the user chooses other types of files? (.exe, .pdf) I tried choosing an .exe file, but it creates a .txt file, using notepad to read the .exe file. :( Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\" + dlg.FileName) EndIf I tried this, but it gives an error indicating "The given path's format is not supported".

                      Y 1 Reply Last reply
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                      • N Nguyen Dinh Quy

                        Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\a.txt") EndIf Yah, this will create a text file in C:\...but what if the user chooses other types of files? (.exe, .pdf) I tried choosing an .exe file, but it creates a .txt file, using notepad to read the .exe file. :( Dim dlg as New OpenFileDialog if dlg.ShowDialog() = DialogResult.Ok then System.IO.File.Copy(dlg.FileName, "C:\" + dlg.FileName) EndIf I tried this, but it gives an error indicating "The given path's format is not supported".

                        Y Offline
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                        Yulianto
                        wrote on last edited by
                        #13

                        Nguyen Dinh Quy wrote: Yah, this will create a text file in C:\...but what if the user chooses other types of files? (.exe, .pdf) I tried choosing an .exe file, but it creates a .txt file, using notepad to read the .exe file. Do you mean you want do copy file dynamically, where the user can choose which file to be copied and where? :confused: If the file you want to copy is .exe then the destination file must also .exe so System.IO.File.Copy(dlg.FileName, "C:\a.exe") Nguyen Dinh Quy wrote: ystem.IO.File.Copy(dlg.FileName, "C:\" + dlg.FileName) if dlg.FileName="c:\a.txt" then the destination file you use "C:\" + dlg.FileName, then it means that you're copying file to "C:\c:\a.txt".


                        Work hard, Work effectively.

                        N 1 Reply Last reply
                        0
                        • Y Yulianto

                          Nguyen Dinh Quy wrote: Yah, this will create a text file in C:\...but what if the user chooses other types of files? (.exe, .pdf) I tried choosing an .exe file, but it creates a .txt file, using notepad to read the .exe file. Do you mean you want do copy file dynamically, where the user can choose which file to be copied and where? :confused: If the file you want to copy is .exe then the destination file must also .exe so System.IO.File.Copy(dlg.FileName, "C:\a.exe") Nguyen Dinh Quy wrote: ystem.IO.File.Copy(dlg.FileName, "C:\" + dlg.FileName) if dlg.FileName="c:\a.txt" then the destination file you use "C:\" + dlg.FileName, then it means that you're copying file to "C:\c:\a.txt".


                          Work hard, Work effectively.

                          N Offline
                          N Offline
                          Nguyen Dinh Quy
                          wrote on last edited by
                          #14

                          Ahh i guess my first post was horribly phased ;P Anyway yah, users can choose from the dialog box which files to be copied, but the location where the files are copied to is fixed. With "Files", i mean any types of files.. e.g. User chooses Example.txt from C:\, and it is copied to D:\.So now there's C:\Example.txt and D:\Example.txt. The 1st question here is that i don't know how to do it.

                          A 1 Reply Last reply
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                          • N Nguyen Dinh Quy

                            Ahh i guess my first post was horribly phased ;P Anyway yah, users can choose from the dialog box which files to be copied, but the location where the files are copied to is fixed. With "Files", i mean any types of files.. e.g. User chooses Example.txt from C:\, and it is copied to D:\.So now there's C:\Example.txt and D:\Example.txt. The 1st question here is that i don't know how to do it.

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                            Anonymous
                            wrote on last edited by
                            #15

                            The following code will handle copying of any file extension. Dim FileDialog As New OpenFileDialog 'FileInfo class has Dim myFile as System.IO.FileInfo With FileDialog if .ShowDialog() = DialogResults.Ok then 'create FileInfo instance where .Filename is the fullpath of source 'file. For example: C:\Example.txt myFile = new System.IO.FileInfo(.Filename) 'FileInfo.Name property indicates the file name 'For example: If the source file is C:\Example.txt than the FileInfo.Name 'property is Example.txt myFile.CopyTo("D:\" & myFile.Name, false) 'syntax: FileInfo.CopyTo(destfilename as String, forceoverwrite as Boolean) End If End With I hope this makes sense :-D A.S.

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