(FileUpload)FormView1.FindControl(FileUpload1).FileName

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  • Dave
    #1

    (FileUpload)FormView1.FindControl(FileUpload1).FileName

    If you had a FileUpload control inside of a FormView...how would you
    use FindControl to access the FileUpload properties? Let me just say
    that (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName doesn't
    work.

    The purpose is to insert a graphic in the InsertItemTempl ate section of
    the form. I'm using the expample that's in the online help for the
    post back.

    protected void Page_Load(objec t sender, EventArgs e)
    {
    Profile.UserID = User.Identity.N ame;
    if (IsPostBack)
    {
    Boolean fileOK = false;
    String path = Server.MapPath( "~/images/entry/");
    if
    ((FileUpload)Fo rmView1.Row.Fin dControl(FileUp load1).HasFile)
    {
    String fileExtension =

    System.IO.Path. GetExtension((F ileUpload)FormV iew1.Row.FindCo ntrol(FileUploa d1.FileName)).T oLower();
    String[] allowedExtensio ns =
    { ".gif", ".png", ".jpeg", ".jpg" };
    for (int i = 0; i < allowedExtensio ns.Length; i++)
    {
    if (fileExtension == allowedExtensio ns[i])
    {
    fileOK = true;
    }
    }
    }

    if (fileOK)
    {
    try
    {
    FileUpload1.Pos tedFile.SaveAs( path
    +
    (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName);
    Label1.Text = "File uploaded!";
    }
    catch (Exception ex)
    {
    Label1.Text = "File could not be uploaded.";
    }
    }
    else
    {
    Label1.Text = "Cannot accept files of this type.";
    }
    }
    }


    If there's a better way to do this...I'm open for suggestions. I would
    still like to know the trick behind the FindContol issue though.

    Thanks

    Tags: None
    • Garfilone
      #2
      Re: (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName

      FindControl() method just returns a handle of the control to you,and
      you have to new a same-type instance referring to the control on the
      page.You can modify your code & try

      GOod Luck :-)

      Dave ¼g¹D¡G
      If you had a FileUpload control inside of a FormView...how would you
      use FindControl to access the FileUpload properties? Let me just say
      that (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName doesn't
      work.
      >
      The purpose is to insert a graphic in the InsertItemTempl ate section of
      the form. I'm using the expample that's in the online help for the
      post back.
      >
      protected void Page_Load(objec t sender, EventArgs e)
      {
      Profile.UserID = User.Identity.N ame;
      if (IsPostBack)
      {
      Boolean fileOK = false;
      String path = Server.MapPath( "~/images/entry/");
      if
      ((FileUpload)Fo rmView1.Row.Fin dControl(FileUp load1).HasFile)
      {
      String fileExtension =
      >
      System.IO.Path. GetExtension((F ileUpload)FormV iew1.Row.FindCo ntrol(FileUploa d1.FileName)).T oLower();
      String[] allowedExtensio ns =
      { ".gif", ".png", ".jpeg", ".jpg" };
      for (int i = 0; i < allowedExtensio ns.Length; i++)
      {
      if (fileExtension == allowedExtensio ns[i])
      {
      fileOK = true;
      }
      }
      }
      >
      if (fileOK)
      {
      try
      {
      FileUpload1.Pos tedFile.SaveAs( path
      +
      (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName);
      Label1.Text = "File uploaded!";
      }
      catch (Exception ex)
      {
      Label1.Text = "File could not be uploaded.";
      }
      }
      else
      {
      Label1.Text = "Cannot accept files of this type.";
      }
      }
      }
      >
      >
      If there's a better way to do this...I'm open for suggestions. I would
      still like to know the trick behind the FindContol issue though.

      Thanks

        Comment

        • Dave
          #3
          Re: (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName

          Wow...you totally lost me on that one. :-s



          Garfilone wrote:
          FindControl() method just returns a handle of the control to you,and
          you have to new a same-type instance referring to the control on the
          page.You can modify your code & try
          >
          GOod Luck :-)
          >
          Dave ¼g¹D¡G
          >
          If you had a FileUpload control inside of a FormView...how would you
          use FindControl to access the FileUpload properties? Let me just say
          that (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName doesn't
          work.

          The purpose is to insert a graphic in the InsertItemTempl ate section of
          the form. I'm using the expample that's in the online help for the
          post back.

          protected void Page_Load(objec t sender, EventArgs e)
          {
          Profile.UserID = User.Identity.N ame;
          if (IsPostBack)
          {
          Boolean fileOK = false;
          String path = Server.MapPath( "~/images/entry/");
          if
          ((FileUpload)Fo rmView1.Row.Fin dControl(FileUp load1).HasFile)
          {
          String fileExtension =

          System.IO.Path. GetExtension((F ileUpload)FormV iew1.Row.FindCo ntrol(FileUploa d1.FileName)).T oLower();
          String[] allowedExtensio ns =
          { ".gif", ".png", ".jpeg", ".jpg" };
          for (int i = 0; i < allowedExtensio ns.Length; i++)
          {
          if (fileExtension == allowedExtensio ns[i])
          {
          fileOK = true;
          }
          }
          }

          if (fileOK)
          {
          try
          {
          FileUpload1.Pos tedFile.SaveAs( path
          +
          (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName);
          Label1.Text = "File uploaded!";
          }
          catch (Exception ex)
          {
          Label1.Text = "File could not be uploaded.";
          }
          }
          else
          {
          Label1.Text = "Cannot accept files of this type.";
          }
          }
          }


          If there's a better way to do this...I'm open for suggestions. I would
          still like to know the trick behind the FindContol issue though.

          Thanks

            Comment

            • Dave
              #4
              Re: (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName

              Wow...you totally lost me on that one. :-s



              Garfilone wrote:
              FindControl() method just returns a handle of the control to you,and
              you have to new a same-type instance referring to the control on the
              page.You can modify your code & try
              >
              GOod Luck :-)
              >
              Dave ¼g¹D¡G
              >
              If you had a FileUpload control inside of a FormView...how would you
              use FindControl to access the FileUpload properties? Let me just say
              that (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName doesn't
              work.

              The purpose is to insert a graphic in the InsertItemTempl ate section of
              the form. I'm using the expample that's in the online help for the
              post back.

              protected void Page_Load(objec t sender, EventArgs e)
              {
              Profile.UserID = User.Identity.N ame;
              if (IsPostBack)
              {
              Boolean fileOK = false;
              String path = Server.MapPath( "~/images/entry/");
              if
              ((FileUpload)Fo rmView1.Row.Fin dControl(FileUp load1).HasFile)
              {
              String fileExtension =

              System.IO.Path. GetExtension((F ileUpload)FormV iew1.Row.FindCo ntrol(FileUploa d1.FileName)).T oLower();
              String[] allowedExtensio ns =
              { ".gif", ".png", ".jpeg", ".jpg" };
              for (int i = 0; i < allowedExtensio ns.Length; i++)
              {
              if (fileExtension == allowedExtensio ns[i])
              {
              fileOK = true;
              }
              }
              }

              if (fileOK)
              {
              try
              {
              FileUpload1.Pos tedFile.SaveAs( path
              +
              (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName);
              Label1.Text = "File uploaded!";
              }
              catch (Exception ex)
              {
              Label1.Text = "File could not be uploaded.";
              }
              }
              else
              {
              Label1.Text = "Cannot accept files of this type.";
              }
              }
              }


              If there's a better way to do this...I'm open for suggestions. I would
              still like to know the trick behind the FindContol issue though.

              Thanks

                Comment

                • treefort.jesse@gmail.com
                  #5
                  Re: (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName

                  I am in the middle of the exact same situation. did you find any
                  resolution here?
                  --
                  Jesse Gavin



                  On Nov 19, 6:13 am, "Dave" <KillnComput... @Verizon.Netwro te:
                  Wow...you totally lost me on that one. :-s
                  >
                  Garfilone wrote:
                  FindControl() method just returns a handle of the control to you,and
                  you have to new a same-type instance referring to the control on the
                  page.You can modify your code & try
                  >
                  GOod Luck :-)
                  >
                  Dave 寫道：
                  >
                  If you had a FileUpload control inside of a FormView...how would you
                  use FindControl to access the FileUpload properties? Let me just say
                  that (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName doesn't
                  work.
                  >
                  The purpose is to insert a graphic in the InsertItemTempl ate section of
                  the form. I'm using the expample that's in the online help for the
                  post back.
                  >
                  protected void Page_Load(objec t sender, EventArgs e)
                  {
                  Profile.UserID = User.Identity.N ame;
                  if (IsPostBack)
                  {
                  Boolean fileOK = false;
                  String path = Server.MapPath( "~/images/entry/");
                  if
                  ((FileUpload)Fo rmView1.Row.Fin dControl(FileUp load1).HasFile)
                  {
                  String fileExtension =
                  >
                  System.IO.Path. GetExtension((F ileUpload)FormV iew1.Row.FindCo ntrol(FileUploa d1.FileName)).T oLower();
                  String[] allowedExtensio ns =
                  { ".gif", ".png", ".jpeg", ".jpg" };
                  for (int i = 0; i < allowedExtensio ns.Length; i++)
                  {
                  if (fileExtension == allowedExtensio ns[i])
                  {
                  fileOK = true;
                  }
                  }
                  }
                  >
                  if (fileOK)
                  {
                  try
                  {
                  FileUpload1.Pos tedFile.SaveAs( path
                  +
                  (FileUpload)For mView1.FindCont rol(FileUpload1 ).FileName);
                  Label1.Text = "File uploaded!";
                  }
                  catch (Exception ex)
                  {
                  Label1.Text = "File could not be uploaded.";
                  }
                  }
                  else
                  {
                  Label1.Text = "Cannot accept files of this type..";
                  }
                  }
                  }
                  >
                  If there's a better way to do this...I'm open for suggestions. I would
                  still like to know the trick behind the FindContol issue though.
                  Thanks

                    Comment

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