How to show the existing mdi chid form

hi
I have created a form and added a menu item and set the ismdiparent property to true i have also two other forms which i call using the menu in the mdi form
i use the following code to show the child form

in the menu click event
dim frm as new frmexample
frm.show

the problem is when ever i clicked the menu button new instance of the form is created and displayed i dont want this i want the same form to be shown again. say if opened two diff child forms when i clicked menu button of the existing form which is currently not in focus should come in front how to do this

LUV
PurpleHaze

Comments

  • Add a new menu item called something like "Window".
    For this menuitem set MdiList property to true.
    Once this is done ,at run time clicking the "Window" item will
    show a list of all open child windows using the Text property of the child window.
    For example,assume that one child window has text property of Child1 ,and another has text property of child3.And both the forms are loaded then clicking the 'Window menu' at runtime will show a list containing Child1,Child2.Then of course you can
    click Child1 or child2 ,to go the respective form.

    Also you can add menuitems like Cascade,TileHorizontal,TileVertical,ArrangeIcons to this 'Window' menuitem.
    In the click event of Cascade write LayoutMdi(MdiLayout.Cascade).This will cascade the open child windows.
    Similarly you can use LayoutMdi(MdiLayout.TileHorizontal),LayoutMdi(MdiLayout.TileVertical),LayoutMdi(MdiLayout.ArrangeIcons).
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