Make a Button to visible using DataGridView's CheckBox to check

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I have one problem
How i make a button visible or Un Visible using the DataGridView's Checkbox....
Please any one help me......?

i used a v s 2008 (windows Form)

Posted 7-Dec-10 23:59pm

Neeil

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Manoj from Pune · Answer 1 · 2010-12-08T00:03:00

Solution 1

On checkedchanged event of that checkbox write code like
buttonname.visible = true -- To make button visible
buttonname.visible = false -- TO make button invisible

Posted 8-Dec-10 0:03am

Manoj from Pune

Comments

Neeil 8-Dec-10 6:17am

@ manoj from pune......ok , but how can i write that is useful hen i make checkd a checkbox of datagridview......and imlies it a button make visible

Manoj from Pune 8-Dec-10 6:37am

Please be claer.. M not getting nything from ur commment

Neeil 8-Dec-10 6:49am

when i change a checkbox to check in the datagridview.....the button is visible other-wise unvisible.......Can u give me code...

Manoj from Pune 8-Dec-10 7:01am

Private Sub checkbox_CheckedChanged(ByVal sender As Object, ByVal e As System.EventArgs) Handles checkbox.CheckedChanged
if checkbox.checked = true then
button.visible = true
else if checkbox.checked = false then
button.visible = false
End Sub

vivekse · Answer 2 · 2010-12-08T00:04:00

Hi Neeil,

You can set visible a column if button in the grid.

dgvTest.Columns("Column").Visible = True

alternate if you want to visible a button on form then,

on cell content click event of grid code like this . I am giving only steps.

1. check e.colimnindex and e. rowindex greater than -1.
2. check if it is desired column or not.
3. then check check box is checked or not.

If e.RowIndex <> -1 Then
If e.ColumnIndex <> -1 Then
If dgvTest.Columns(e.ColumnIndex).Name = "testcolumn" Then
if Convert.ToBoolen(dgvTest(e.RowIndex)(e.ColumnIndex).Value) = Ture Then
btntest.visible= True
Else
btntest.visible= False

EndIF
EndIF
EndIF