DataGrid Binary Column

[Art]

Hi,

I need some advice with a binary column in a DataGrid. Everything seems
fine except for one thing. In order to check a box, I have to click on it
once to select it, and then click again to check it. Somehow this doesn't
seem right. Does anyone know if this is normal behavior for a DataGrid, of if
there's something I need to know?

thanks,

Art

[Cor Ligthert [MVP]]

Re: DataGrid Binary Column

Art,

I assume that you mean a boolean column. I did test it and had not your
behaviour. Can you tell tell us how you made that column

I hope this helps something,

Cor

[Art]

Re: DataGrid Binary Column

Cor,

Yes, sorry -- it is a boolean column:

I have a GridStyle:

Public ReadOnly Property Analysis_Select ion_Style() As DataGridTableSt yle
Get
Dim TblStyle As New DataGridTableSt yle
With TblStyle
With .GridColumnStyl es
Dim High As New DataGridBoolCol umn
High.MappingNam e = "High"
High.HeaderText = "High"
High.Width = 50
High.AllowNull = False
High.ReadOnly = False
Dim Low As New DataGridBoolCol umn
Low.MappingName = "Low"
Low.HeaderText = "Low"
Low.Width = 50
Low.AllowNull = False
Low.ReadOnly = False


.Add(TextBoxCol umn(ColumnForma t.Dec0, "ID", 50, True))
.Add(High)
.Add(Low)
.Add(TextBoxCol umn(ColumnForma t.Text, "Name", 120, True))
.Add(TextBoxCol umn(ColumnForma t.DateTime, "RunDate", 60,
True))
.Add(TextBoxCol umn(ColumnForma t.Text, "Peril", 60, True))
.Add(BooleanCol umn("IsGroup", 60, True))
.Add(TextBoxCol umn(ColumnForma t.Text, "CEDANTID", 40,
True))
.Add(TextBoxCol umn(ColumnForma t.Text, "USER1", 60, True))
.Add(TextBoxCol umn(ColumnForma t.Text, "USER2", 60, True))
.Add(TextBoxCol umn(ColumnForma t.Text, "USER3", 60, True))
.Add(TextBoxCol umn(ColumnForma t.Text, "USER4", 60, True))
End With
.AlternatingBac kColor = Color.AliceBlue
End With
Return TblStyle
End Get
End Property


Private ReadOnly Property BooleanColumn(B yVal mName As String, _
ByVal mWidth As Int32, ByVal mReadOnly As Boolean) As DataGridBoolCol umn
Get
Dim x As New DataGridBoolCol umn
x.MappingName = mName
x.ReadOnly = mReadOnly
x.HeaderText = mName
x.Width = mWidth
x.Alignment = HorizontalAlign ment.Center
Return x
End Get
End Property

Then, when the form containing the grid is loaded:
Private Sub AddAnalyses_For m_Load(ByVal sender As Object, _
ByVal e As System.EventArg s) Handles MyBase.Load
Dim mGridStyles As MyGridStyles
mGridStyles = mGridStyles.Ins tantiate()

Dim mQuery As New d.mdb_Query("_0 1_06_RefreshTre atySelections")
mQuery.RunQuery ()
mQuery = Nothing

mTreatyTable = New d.MyDataTable
mTreatyTable.So urceDB = g.ProjectDB
mTreatyTable.So urceTable = "dbo.mk_TreatyS elections"
Me.SetUpTable(m TreatyTable)

Me.cAvailableTr eatyGrid.TableS tyles.Add(mGrid Styles.Analysis _Selection_Styl e)
Me.cAvailableTr eatyGrid.DataSo urce = mTreatyTable

I think that's about it. While you're at it (should you decide to read
through all of my stuff -- which I again appreciate: I would like to capture
data entry in a text box (another grid). Basically I want to divide by 100
when the user makes an entry since he will be entering a percentage. Can I
do this?

Thank you very much for helping

Art

"Cor Ligthert [MVP]" wrote:
[color=blue]
> Art,
>
> I assume that you mean a boolean column. I did test it and had not your
> behaviour. Can you tell tell us how you made that column
>
> I hope this helps something,
>
> Cor
>
>
>[/color]

[Cor Ligthert [MVP]]

Re: DataGrid Binary Column

Art,

It is difficult to read, however if you need a column divided by 100 than
why not use an expression.

http://www.vb-tips.com/default.aspx?ID=76a81eb8-ea2d-48f4-99c3-a3539697edbd

Or maybe this one

http://www.vb-tips.com/default.aspx?ID=30d9d2fd-9f10-4928-b7c8-1def3152436f

I did not use styles with this, however that can of course as well.

Cor