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Display or not a button in DataGrid-ButtonColumn

Posted on 2013-06-04
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Last Modified: 2013-06-11
I have a DataGrid with a column as a ButtonColumn. I need to display or not the button on a row-by-row basis based on data in the data binding. How do I do this?
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Question by:allelopath
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6 Comments
 
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Accepted Solution

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Kyle Abrahams earned 500 total points
ID: 39218952
http://stackoverflow.com/questions/6834634/enabling-and-disabling-buttons-in-gridview

change it to a template field.

instead of enabled just set the visible property.
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Author Comment

by:allelopath
ID: 39219572
I have a DataGrid, not GridView. Will this work the same?
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Expert Comment

by:Kyle Abrahams
ID: 39220128
Yep.  Will work exactly the same.
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Author Comment

by:allelopath
ID: 39230284
Working on it. I have the asp setup, but not sure what to do with the btnDelete_DataBinding() method. The DataGrid data source is bound to a System.Collections.Generic.List<MyObjects>, so somehow this gets used in the method:
protected void btnDelete_DataBinding(object sender, System.EventArgs e)
{
    Button btn = (Button)(sender);
    btn.Entable = true;
    if (myObjectList (current element?).ACertainField.toString().Equals("someState")) {
        btn.Entable = false;
    }
}

Open in new window

the if statement above being the critical part. Can you suggest what to do ?
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Author Comment

by:allelopath
ID: 39234791
Also I get this run-time error:
Parser Error Message: System.Web.UI.WebControls.DataGridColumnCollection must have items of type 'System.Web.UI.WebControls.DataGridColumn'. 'asp:TemplateField' is of type 'System.Web.UI.WebControls.TemplateField'.
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Author Comment

by:allelopath
ID: 39239030
It does not work exactly the same. With a DataGrid, one must use <asp:TemplateColumn>
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