You are very welcome, pal! :D

1:Add textbox items in dataTable 2:Save dataTable in Session 3:Set GridView datasource as dataTable from Session check this sample code protected void Button1_Click(object sender, EventArgs e) { DataTable dt; DataRow drow; if (Session["datatable"] == null) { dt = new DataTable(); DataColumn dcol = new DataColumn("Column1", typeof(string)); //adding column to datatable dt.Columns.Add(dcol); int rowcount = 0; drow = dt.NewRow(); dt.Rows.Add(drow); dt.Rows\[rowcount\]\[dcol\] = TextBox1.Text; //add datatable to session Session.Add("datatable", dt); } else { dt = (DataTable)Session\["datatable"\]; drow = dt.NewRow(); dt.Rows.Add(drow); dt.Rows\[dt.Rows.Count-1\]\["Column1"\] = TextBox1.Text; //add datatable to session Session.Add("datatable", dt); } //bind gridview GridView1.DataSource = dt; GridView1.DataBind(); } makhaai modified on Thursday, June 10, 2010 5:06 PM

Here's how I would do this: 1. Make a binary tree. Leaf nodes stand for your variables (A, B, f, g...). Interior nodes stand for an operation on the values of the interior node's two sons. 2. Go through your statement and put all tokens into a list of leaf nodes. (The operations will eventually be transferred to interior nodes.) 3. Find the range of nodes where the parentheses are most deeply nested. 4. If you have the pattern [left parenthesis] [variable] [right parenthesis], delete the two parentheses nodes. (The variable could also be an interior node.) 5. In this deeply-nested region, replace the pattern [variable] [high-precedence-operation] [variable] with an interior node for the operation, with the two variables as sons. (The variables could also be other interior nodes.) 6. Repeat Step 5 for the low-precedence operations. You should now be able to apply Step 4 on the remaining interior node left in the parentheses. 7. To evaluate the expression, do a post-order traversal on the binary tree, getting the variable values for leaf nodes, and applying the operation to the two values for interior nodes. The result of the evaluation will be left in your root node. I find this approach simpler than the recursive algorithms. modified on Friday, March 27, 2009 10:10 AM