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PODataTable: Get Max Value Using LINQ With Criteria Field (GroupBy)
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  1. PODataTable: Get Max Value Using LINQ With Criteria Field (GroupBy)
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    copied!<p>I have a DataTable structured like this: </p> <blockquote> <p>username | price </p> <p>"jack"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.01</p> <p>"jack"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.02</p> <p>"mary"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.03</p> <p>"mary"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.04</p> </blockquote> <p>How can I use LINQ on the DataTable to return the MAX PRICE FOR JACK (ex: .02)? </p> <p>Code to setup the table (can ignore). </p> <pre><code>DataTable tbl = new DataTable(); tbl.Columns.Add("username"); tbl.Columns["username"].DataType = typeof(string); tbl.Columns.Add("price"); tbl.Columns["price"].DataType = typeof(double); DataRow r = tbl.NewRow(); r["username"] = "jack"; r["price"] = .01; tbl.Rows.Add(r); r = tbl.NewRow(); r["username"] = "jack"; r["price"] = .02; tbl.Rows.Add(r); r = tbl.NewRow(); r["username"] = "mary"; r["price"] = .03; tbl.Rows.Add(r); r = tbl.NewRow(); r["username"] = "mary"; r["price"] = .04; tbl.Rows.Add(r); </code></pre> <p>This is where I get stuck. Want to return one row for Jack with his highest price (ex: ".02"). </p> <pre><code>var result = from row in tbl.AsEnumerable() where (string) row["username"] == "jack" group row by new {usernameKey = row["username"]} into g select new { //WHAT IS THE LINQ SYNTAX TO USE HERE? jackHighestPrice = g.Max(x =&gt; x.price); //This doesn't work, VS doesn't see the "price" field as strongly typed (not sure why). }; //This should display Jack's highest price. MessageBox.Show(result.First().jackHighestPrice.ToString()); </code></pre> <p>I'm not sure how to get Visual Studio to recognize the "Price" field as strongly typed. Guessing it's related to the issue. </p> <p>Of course, two queries will work (one to filter by username, then another to select Max but it's not as elegant. </p> <p>Related to <a href="https://stackoverflow.com/a/3299573/722945">this answer</a>. Tried just about everything/looked all over but no luck. </p> <p>Thanks.</p>
 

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