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Home/ Questions/Q 6561923
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T13:38:52+00:00 2026-05-25T13:38:52+00:00

I have a method that determines the min and max of a column in

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I have a method that determines the min and max of a column in a DataTable:

public void GetMinMaxRange( DataTable data, string valueColumnName )
{
   var min = data.AsEnumerable().Min(m => m.Field<double>(valueColumnName));
   var max = data.AsEnumerable().Max(m => m.Field<double>(valueColumnName));
}

I want to refactor this to be:

public void GetMinMaxRange( DataTable data, string valueColumnName )
{
   DataColumn column = data.Columns[valueColumnName];
   var min = data.AsEnumerable().Min(m => m.Field<column.DataType>(valueColumnName));
   var max = data.AsEnumerable().Max(m => m.Field<column.DataType>(valueColumnName));
}

I need to determine the datatype and use it instead of hard-coding m.Field<double>. How to do this?

UPDATE
As to why I want to calculate the difference between the min and the max

    public static double/decimal/int GetMinMaxRange<T>(DataTable data, 
          string valueColumnName) where T : IComparable<T>
    {
        DataColumn column = data.Columns[valueColumnName];
        var min = data.AsEnumerable().Min(m => m.Field<T>(valueColumnName));
        var max = data.AsEnumerable().Max(m => m.Field<T>(valueColumnName)); ;
        return max - min;
    }
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T13:38:53+00:00Added an answer on May 25, 2026 at 1:38 pm

    Simply creating this as a generic should work:

    public void GetMinMaxRange<T>( DataTable data, string valueColumnName ) 
                                                                 where T : IComparable<T>
    {
       DataColumn column = data.Columns[valueColumnName];
       var min = data.AsEnumerable().Min(m => m.Field<T>(valueColumnName));
       var max = data.AsEnumerable().Max(m => m.Field<T>(valueColumnName));
    }
    

    Which you would then use as:

    GetMinMaxRange<MyType>(dataTable, valueColumnName);
    
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