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How to sort a List<T> by double value?

开发者 https://www.devze.com 2022-12-18 06:09 出处:网络
This sound simple but开发者_开发知识库 it not that much. I want to order a List based on one of the properties of T, which is double type. If you know the propertyname before compilation:

This sound simple but开发者_开发知识库 it not that much.

I want to order a List based on one of the properties of T, which is double type.


If you know the propertyname before compilation:

myList = myList.OrderBy(a=>a.propertyName).ToList();

or

myList = (from m in myList order by m.propertyName).ToList();

If you don't have the property at compile time (e.g. dynamic sorting in a grid or something); try the following extension methods:

static class OrderByExtender
{
    public static IOrderedEnumerable<T> OrderBy<T>(this IEnumerable<T> collection, string key, string direction)
    {
        LambdaExpression sortLambda = BuildLambda<T>(key);

        if(direction.ToUpper() == "ASC")
            return collection.OrderBy((Func<T, object>)sortLambda.Compile());
        else
            return collection.OrderByDescending((Func<T, object>)sortLambda.Compile());
    }

    public static IOrderedEnumerable<T> ThenBy<T>(this IOrderedEnumerable<T> collection, string key, string direction)
    {
        LambdaExpression sortLambda = BuildLambda<T>(key);

        if (direction.ToUpper() == "ASC")
            return collection.ThenBy((Func<T, object>)sortLambda.Compile());
        else
            return collection.ThenByDescending((Func<T, object>)sortLambda.Compile());
    }

    private static LambdaExpression BuildLambda<T>(string key)
    {
        ParameterExpression TParameterExpression = Expression.Parameter(typeof(T), "p");
        LambdaExpression sortLambda = Expression.Lambda(Expression.Convert(Expression.Property(TParameterExpression, key), typeof(object)), TParameterExpression);
        return sortLambda;
    }
}

Then order like

myList = myList.OrderBy("propertyName", "ASC").ToList();


var list = (from t in list
            orderby t.doubleVal).ToList();
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