NameValue, now with Distinct() support

One of the more useful little classes in my utility box is NameValue. It’s so veryeasy to fill with useful data from a Linq query, and once filled so easy to move betweenlayers, in a way that a projection simply can’t be used. The simple little NameValueclass bind just fine to a drop down menu, and will bring you tea after you finishwith that stupid girl that didn’t like your pony/monkey chimera.

I recently had to add equality checks and an override for == and != in order to makemerging two lists of name value pairs together possible. Now .Distinct() works ona List<NameValue>! There’s some goodmaterial out there for how to make a quality equality implementation.

[Serializable] public class NameValue{ public object Name{ get { return _Name;} set {_Name = value;} } protected virtual object _Name{ get; set;} public object Value{ get { return _Value;} set {_Value = value;} } protected virtual object _Value{ get; set;} public override bool Equals(object obj){ if (obj == null) return false; if (this.GetType() != obj.GetType()) return false; // safebecause of the GetType check NameValuenv = (NameValue)obj; // usethis pattern to compare reference members if (!Object.Equals(this.Value,nv.Value)) return false; if (!Object.Equals(this.Name,nv.Name)) return false; return true;} public override int GetHashCode(){ return Name.GetHashCode() ^ Value.GetHashCode();} public static bool operator ==(NameValuen1, NameValue n2) { return n1.Equals(n2);} public static bool operator !=(NameValuen1, NameValue n2) { return !n1.Equals(n2);} }
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