C# net 遍历枚举和特性(Attribute)

C# net 遍历 枚举 特性 Attribute 

C# net 遍历 枚举 enum 特性 Attribute 

net 遍历 枚举 enum 特性 Attribute 

假如我们有如下枚举

    /// <summary>
    /// 角色
    /// </summary>
    public enum Role
    {
        /// <summary>
        /// 超级管理员
        /// </summary>
        [Description("超级管理员")]
        Admin = 0,
        /// <summary>
        /// 租借用户
        /// </summary>
        [Description("租借用户")]
        Lease = 1,
        /// <summary>
        /// 普通购买用户
        /// </summary>
        [Description("普通购买用户")]
        Money = 2,
    }

我们想获取到 3个参数  {["Admin",0,"超级管理员"],["Lease",...]} 怎么办呢?

代码如下

        public static void Gets(Type type)
        {
            if (type.IsEnum)
            {
                var fields = type.GetFields(BindingFlags.Static | BindingFlags.Public) ?? new FieldInfo[] { };
                foreach (var field in fields)
                {
                    //取到:Admin
                    var name = field.Name;
                    //取到:0
                    var val = (int)field.GetValue(null);

                    var atts = field.GetCustomAttributes(typeof(DescriptionAttribute), false);
                    if (atts != null && atts.Length > 0)
                    {
                        //取到:超级管理员
                        var att = ((DescriptionAttribute[])atts)[0];
                        var des = att.Description;
                    }
                }
            }
            else
            {

            }
        }

调用方法,如下:

Gets(typeof(Role));

完成!

补充:如果需要获取单个枚举的值,参考 https://www.cnblogs.com/ping9719/p/15699109.html

如有问题请联系QQ: var d=["1","2","3","4","5","6","7","8","9"]; var pass=d[8]+d[6]+d[0]+d[8]+d[2]+d[0]+d[4]+d[3]+d[2];
原文地址:https://www.cnblogs.com/ping9719/p/15698921.html