以强类型方式获取属性的[DisplayName]属性

时间：2011-03-29 14:47:49

标签： asp.net-mvc asp.net-mvc-2 data-annotations

美好的一天！

我有这样的方法来获取属性的[DisplayName]属性值（直接附加或使用[MetadataType]属性）。我在极少数情况下使用它，我需要在控制器代码中获得[DisplayName]。

public static class MetaDataHelper
{
    public static string GetDisplayName(Type dataType, string fieldName)
    {       
        // First look into attributes on a type and it's parents
        DisplayNameAttribute attr;
        attr = (DisplayNameAttribute)dataType.GetProperty(fieldName).GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();

        // Look for [MetadataType] attribute in type hierarchy
        // http://stackoverflow.com/questions/1910532/attribute-isdefined-doesnt-see-attributes-applied-with-metadatatype-class
        if (attr == null)
        {
            MetadataTypeAttribute metadataType = (MetadataTypeAttribute)dataType.GetCustomAttributes(typeof(MetadataTypeAttribute), true).FirstOrDefault();
            if (metadataType != null)
            {
                var property = metadataType.MetadataClassType.GetProperty(fieldName);
                if (property != null)
                {
                    attr = (DisplayNameAttribute)property.GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();
                }
            }
        }
        return (attr != null) ? attr.DisplayName : String.Empty;
    }
}

它有效，但它有两个缺点：

  • 它需要字段名称为字符串
  • 如果我想获得财产的财产
    • ，它就无效

是否有可能使用lambdas克服这两个问题，就像我们在ASP.NET MVC中所做的那样：

Html.LabelFor(m => m.Property.Can.Be.Very.Complex.But.Strongly.Typed);

更新

以下是 BuildStarted 解决方案的更新和检查版本。它被修改为使用DisplayName属性（如果您使用它，可以修改回Display属性）。并修复了小错误以获取嵌套属性的属性。

public static string GetDisplayName<TModel>(Expression<Func<TModel, object>> expression)
{
    Type type = typeof(TModel);

    string propertyName = null;
    string[] properties = null;
    IEnumerable<string> propertyList;
    //unless it's a root property the expression NodeType will always be Convert
    switch (expression.Body.NodeType)
    {
        case ExpressionType.Convert:
        case ExpressionType.ConvertChecked:
            var ue = expression.Body as UnaryExpression;
            propertyList = (ue != null ? ue.Operand : null).ToString().Split(".".ToCharArray()).Skip(1); //don't use the root property
            break;
        default:
            propertyList = expression.Body.ToString().Split(".".ToCharArray()).Skip(1);
            break;
    }

    //the propert name is what we're after
    propertyName = propertyList.Last();
    //list of properties - the last property name
    properties = propertyList.Take(propertyList.Count() - 1).ToArray(); //grab all the parent properties

    foreach (string property in properties)
    {
        PropertyInfo propertyInfo = type.GetProperty(property);
        type = propertyInfo.PropertyType;
    }

    DisplayNameAttribute attr;
    attr = (DisplayNameAttribute)type.GetProperty(propertyName).GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();

    // Look for [MetadataType] attribute in type hierarchy
    // http://stackoverflow.com/questions/1910532/attribute-isdefined-doesnt-see-attributes-applied-with-metadatatype-class
    if (attr == null)
    {
        MetadataTypeAttribute metadataType = (MetadataTypeAttribute)type.GetCustomAttributes(typeof(MetadataTypeAttribute), true).FirstOrDefault();
        if (metadataType != null)
        {
            var property = metadataType.MetadataClassType.GetProperty(propertyName);
            if (property != null)
            {
                attr = (DisplayNameAttribute)property.GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();
            }
        }
    }
    return (attr != null) ? attr.DisplayName : String.Empty;
}

8 个答案:

答案 0 :(得分：37)

有两种方法可以做到这一点：

Models.Test test = new Models.Test();
string DisplayName = test.GetDisplayName(t => t.Name);

string DisplayName = Helpers.GetDisplayName<Models.Test>(t => t.Name);

第一个工作原理是将通用扩展方法写入任何TModel（所有类型）。这意味着它可以在任何对象上使用，而不仅仅是您的模型。不是真的推荐，但很好，因为它的语法简洁。

第二种方法要求您传入模型的类型 - 您已经在做的但是作为参数。这个方法需要通过泛型来定义类型，因为Func期望它。

以下是您查看的方法。

所有对象的静态扩展方法

public static string GetDisplayName<TModel, TProperty>(this TModel model, Expression<Func<TModel, TProperty>> expression) {

    Type type = typeof(TModel);

    MemberExpression memberExpression = (MemberExpression)expression.Body;
    string propertyName = ((memberExpression.Member is PropertyInfo) ? memberExpression.Member.Name : null);

    // First look into attributes on a type and it's parents
    DisplayAttribute attr;
    attr = (DisplayAttribute)type.GetProperty(propertyName).GetCustomAttributes(typeof(DisplayAttribute), true).SingleOrDefault();

    // Look for [MetadataType] attribute in type hierarchy
    // http://stackoverflow.com/questions/1910532/attribute-isdefined-doesnt-see-attributes-applied-with-metadatatype-class
    if (attr == null) {
        MetadataTypeAttribute metadataType = (MetadataTypeAttribute)type.GetCustomAttributes(typeof(MetadataTypeAttribute), true).FirstOrDefault();
        if (metadataType != null) {
            var property = metadataType.MetadataClassType.GetProperty(propertyName);
            if (property != null) {
                attr = (DisplayAttribute)property.GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();
            }
        }
    }
    return (attr != null) ? attr.Name : String.Empty;


}

特定于类型的方法的签名 - 与上面相同的代码只是不同的调用

public static string GetDisplayName<TModel>(Expression<Func<TModel, object>> expression) { }

你不能仅仅使用Something.GetDisplayName(t => t.Name)的原因是因为在Razor引擎中你实际上传递了HtmlHelper<TModel>的实例化对象，这就是第一种方法需要实例化对象的原因 - 只有编译器才需要推断哪些类型属于哪个通用名称。

使用递归属性更新

public static string GetDisplayName<TModel>(Expression<Func<TModel, object>> expression) {

    Type type = typeof(TModel);

    string propertyName = null;
    string[] properties = null;
    IEnumerable<string> propertyList;
    //unless it's a root property the expression NodeType will always be Convert
    switch (expression.Body.NodeType) {
        case ExpressionType.Convert:
        case ExpressionType.ConvertChecked:
            var ue = expression.Body as UnaryExpression;
            propertyList = (ue != null ? ue.Operand : null).ToString().Split(".".ToCharArray()).Skip(1); //don't use the root property
            break;
        default:
            propertyList = expression.Body.ToString().Split(".".ToCharArray()).Skip(1);
            break;
    }

    //the propert name is what we're after
    propertyName = propertyList.Last();
    //list of properties - the last property name
    properties = propertyList.Take(propertyList.Count() - 1).ToArray(); //grab all the parent properties

    Expression expr = null;
    foreach (string property in properties) {
        PropertyInfo propertyInfo = type.GetProperty(property);
        expr = Expression.Property(expr, type.GetProperty(property));
        type = propertyInfo.PropertyType;
    }

    DisplayAttribute attr;
    attr = (DisplayAttribute)type.GetProperty(propertyName).GetCustomAttributes(typeof(DisplayAttribute), true).SingleOrDefault();

    // Look for [MetadataType] attribute in type hierarchy
    // http://stackoverflow.com/questions/1910532/attribute-isdefined-doesnt-see-attributes-applied-with-metadatatype-class
    if (attr == null) {
        MetadataTypeAttribute metadataType = (MetadataTypeAttribute)type.GetCustomAttributes(typeof(MetadataTypeAttribute), true).FirstOrDefault();
        if (metadataType != null) {
            var property = metadataType.MetadataClassType.GetProperty(propertyName);
            if (property != null) {
                attr = (DisplayAttribute)property.GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();
            }
        }
    }
    return (attr != null) ? attr.Name : String.Empty;



}

答案 1 :(得分：30)

比赛结束，但是......

我使用像@Daniel提到的ModelMetadata创建了一个帮助方法，我想我会分享它：

public static string GetDisplayName<TModel, TProperty>(
      this TModel model
    , Expression<Func<TModel, TProperty>> expression)
{
    return ModelMetadata.FromLambdaExpression<TModel, TProperty>(
        expression,
        new ViewDataDictionary<TModel>(model)
        ).DisplayName;
}

示例用法：

Models：

public class MySubObject
{
    [DisplayName("Sub-Awesome!")]
    public string Sub { get; set; }
}

public class MyObject
{
    [DisplayName("Awesome!")]
    public MySubObject Prop { get; set; }
}

Use：

HelperNamespace.GetDisplayName(Model, m => m.Prop) // "Awesome!"
HelperNamespace.GetDisplayName(Model, m => m.Prop.Sub) // "Sub-Awesome!"

答案 2 :(得分：4)

这样做：

using System.ComponentModel;
using System.Linq;
using System.Reflection;

namespace yournamespace
{
    public static class ExtensionMethods
    {
        public static string GetDisplayName(this PropertyInfo prop)
        {
            if (prop.CustomAttributes == null || prop.CustomAttributes.Count() == 0)
                return prop.Name;

            var displayNameAttribute = prop.CustomAttributes.Where(x => x.AttributeType == typeof(DisplayNameAttribute)).FirstOrDefault();

            if (displayNameAttribute == null || displayNameAttribute.ConstructorArguments == null || displayNameAttribute.ConstructorArguments.Count == 0)
                return prop.Name;

            return displayNameAttribute.ConstructorArguments[0].Value.ToString() ?? prop.Name;
        }
    }
}

请求示例：

var props = typeof(YourType).GetProperties(BindingFlags.Public | BindingFlags.Instance).Where(p => p.CanRead);

var propFriendlyNames = props.Select(x => x.GetDisplayName());

答案 3 :(得分：2)

我完全同意BuildStarted提供的解决方案。我唯一要改变的是ExtensionsMethode不支持翻译。为了支持这一点，需要进行简单的小改动。我会把它放在评论中，但我没有足够的分数这样做。寻找方法中的最后一行。

扩展方法 

public static string GetDisplayName<TModel, TProperty>(this TModel model, Expression<Func<TModel, TProperty>> expression)
{
        Type type = typeof(TModel);
        IEnumerable<string> propertyList;

        //unless it's a root property the expression NodeType will always be Convert
        switch (expression.Body.NodeType)
        {
            case ExpressionType.Convert:
            case ExpressionType.ConvertChecked:
                var ue = expression.Body as UnaryExpression;
                propertyList = (ue != null ? ue.Operand : null).ToString().Split(".".ToCharArray()).Skip(1); //don't use the root property
                break;
            default:
                propertyList = expression.Body.ToString().Split(".".ToCharArray()).Skip(1);
                break;
        }

        //the propert name is what we're after
        string propertyName = propertyList.Last();
        //list of properties - the last property name
        string[] properties = propertyList.Take(propertyList.Count() - 1).ToArray();

        Expression expr = null;
        foreach (string property in properties)
        {
            PropertyInfo propertyInfo = type.GetProperty(property);
            expr = Expression.Property(expr, type.GetProperty(property));
            type = propertyInfo.PropertyType;
        }

        DisplayAttribute attr = (DisplayAttribute)type.GetProperty(propertyName).GetCustomAttributes(typeof(DisplayAttribute), true).SingleOrDefault();

        // Look for [MetadataType] attribute in type hierarchy
        // http://stackoverflow.com/questions/1910532/attribute-isdefined-doesnt-see-attributes-applied-with-metadatatype-class
        if (attr == null)
        {
            MetadataTypeAttribute metadataType = (MetadataTypeAttribute)type.GetCustomAttributes(typeof(MetadataTypeAttribute), true).FirstOrDefault();
            if (metadataType != null)
            {
                var property = metadataType.MetadataClassType.GetProperty(propertyName);
                if (property != null)
                {
                    attr = (DisplayAttribute)property.GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();
                }
            }
        }
        //To support translations call attr.GetName() instead of attr.Name
        return (attr != null) ? attr.GetName() : String.Empty;
 }

答案 4 :(得分：1)

我做了一点改变，你是使用资源来获取DisplayName 

    public static string GetDisplayName<TModel>(Expression<Func<TModel, object>> expression)
  {

     string _ReturnValue = string.Empty;

     Type type = typeof(TModel);

     string propertyName = null;
     string[] properties = null;
     IEnumerable<string> propertyList;
     //unless it's a root property the expression NodeType will always be Convert
     switch (expression.Body.NodeType)
     {
        case ExpressionType.Convert:
        case ExpressionType.ConvertChecked:
           var ue = expression.Body as UnaryExpression;
           propertyList = (ue != null ? ue.Operand : null).ToString().Split(".".ToCharArray()).Skip(1); //don't use the root property
           break;
        default:
           propertyList = expression.Body.ToString().Split(".".ToCharArray()).Skip(1);
           break;
     }

     //the propert name is what we're after
     propertyName = propertyList.Last();
     //list of properties - the last property name
     properties = propertyList.Take(propertyList.Count() - 1).ToArray(); //grab all the parent properties

     Expression expr = null;
     foreach (string property in properties)
     {
        PropertyInfo propertyInfo = type.GetProperty(property);
        expr = Expression.Property(expr, type.GetProperty(property));
        type = propertyInfo.PropertyType;
     }

     DisplayAttribute attr;
     attr = (DisplayAttribute)type.GetProperty(propertyName).GetCustomAttributes(typeof(DisplayAttribute), true).SingleOrDefault();

     // Look for [MetadataType] attribute in type hierarchy
     // http://stackoverflow.com/questions/1910532/attribute-isdefined-doesnt-see-attributes-applied-with-metadatatype-class
     if (attr == null)
     {
        MetadataTypeAttribute metadataType = (MetadataTypeAttribute)type.GetCustomAttributes(typeof(MetadataTypeAttribute), true).FirstOrDefault();
        if (metadataType != null)
        {
           var property = metadataType.MetadataClassType.GetProperty(propertyName);
           if (property != null)
           {
              attr = (DisplayAttribute)property.GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();
           }
        }
     }

     if (attr != null && attr.ResourceType != null)
        _ReturnValue = attr.ResourceType.GetProperty(attr.Name).GetValue(attr).ToString();
     else if (attr != null)
        _ReturnValue = attr.Name;

     return _ReturnValue;
  }

快乐编码

答案 5 :(得分：1)

我找到了另一个不错的代码段here，我稍微修改了一下'DisplayName'目的

    public static string GetDisplayName<TSource, TProperty>(Expression<Func<TSource, TProperty>> expression)
    {
        var attribute = Attribute.GetCustomAttribute(((MemberExpression)expression.Body).Member, typeof(DisplayNameAttribute)) as DisplayNameAttribute;

        if (attribute == null)
        {
            throw new ArgumentException($"Expression '{expression}' doesn't have DisplayAttribute");
        }

        return attribute.DisplayName;
    }

和用法

GetDisplayName<ModelName, string>(i => i.PropertyName)

答案 6 :(得分：0)

另一个代码.Net的代码片段用于执行此操作

public static class WebModelExtensions
{
    public static string GetDisplayName<TModel, TProperty>(
      this HtmlHelper<TModel> html, 
      Expression<Func<TModel, TProperty>> expression)
    {
        // Taken from LabelExtensions
        var metadata = ModelMetadata.FromLambdaExpression(expression, html.ViewData);

        string displayName = metadata.DisplayName;
        if (displayName == null)
        {
            string propertyName = metadata.PropertyName;
            if (propertyName == null)
            {
                var htmlFieldName = ExpressionHelper.GetExpressionText(expression);
                displayName = ((IEnumerable<string>) htmlFieldName.Split('.')).Last<string>();
            }
            else
                displayName = propertyName;
        }

        return displayName;
    }
}
// Usage
Html.GetDisplayName(model => model.Password)

答案 7 :(得分：0)

@Buildstarted答案有效，但是我想通过属性名称而不是使用linq表达式来获取DisplayName，所以我做了一些改动以节省您的时间

public static string GetDisplayNameByMemberName<TModel>(string memberName)
    {
        Type type = typeof(TModel);
        string propertyName = memberName;

        DisplayAttribute attr;
        attr = (DisplayAttribute)type.GetProperty(propertyName).GetCustomAttributes(typeof(DisplayAttribute), true).SingleOrDefault();

        //Look for [MetadataType] attribute in type hierarchy
        //http://stackoverflow.com/questions/1910532/attribute-isdefined-doesnt-see-attributes-applied-with-metadatatype-class
        if (attr == null)
            {
                MetadataTypeAttribute metadataType = (MetadataTypeAttribute)type.GetCustomAttributes(typeof(MetadataTypeAttribute), true).FirstOrDefault();
                if (metadataType != null)
                {
                    var property = metadataType.MetadataClassType.GetProperty(propertyName);
                    if (property != null)
                    {
                        attr = (DisplayAttribute)property.GetCustomAttributes(typeof(DisplayNameAttribute), true).SingleOrDefault();
                    }
                }
            }
        return (attr != null) ? attr.Name : String.Empty;
    }

我还想获取资源（.resx文件）中的本地化值，所以：

string displayName = GeneralHelper.GetDisplayNameByMemberName<ViewModels.ProductVM>(memberName);
string displayNameTranslated = resourceManager.GetString(displayName, MultiLangHelper.CurrentCultureInfo);

这是一个帮助程序函数，用于返回“ CultureInfo”类型的对象，创建某种语言的区域性信息或传递当前语言

//MultiLangHelper.CurrentCultureInfo
return System.Threading.Thread.CurrentThread.CurrentCulture;
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