使用C＃从强类型类动态构建对象？

目前，我正在手动将属性和值添加到对象，如此示例 ，并发送到Dapper.SimpleCRUD以从Dapper Orm获取数据。 这是我想要达到的理想输出 。

object whereCriteria = null; whereCriteria = new { CountryId = 2, CountryName = "Anywhere on Earth", CountryCode = "AOE", IsActive = true }; 

以下类应以上述格式构建对象并返回现成的对象。

 public static class WhereClauseBuilder { public static object BuildWhereClause(object model) { object whereObject = null; var properties = GetProperties(model); foreach (var property in properties) { var value = GetValue(property, model); //Want to whereObject according to the property and value. Need help in this part!!! } return whereObject; } private static object GetValue(PropertyInfo property, object model) { return property.GetValue(model); } private static IEnumerable GetProperties(object model) { return model.GetType().GetProperties(); } } 

这个函数WhereClauseBuilder.BuildWhereClause(object model)应该以预期的格式返回对象（如上所述） 。 以下是我想要使用的实现方式。

 public sealed class CountryModel { public int CountryId { get; set; } public string CountryName { get; set; } public string CountryCode { get; set; } public bool IsActive { get; set; } } public class WhereClauseClass { public WhereClauseClass() { var model = new CountryModel() { CountryCode = "AOE", CountryId = 2, CountryName = "Anywhere on Earth", IsActive = true }; //Currently, won't return the correct object because the implementation is missing. var whereClauseObject = WhereClauseBuilder.BuildWhereClause(model); } } 

Answer 1

也许是这样的：

 private const string CodeTemplate = @" namespace XXXX { public class Surrogate { ##code## } }"; public static Type CreateSurrogate(IEnumerable properties) { var compiler = new CSharpCodeProvider(); var compilerParameters = new CompilerParameters { GenerateInMemory = true }; foreach (var item in AppDomain.CurrentDomain.GetAssemblies().Where(x => !x.IsDynamic)) { compilerParameters.ReferencedAssemblies.Add(item.Location); } var propertiesCode = string.join("\n\n", from pi in properties select "public " + pi.PropertyType.Name + " " + pi.Name + " { get; set; }"); var source = CodeTemplate.Replace("##code##", propertiesCode); var compilerResult = compiler.CompileAssemblyFromSource(compilerParameters, source); if (compilerResult.Errors.HasErrors) { throw new InvalidOperationException(string.Format("Surrogate compilation error: {0}", string.Join("\n", compilerResult.Errors.Cast()))); } return compilerResult.CompiledAssembly.GetTypes().First(x => x.Name == "Surrogate"); } 

现在使用它：

 public static object BuildWhereClause(object model) { var properties = GetProperties(model); var surrogateType = CreateSurrogate(properties); var result = Activator.CreateInstance(surrogateType); foreach (var property in properties) { var value = GetValue(property, model); var targetProperty = surrogateType.GetProperty(property.Name); targetProperty.SetValue(result, value, null); } return result; } 

我没有编译。 它只写在这里。 也许有一些错误。 🙂

编辑：

要使用ExpandoObject，您可以尝试：

 public static object BuildWhereClause(object model) { var properties = GetProperties(model); var result = (IDictionary)new ExpandoObject(); foreach (var property in properties) { var value = GetValue(property, model); result.Add(property.Name, value); } return result; } 

但我不知道这是否适合你。