我有一个通常包含字段,属性的类。我要实现的不是此:
class Example
{
public string Field = "EN";
public string Name { get; set; }
public int? Age { get; set; }
public List<string> A_State_of_String { get; set; }
}
public static void Test()
{
var c1 = new Example
{
Name = "Philip",
Age = null,
A_State_of_String = new List<string>
{
"Some Strings"
}
};
var c2 = new Example();
//Instead of doing that
c2.Name = string.IsNullOrEmpty(c1.Name) ? "" : c1.Name;
c2.Age = c1.Age ?? 0;
c2.A_State_of_String = c1.A_State_of_String ?? new List<string>();
//Just do that
c1.CopyEmAll(c2);
}
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我想出了什么,但没有按预期工作。
public static void CopyEmAll(this object src, object dest)
{
if (src == null) {
throw new ArgumentNullException("src");
}
foreach (PropertyDescriptor item in TypeDescriptor.GetProperties(src)) {
var val = item.GetValue(src);
if (val == null) {
continue;
}
item.SetValue(dest, val);
}
}
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问题:
笔记:
AutoMapper一些技术问题。基于Leo的答案,但使用泛型并复制字段:
public void CopyAll<T>(T source, T target)
{
var type = typeof(T);
foreach (var sourceProperty in type.GetProperties())
{
var targetProperty = type.GetProperty(sourceProperty.Name);
targetProperty.SetValue(target, sourceProperty.GetValue(source, null), null);
}
foreach (var sourceField in type.GetFields())
{
var targetField = type.GetField(sourceField.Name);
targetField.SetValue(target, sourceField.GetValue(source));
}
}
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然后:
CopyAll(f1, f2);
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