本文介绍了复杂的属性类型的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我有一个名为TerminalConnectionParameters的类，它包含一些不同类型的字段，如Hostname，IPAddress，UserName，Password等。 我想用这个类键入Dummy类的属性，例如： public class Dummy { private TerminalConnectionParameters _parameters; public TerminalConnectionParameters参数 { get { // 在这里做一些花哨的计算 返回 _parameters; } set { _parameters = value ; // 在这里做一些花哨的计算 } } } 这确保我可以获取或设置Dummy对象的整个Parameters类。 到目前为止一切都那么好，但是，如果我想使用这个属性如下： objectDummy.Parameters.Hostname = dummyRouter; 当然，这个赋值语句不会设置dummyObject的内部_parameters字段的值。那么呢？什么是实现这样的属性编写器的正确方法，而不必独特地创建TerminalConnectionParameters的每个字段的属性访问器？ 我知道如下的简单方法，但没有更好的解决方案？ TerminalConnectionParameters temp = objectDummy.Parameters; temp.HostName = dummyRouter; objectDummy.Parameters = temp; 这个问题背后的原因是我想用PropertyGrid来处理表格的内部_parameters对象。 我尝试过： 如上例所示。解决方案 实际上，如果您的_parameters成员已经初始化，那么您的第一个赋值语句很好 - 您可以在声明中执行此操作 私人 TerminalConnectionParameters _parameters = new TerminalConnectionParameters（）; 或在getter中 private TerminalConnectionParameters _parameters = null ; public TerminalConnectionParameters参数 { get { 如果（_parameters - null ） _parameters = new TerminalConnectionParameters（）; // 在这里做一些花哨的计算 返回 _parameters; 现在，一旦你是确保初始化，您需要做的就是用以下属性装饰参数属性，设计师将神奇地处理其余的事情： [DesignerSerializationVisibility（DesignerSerializationVisibility.Content）] public TerminalConnectionParameters参数 { 您可以序列化虚拟对象中的参数，并在宿主对象中反序列化。 谢谢Midi_Mick，就是这样。不知怎的，我觉得它必须简单......： - ） I have a class named TerminalConnectionParameters which contains a few differently typed fields, like Hostname, IPAddress, UserName, Password, etc.I want to used this class type as a property of a Dummy class, like :public class Dummy{ private TerminalConnectionParameters _parameters; public TerminalConnectionParameters Parameters { get { // do some fancy calculation here return _parameters; } set { _parameters = value; // do some fancy calculation here } }}This ensures that I can get or set the whole Parameters class of a Dummy object.So far so good, but what, if I want to use this property like :objectDummy.Parameters.Hostname= "dummyRouter";For sure, this assignment statement will not set the value of the dummyObject's internal _parameters field. But what then ? And what is the correct way to implenment such property writer without having to create a property accessor of each field of the TerminalConnectionParameters uniquely ?I am aware of the trivial way as below, but is there no better solution ?TerminalConnectionParameters temp = objectDummy.Parameters;temp.HostName = "dummyRouter";objectDummy.Parameters = temp;And the reason behind this question is that I want to use PropertyGrid to deal with the internal _parameters object of a Form.What I have tried:As outlined in the example above. 解决方案 Actually, your first assignment statement is good providing your _parameters member has been initialised - you can either do that in the declarationprivate TerminalConnectionParameters _parameters = new TerminalConnectionParameters();or in the getterprivate TerminalConnectionParameters _parameters = null; public TerminalConnectionParameters Parameters { get { if (_parameters - null) _parameters = new TerminalConnectionParameters(); // do some fancy calculation here return _parameters;Now, once you are ensured of initialisation, all you need to do is decorate the Parameters property with the following attribute, and the designer will magically take care of the rest:[DesignerSerializationVisibility(DesignerSerializationVisibility.Content)]public TerminalConnectionParameters Parameters{You could serialize the parameters in the dummy object, and deserialize it in the host object.Thank you Midi_Mick, that was it. Somehow I felt it must be simple...:-) 这篇关于复杂的属性类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！