Question

我有一个电子邮件服务，它通过构造函数传递电子邮件模型，并且都使用相同的基类。例如，一个模型可用于身份验证，另一个模型用于密码重置。我的问题是如何允许服务传递具有相同基类的匿名模型。

示例代码：

public class EmailService<T> : IEmailService 
{
    private readonly T _emailModel;
    private readonly EmailType _emailType;
    private readonly IEmailRepository _emailRepository;
    private MailBuilder _mailBuilder;
    private EmailTemplates _message;


    public EmailService(T emailModel, EmailType emailType, IEmailRepository emailRepository)
    {
        _emailModel = emailModel;
        _emailType = emailType;
        _emailRepository = emailRepository;

        getMessage();
        constructEmail();
    }

    private void getMessage()
    {
        _message = _emailRepository.GetTemplateByUser((int)_emailType, _emailModel.UserTypeId);
    }

    private void constructEmail()
    {
        _mailBuilder = new MailBuilder(_message, _emailType, ObjectConverters.ConvertProperiesToDictionary(_emailModel)); 
    }

    public void Send()
    {
        EmailSettings emailSettings = SiteSettingsService.SiteConfiguration.EmailSettings;

        MailSettings settings = new MailSettings
        {
            MailFrom = emailSettings.MailFrom,
            MailSmtpHost = emailSettings.SmtpHost,
            MailSmtpPort = emailSettings.SmtpPort,
            EnableSsi = emailSettings.EnableSsi,
            MailSmtpUsername = emailSettings.SmtpUsername,
            MailSmtpPassword = emailSettings.SmtpPassword
        };

        new EmailSender(settings).SendEmail(_emailModel.EmailAddress, _message.Title, _mailBuilder.HtmlTemplate);
    }
}

Answer 1

您可以让派生类都实现一个新的IModel接口，然后您的EmailService类通过ctor接受。

public class AuthenticationModel : BaseModel, IModel
{
    public void PerformMainFunction()
    {
        // authenticate
    }
}

public class PasswordResetModel : BaseModel, IModel
{
    public void PerformMainFunction()
    {
        // reset password
    }
}

public class BaseModel
{
    public int UserTypeId { get; set; }
}

public interface IModel
{
    void PerformMainFunction();

    int UserTypeId { get; set; }
}

public class EmailService : IEmailService
{
    private readonly IModel _emailModel;
    ...

    public EmailService(IModel emailModel, EmailType emailType, IEmailRepository emailRepository)
    {
        _emailModel = emailModel;
       ...

您可以使用类似Unity的内容来注入依赖项，或者，现在，只需执行：

IModel model = new AuthenticationModel();

var eServ = new EmailService(model, ....

Answer 2

我想到了一个更好的方法。我们总是将对象转换为服务中的字典。也许我们从调用电子邮件服务的服务转换它，所以我们总是传递字典而不是接口，这样我们就不需要知道结构了。字典将始终是字符串和字符串的键值。

传递具有相同基类的任何模型

2 个答案: