生成器模式的抽象类

Question

我有一些始终使用构建器模式的现有模型类，如下所示：

public class Model {
    public static class Builder {
        private boolean isValid;
        private List<String> errorMessagesOrNull;

        public Builder setIsValid(final boolean isValid) {
            this.isValid = isValid;
            return this;
        }

        public Builder setErrorMessages(final List<String> errorMessages) {
            this.errorMessagesOrNull = errorMessages;
            return this;
        }

        public List<String> getErrorMessages() {
            return this.errorMessagesOrNull == null ? new ArrayList<>() : this.errorMessagesOrNull;
        }

        public Model Build() {
            return new Model(this);
        }
    }

    private boolean isValid;
    private List<String> errorMessages;

    private Model(final Builder builder) {
        this.isValid = builder.isValid;
        this.errorMessages = builder.getErrorMessages();
    }

    public boolean getIsValid() {
        return isValid;
    }

    public List<String> getErrorMessages() {
        return errorMessages;
    }
}

如您所见，模型类始终具有isValid和errorMessages 。 我想编写一个抽象类，以最小化这些模型类的重复逻辑。

所以我像这样的抽象类来了：

public abstract class AbstractModel<T extends AbstractModel<T>> {

    public static abstract class Builder<T> {
        private boolean isValid;
        private List<String> errorMessagesOrNull;

        public Builder<T> setIsValid(final boolean isValid) {
            this.isValid = isValid;
            return this;
        }

        public Builder<T> setErrorMessages(final List<String> errorMessages) {
            this.errorMessagesOrNull = errorMessages;
            return this;
        }

        public List<String> getErrorMessages() {
            return this.errorMessagesOrNull == null ? new ArrayList<>() : this.errorMessagesOrNull;
        }

        public abstract T Build();
    }

    private boolean isValid;
    private List<String> errorMessages;

    private AbstractModel(final Builder<T> builder) {
        this.isValid = builder.isValid;
        this.errorMessages = builder.getErrorMessages();
    }

    public boolean getIsValid() {
        return isValid;
    }

    public List<String> getErrorMessages() {
        return errorMessages;
    }
}

但这并没有按照我的预期工作。 当我扩展抽象类时：

public class Model extends AbstractModel<Model> {
    // Empty here since all fields are extended
}

我不能做这样的事情：

Model model = new Model.Builder.setIsValid(true).Build();

我希望抽象类具有Builder静态类，这样我就不必每次都写静态类Builder 。

请指教。

Answer 1

我认为您的逻辑存在巨大缺陷。 该程序本身根本没有任何意义。 为什么首先要使用Builder类构建Model ？ 我认为最好向您展示如何编写程序，而不是仅将其“结合”在一起。 好吧，让我们从Model类开始。

假设Model类不能在没有Builder情况下构造。 然后将Builder类添加到Model类是否有意义？ 简短的回答：不，不会。 而是， Builder类应包含Model类作为非静态内部类。

/**
 * The {@code Builder} can construct new instances of the {@code Model} class.
 *
 * @see Model
 */
public class Builder
{
    private final String[] log;

    /**
     * The {@code Model} class can do something. You can only construct it through a {@code Builder}.
     *
     * @see Builder
     */
    public class Model
    {
        private final Builder builder;

        /**
         * Constructs a new {@code Model} with the specified argument.
         *
         * @param builder the {@code Builder} that constructed the model.
         */
        public Model(final Builder builder)
        {
            this.builder = builder;
        }

        /**
         * Returns the associated {@code Builder}.
         *
         * @return the builder that constructed the model.
         */
        public Builder getBuilder()
        {
            return this.builder;
        }
    }

    /**
     * Constructs a new instance of the {@code Builder} class with the specified argument.
     *
     * @param log the log of the {@code Builder}.
     */
    public Builder(final String... log)
    {
        this.log = log;
    }

    /**
     * Tries to {@code build} a new instance of the {@code Model} class.
     *
     * @return the constructed {@code Model}.
     */
    public Model build()
    {
        return new Model(this);
    }

    /**
     * Returns the log of the {@code Builder}.
     *
     * @return an log.
     */
    public String[] getLog()
    {
        return this.log;
    }

    /**
     * Determines whether or not the {@code Builder} is valid.
     *
     * @return {@code true} when the specified {@code log} is not {@code null}; {@code false} otherwise.
     */
    public boolean isValid()
    {
        return this.log != null;
    }
}

除了Builder之外，没有其他类可以构造Model 。 但是，如果您构造了Builder类的新实例并获得了调用build方法的结果，则可以访问所有public变量和方法。

如果您知道要构造一个Model ，则可以这样做：

Builder.Model model = new Builder().build();

如果您不想要Builder. 前缀，只需添加导入Model类的import语句。

import organisation.projectname.pathToBuilder.Builder.Model;

Answer 2

您还需要实现Builder 。

public class Model extends AbstractModel<Model>{


    private Model(final Builder builder) {
        super(builder);
    }

    public static class Builder2 extends AbstractModel.Builder<Model> {

        @Override
        public Model Build() {
            return new Model(this);
        }
    }
}

然后可以打电话

Model model = new Model.Builder2().Build();

编辑

同样，AbstractBuilder的构造函数也必须protected 。

  protected AbstractModel(final Builder<? extends Builder<T>> builder) {
        this.isValid = builder.isValid;
        this.errorMessages = builder.getErrorMessages();
    }