特征可以从基础实现抽象方法 class

Question

我有一个 ContactableMemberHandler，即现在扩展了一些 RecordListHandler，但我还需要一个扩展 SingleRecordHandler 的可联系成员处理程序版本。当然，该版本应该包含原始版本中的业务问题...

BaseRecordHandler 看起来像这样：

class BaseRecordHandler<T> { 
    // a lot of implementation details irrelevant to this question...

    /**
     * 
     * @param records
     * @param onSubmit
     * @param onCheckDuplicate
     * @param onRetry
     */
    public void handle(List<T> records, Closure onSubmit, Closure onCheckDuplicate, Closure<T> onRetry) {
        boolean keepTrying = true;

        while(keepTrying) {
            try {
                onSubmit();
                keepTrying = false;
            } catch (StepFailedException ex) {
                this.handleFailure(ex, onSubmit, onCheckDuplicate, onRetry, records);
            }
        }

        for (T n : records)
            this.usedRecords.add(n);
    }

    protected abstract void handleFailure(StepFailedException ex, Closure onSubmit, Closure onCheckDuplicate, Closure<T> onRetry, List<T> newRecords) throws StepFailedException;

    // more irrelevant business logic
}

SingleRecordHandler 看起来像这样：


@InheritConstructors
public abstract class SingleRecordHandler<T> extends BaseRecordHandler<T> {

    /**
     * @param record
     * @param onSubmit @throws StepFailedException
     * @param onCheckDuplicate
     * @param onRetry
     */
    public void handle(T record, Closure onSubmit, Closure<Boolean> onCheckDuplicate, Closure<T> onRetry) {
        this.handle([record], onSubmit, onCheckDuplicate, onRetry);
    }

    @Override
    protected void handleFailure(StepFailedException ex, Closure onSubmit, Closure onCheckDuplicate, Closure<T> onRetry,
            List<T> newRecords) throws StepFailedException {
        if (onCheckDuplicate(ex)) {
            newRecords.add(onRetry());
        }
        else {
            newRecords.pop();
            throw ex;
        }
    }
}

和 RecordListHandler 看起来像这样：

@InheritConstructors
public abstract class RecordListHandler<T> extends BaseRecordHandler<T> {

    @Override
    protected void handleFailure(StepFailedException ex, Closure onSubmit, Closure onCheckDuplicate, Closure<T> onRetry,
            List<T> newRecords) throws StepFailedException {
        int duplicateRecordIdx = onCheckDuplicate(ex);

        if (duplicateRecordIdx > -1) {
            newRecords.push(onRetry(duplicateRecordIdx));
        } else {
            newRecords.pop();
            throw ex;
        }
    }
}

...但是如果我们使用特征呢？

听起来将 SingleRecordHandler、RecordListHandler 转换为特征可以解决这个问题...

...但是当我们像这样使用它们时：

public class ContactableMemberListHandler extends BaseRecordHandler<ContactableMemberModel> implements RecordListHandler<ContactableMemberModel> { 
   // ... business logic
}

特征会handleFailure()实现吗？

Answer 1

显然，确实如此。但是 trait 方法只有 private 和 public，因此您必须使方法 public.

一个trait可以扩展一个class，但是当你要使用特性时你仍然需要扩展那个基础class...

奇怪...

特征可以从基础实现抽象方法 class

Can a trait implement an abstract method from a base class

groovy