以上方法如何写路由函数，使其传输json格式数据，请改正并优化代码：@films_bp.route('/get_top_10_directors', methods=['GET']) def get_top_10_directors(): #返回按年统计的每年电影的平均得分，,返回的json数据格式如下所示 try: result = FilmInfo.get_top_10_directors() #直接调用FilmInfo类中构建的静态方法count_by_year() logging.info('获得各年份电影数量序列') return jsonify(result) except Exception as e: logging.error('Error occurred while retrieving students from the database. Error message: {}'.format(str(e))) return jsonify({"error": str(e)})

可以将代码优化如下：

@films_bp.route('/get_top_10_directors', methods=['GET'])
def get_top_10_directors():
    try:
        result = FilmInfo.get_top_10_directors() # 直接调用FilmInfo类中构建的静态方法count_by_year()
        logging.info('获取各年份电影数量序列')
        return result.to_json(orient='records') # 将DataFrame对象转换为JSON格式并返回
    except Exception as e:
        logging.error('从数据库中检索信息时出现错误，错误信息：{}'.format(str(e)))
        return jsonify({"error": str(e)})

这里使用了`to_json()`方法将DataFrame对象转换为JSON格式，同时将`jsonify(result)`改为`result.to_json(orient='records')`，以避免重复转换。另外，日志信息也进行了优化。

相关问题

@films_bp.route('/avg_rating_num_by_avg_rates', methods=['GET']) def avg_rating_num_by_avg_rates(): #返回按年统计的每年电影的平均得分，,返回的json数据格式如下所示 try: conn = pymysql.Connection( host='localhost', port=3306, user='root', passwd='21591314cxk' ) cur = conn.cursor() cur.execute('select rate,rating_num from movie.movieinfo where rating_num > 500') dic = cur.fetchall() conn.commit() cur.close() return jsonify(dict(dic)) except Exception as e: logging.error('Error occurred while retrieving students from the database. Error message: {}'.format(str(e))) return jsonify({"error": str(e)})这段代码只返回一个评分结果对应的数据，一旦查找到一个评分就不会继续查找同样评分对应的数据，而我想获取所有数据，请改正并优化代码。

可以将查询结果按照评分进行分组，然后再遍历每个分组获取对应的评分和评分数量，最后组成字典返回即可。以下是修改后的代码示例：

@films_bp.route('/avg_rating_num_by_avg_rates', methods=['GET'])
def avg_rating_num_by_avg_rates():
    try:
        conn = pymysql.Connection(
            host='localhost',
            port=3306,
            user='root',
            passwd='21591314cxk'
        )
        cur = conn.cursor()
        cur.execute('SELECT rate, COUNT(*) as rating_num FROM movie.movieinfo WHERE rating_num > 500 GROUP BY rate')
        rows = cur.fetchall()
        conn.commit()
        cur.close()
        result = {}
        for row in rows:
            result[row[0]] = row[1]
        return jsonify(result)
    except Exception as e:
        logging.error('Error occurred while retrieving students from the database. Error message: {}'.format(str(e)))
        return jsonify({"error": str(e)})

这样修改后，查询结果会按照评分进行分组，返回的字典中每个键值对对应一个评分和该评分对应的数量。

请问为什么下列代码会报错，请修改并优化： @staticmethod def good_actors(): actors_info = {} @films_bp.route('/good_actors', methods=['GET']) def good_actors(): try: result = FilmInfo.good_actors() #直接调用Fileinfo类中构建的静态方法good_actors() logging.info('获得各年份电影数量序列') return jsonify(result) except Exception as e: logging.error('Error occurred while retrieving students from the database. Error message: {}'.format(str(e))) return jsonify({"error": str(e)}) # 查询满足条件的电影信息 films = FilmInfo.query.all() for film in films: actors_str = film.actor.strip("[]") # 去除方括号和单引号 actors = [actor.strip("'") for actor in actors_str.split(",")] # 统计演员参与的电影数量 for actor in actors: info = actors_info.get(actor, {'total_films': 0, 'best_films': 0, 'worst_films': 0, 'average_films': 0}) info['total_films'] += 1 actors_info[actor] = info # 判断电影评分并更新好片、烂片和一般片数量 if film.rate is not None: for actor in actors: info = actors_info.get(actor, {'total_films': 0, 'best_films': 0, 'worst_films': 0, 'average_films': 0}) if film.rate >= 8.5: info['best_films'] += 1 elif film.rate < 6: info['worst_films'] += 1 else: info['average_films'] += 1 actors_info[actor] = info # 计算好片占比并排序 good_actors = sorted(actors_info.items(), key=lambda x: x[1]['best_films'] / x[1]['total_films'], reverse=True)[:10] return good_actors

这段代码会报错的原因是，定义了两个同名的函数good_actors()，一个是FilmInfo类中的静态方法，另一个是films_bp蓝图中的视图函数。这会导致视图函数覆盖了静态方法，导致程序无法正确执行。另外，在视图函数中查询了所有电影信息，这可能会导致性能问题。

可以将FilmInfo类中的静态方法名称修改为其他名称，如get_good_actors()。然后在视图函数中调用该方法，获取演员信息。同时，可以通过在查询时指定只查询演员信息和评分信息来优化查询性能。最终代码如下：

@films_bp.route('/good_actors', methods=['GET'])
def get_good_actors():
    try:
        actors_info = {}
        # 查询演员参与的电影数量
        films = db.session.query(FilmInfo.actor, FilmInfo.rate).filter(FilmInfo.actor != None).all()
        for film in films:
            actors_str = film.actor.strip('[]')
            # 去除方括号和单引号
            actors = [actor.strip("'") for actor in actors_str.split(',')]
            # 统计演员参与的电影数量
            for actor in actors:
                info = actors_info.get(actor, {'total_films': 0, 'best_films': 0, 'worst_films': 0, 'average_films': 0})
                info['total_films'] += 1
                actors_info[actor] = info
            # 判断电影评分并更新好片、烂片和一般片数量
            if film.rate is not None:
                for actor in actors:
                    info = actors_info.get(actor, {'total_films': 0, 'best_films': 0, 'worst_films': 0, 'average_films': 0})
                    if film.rate >= 8.5:
                        info['best_films'] += 1
                    elif film.rate < 6:
                        info['worst_films'] += 1
                    else:
                        info['average_films'] += 1
                    actors_info[actor] = info
        # 计算好片占比并排序
        good_actors = sorted(actors_info.items(), key=lambda x: x[1]['best_films'] / x[1]['total_films'], reverse=True)[:10]
        return jsonify(good_actors)
    except Exception as e:
        logging.error('Error occurred while retrieving students from the database. Error message: {}'.format(str(e)))
        return jsonify({"error": str(e)})

注意，在查询演员信息时，使用了filter()方法指定只查询演员不为空的电影信息，并且只查询演员和评分两个字段，以提高查询性能。