#1 修改测试用例中book_id生成方式,避免book_id重复

def get_book_info(self, start, size) -> [Book]:
        books = []
        conn = sqlite.connect(self.book_db)
        cursor = conn.execute(
            "SELECT id, title, author, "
            "publisher, original_title, "
            "translator, pub_year, pages, "
            "price, currency_unit, binding, "
            "isbn, author_intro, book_intro, "
            "content, tags, picture FROM book ORDER BY id "
            "LIMIT ? OFFSET ?", (size, start))
        for row in cursor:
            book = Book()
            book.id = row[0]
            book.title = row[1]
            book.author = row[2]
            book.publisher = row[3]
            book.original_title = row[4]
            book.translator = row[5]
            book.pub_year = row[6]
            book.pages = row[7]
            book.price = row[8]
            book.currency_unit = row[9]
            book.binding = row[10]
            book.isbn = row[11]
            book.author_intro = row[12]
            book.book_intro = row[13]
            book.content = row[14]
            tags = row[15]

            picture = row[16]

            for tag in tags.split("\\n"):
                if tag.strip() != "":
                    book.tags.append(tag)
            for i in range(0, random.randint(0, 9)):
                if picture is not None:
                    encode_str = base64.b64encode(picture).decode('utf-8')
                    book.pictures.append(encode_str)
            books.append(book)
            # print(tags.decode('utf-8'))

            # print(book.tags, len(book.picture))
            # print(book)
            # print(tags)

        return books

book_db = book.BookDB()
self.books = book_db.get_book_info(0, 2)

# book.id = row[0]
book.id = row[0] + '_' + str(str(uuid.uuid1()))

def book_id_exist(self, store_id, book_id):
        cursor = self.conn.execute("SELECT book_id FROM store WHERE store_id = ? AND book_id = ?;", (store_id, book_id))
        row = cursor.fetchone()
        if row is None:
            return False
        else:
            return True

根据文档设计可知，"商家添加书籍信息"接口要求对Book ID的唯一性作判断，如果book_id已存在则视为错误并返回对应的错误码 (见be/doc/seller.md) 码 | 描述 --- | --- 200 | 添加图书信息成功 5XX | 卖家用户ID不存在 5XX | 商铺ID不存在 5XX | 图书ID已存在 但在该接口的测试用例中，每次测试所使用的图书ID都是一样的，可以看到测试用例中生成输入数据(图书信息)的代码: ```python def get_book_info(self, start, size) -> [Book]: books = [] conn = sqlite.connect(self.book_db) cursor = conn.execute( "SELECT id, title, author, " "publisher, original_title, " "translator, pub_year, pages, " "price, currency_unit, binding, " "isbn, author_intro, book_intro, " "content, tags, picture FROM book ORDER BY id " "LIMIT ? OFFSET ?", (size, start)) for row in cursor: book = Book() book.id = row[0] book.title = row[1] book.author = row[2] book.publisher = row[3] book.original_title = row[4] book.translator = row[5] book.pub_year = row[6] book.pages = row[7] book.price = row[8] book.currency_unit = row[9] book.binding = row[10] book.isbn = row[11] book.author_intro = row[12] book.book_intro = row[13] book.content = row[14] tags = row[15] picture = row[16] for tag in tags.split("\\n"): if tag.strip() != "": book.tags.append(tag) for i in range(0, random.randint(0, 9)): if picture is not None: encode_str = base64.b64encode(picture).decode('utf-8') book.pictures.append(encode_str) books.append(book) # print(tags.decode('utf-8')) # print(book.tags, len(book.picture)) # print(book) # print(tags) return books book_db = book.BookDB() self.books = book_db.get_book_info(0, 2) ``` 可以发现每次使用的都是book.db中id最小的两种书，即id分别为1000067和1000134的《美丽心灵》和《三毛流浪记全集》。 这样一来，由于每次测试用的是相同的图书ID，必须在每次测试前都清空数据库才能保证没有重复的图书id，否则所有测试都将返回错误（book_id已存在）。但每次测试前都清空数据库是不太现实的，这会大大增加测试的代价。 因此我认为正确的测试方法时，每次给输入图书ID也加上一个通过uuid随机生成的后缀，**保证每次测试使用的图书id不一样**，例如修改上述代码中生成book_id的部分，即修改access/book.py ```python # book.id = row[0] book.id = row[0] + '_' + str(str(uuid.uuid1())) ``` 经过测试，按照该方法修改后，参考代码仍然能通过全部测试用例。 另外，参考代码能够通过原测试用例的原因是：其检查book_id是否存在的逻辑是要求同一组store_id和book_id重复出现 ```python def book_id_exist(self, store_id, book_id): cursor = self.conn.execute("SELECT book_id FROM store WHERE store_id = ? AND book_id = ?;", (store_id, book_id)) row = cursor.fetchone() if row is None: return False else: return True ``` 即同时对store_id和book_id同时进行检查，只有该store_id和book_id同时存在才会返回错误。 这样确实可以通过所有的测试用例。但这与文档描述不符，根据文档说明，即使两次添加的store_id不同，只要book_id存在（某一个book_id曾被添加过）就应该返回错误。