Question

我需要处理scrapy的输出并使用mysqldb将其存储到mysql。下面的代码适用于文章表但是我需要访问category_id以作为外键插入articlecategory表。项目[＆＃39;频道＆＃39;]是一个包含类别的列表。但是，下面的输出只返回id：1，它不会退出。

def process_item(self, item, spider):
    try:
        self.cursor.execute(\
            "insert into article (url, pid , date)\
            values (%s, %s, %s)",
            (item['url'], item['pid'], item['pub_date'])
            )
        article_id = self.cursor.lastrowid           
        self.conn.commit()

        for channel in item['channels']:
            self.cursor.execute("insert ignore into category(category_name) values(%s)",([channel]))                
            self.conn.commit()          

            category_id = self.cursor.execute("select id from category where category_name = %s", ([channel]))
            with open("cat.txt", "a") as f:
                f.write("id: %s \n" %  (category_id))

如何访问与类别名称匹配的类别ID？该语法适用于带类别表的插入操作。

Answer 1

解决了这个问题，不得不做fetchall（）

self.cursor.execute("select id from category where category_name = %s", ([channel]))
            category_id = self.cursor.fetchall()
            with open("cat.txt", "a") as f:
                f.write("id: %s \n" %  (category_id[0]))

使用Select Statement for MySQLdb Python的问题

1 个答案: