python - raise err.ProgrammingError("execute() first") pymysql.err.ProgrammingError: execute() first

Question

我有一个返回游标的连接函数，我正在使用 pymysql 连接我的数据库，并且我有另一个文件使用连接函数执行 sql 语句，但是当我获取行时它返回错误raise err.ProgrammingError("execute() first")

这是从我的数据库配置文件扩展而来的连接函数

def kasaa():
    try:
        conn = mysql.connect()
        cursor = conn.cursor(pymysql.cursors.DictCursor)
        return cursor
   except Exception as e:
      print(e)

但是当我在kasaa()内部调用一个员工类时，它首先引发了一个execute()错误

def get_multiple_info(self,employees_ids):
    """Get info of multiple Employees in database"""
    try:
        for employee_id in employees_ids:
            kasaa().execute(
                 "SELECT * FROM ospos_employees WHERE ospos_employees.deleted = 0 AND ospos_employees.person_id = %s ",
                 employee_id
             )
            row = kasaa().fetchone()
            return row()
    except Exception as e:
        print(e)
    finally:
        kasaa().close()

标签： pythonpython-3.x

Answer 1

中的代码kasaa()执行了两次，所以fetchone()实际上是在没有execute(). 获取cursor并使用它来调用其他函数

def get_multiple_info(self, employees_ids):
    """Get info of multiple Employees in database"""
    cursor = None
    try:
        for employee_id in employees_ids:
            cursor = kasaa()
            cursor.execute(
                "SELECT * FROM ospos_employees WHERE ospos_employees.deleted = 0 AND ospos_employees.person_id = %s ",
                employee_id
            )
            row = cursor.fetchone()
            return row()
    except Exception as e:
        print(e)
    finally:
        cursor.close()