为什么 TypeError: __init__() 需要 1 个位置参数，但在运行 sql 命令时给出了 2 个

Question

import mysql.connector

conn = mysql.connector.connect(host='localhost', user='root', password='root', database='nlidb')
cur = conn.cursor()
stmt = "create table statistics(sno int(5), stat_ticker varchar(10) primary key, marketcap varchar(10), enterprise_value varchar(10), return_on_assets varchar(10), total_cash varchar(10), operating_cash_flow varchar(10), levered_free_cash_flow varchar(10), total_debt varchar(10), current_ratio varchar(10), gross_profit varchar(10), profit_margin varchar(10))"
cur.execute(stmt)

我试图在我的数据库中创建一个表，但它给了我这个错误。

Traceback (most recent call last):
  File "G:/python/NLIDB/nlidb/database.py", line 3, in <module>
    conn = mysql.connector.connect('localhost', user='root', password='root', database='nlidb')
  File "G:\python\NLIDB\venv\lib\site-packages\mysql\connector\__init__.py", line 179, in connect
    return MySQLConnection(*args, **kwargs)
  File "G:\python\NLIDB\venv\lib\site-packages\mysql\connector\connection.py", line 57, in __init__
    super(MySQLConnection, self).__init__(*args, **kwargs)
TypeError: __init__() takes 1 positional argument but 2 were given

有人可以帮我解决这个问题吗？？提前致谢。

Answer 1

从我在错误代码中注意到的

    File "G:/python/NLIDB/nlidb/database.py", line 3, in <module>
    conn = mysql.connector.connect('localhost', user='root', password='root', 
    database='nlidb')"

你缺少传递 conn = mysql.connector.connect( host ='localhost',...)

它也将不会与MySQL模块工作，如果用户您刚刚分配值，密码等变量：

    user = 'root'
    password = 'root'
    host = 'localhost'
    database = 'db'

然后尝试像这样运行一行：

    conn = mysql.connector.connect(host, user, password, database)

但是，您可以先为变量赋值，然后编写以下行：

    conn = mysql.connector.connect(host=host, user=user, password=password, database=database)

希望对某人有所帮助:)