荒谬简单的 sql 中的 SQL 语法错误？ [关闭]答案

【问题标题】：SQL syntax error in absurdly simple sql? [closed]荒谬简单的 sql 中的 SQL 语法错误？ [关闭]
【发布时间】：2015-11-13 06:29:45
【问题描述】：

调用：

INSERT IGNORE INTO standard_group (group_name, subject, grade, country, state) VALUES (%(group_name), %(subject), %(grade), %(country), %(state));

与

('Arizona Social Studies Standards', 'Social Studies', '1', 'United States', 'AZ')

例如：

    print group_columns
    print group_sql
    print group
    cur.execute(group_sql, dict(zip(group_columns, group)))

但我总是得到：

['group_name', 'subject', 'standard_category', 'standard_sub_category', 'standard_name', 'standard_description', 'grade', 'country', 'state', 'is_common_core', 'is_state_standard']
INSERT IGNORE INTO standard_group (group_name, subject, grade, country, state) VALUES (%(group_name), %(subject), %(grade), %(country), %(state));
('Arizona Social Studies Standards', 'Social Studies', '1', 'United States', 'AZ')
Traceback (most recent call last):
  File "standards-import/simport.py", line 167, in <module>
    cur.execute(group_sql, dict(zip(group_columns, group)))
  File "/Library/Python/2.7/site-packages/mysql/connector/cursor.py", line 507, in execute
    self._handle_result(self._connection.cmd_query(stmt))
  File "/Library/Python/2.7/site-packages/mysql/connector/connection.py", line 722, in cmd_query
    result = self._handle_result(self._send_cmd(ServerCmd.QUERY, query))
  File "/Library/Python/2.7/site-packages/mysql/connector/connection.py", line 640, in _handle_result
    raise errors.get_exception(packet)
mysql.connector.errors.ProgrammingError: 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '%(group_name), %(subject), %(grade), %(country), %(state))' at line 1

我已经尝试使用命名参数? 和%s，然后得到

并非所有参数都使用
命名参数的参数数量错误
SQL 语法

发生了什么事？？我错过了什么？？？

【问题讨论】：

当您使用%(group_name) 时，替换不应该在字典而不是列表中吗？
dict(zip(group_columns, group)) 将它们变成一个字典（具有正确的键名）。其他问题仍然存在，因为我也不能使用 list + %s 或 ? 格式。
你能粘贴你的python脚本以便我们查看缩进等吗？
源码：pastebin.com/8SwsuNF6

标签： python mysql prepared-statement named-parameters

【解决方案1】：

它看起来像一个小错字。您忘记在参数中添加s

INSERT IGNORE INTO standard_group (group_name, subject, grade, country, state) VALUES (%(group_name)s, %(subject)s, %(grade)s, %(country)s, %(state)s);