Sqlite3 - 使用Python代码更新表 - ％s附近的语法错误

This is my Python code -

这是我的Python代码 -

cursor.execute("""UPDATE tasks SET task_owner=%s,task_remaining_hours=%s,                      task_impediments=%s,task_notes=%s WHERE task_id=%s""",                      (new_task_owner,new_task_remaining_hours,new_task_impediments,
                      new_task_notes,task_id))

This is the SQL statement I try in SQLite3 manager (Firefox extension)

这是我在SQLite3管理器中尝试的SQL语句（Firefox扩展）

UPDATE tasks SET task_owner=%s,task_remaining_hours=%d,task_impediments=%s,task_notes=%s WHERE task_id=%d,("sumod",10,"none","test",1)   

The error I get is -

我得到的错误是 -

sqlite3.OperationalError: near "%": syntax error

I have tried many web searches including SO, tutorials and self-troubleshooting, but this error does not go away. What exactly I am doing wrong here.

我尝试了很多网络搜索，包括SO，教程和自我排除故障，但这个错误并没有消失。究竟我在这里做错了什么。

1 个解决方案

#1

I believe Python's SQLite implementation uses ? placeholders, unlike MySQLdb's %s. Review the documentation.

我相信Python的SQLite实现使用？占位符，与MySQLdb的％s不同。查看文档。

cursor.execute("""UPDATE tasks SET task_owner = ? ,task_remaining_hours = ?,task_impediments = ?,task_notes = ? WHERE task_id= ? """,
  (new_task_owner,new_task_remaining_hours,new_task_impediments,new_task_notes,task_id))