如何解决无法将表名解析为 python sqlite 查询的问题? [英] How do I get around not being able to parse a table name into a python sqlite query?

问题描述

我正在制作一个 python3 程序，它使用带有多个表的 sqlite 数据库，我想创建一个选择器模块以允许我选择从哪个表中提取数据.

I have a python3 program that I'm making which uses a sqlite database with several tables, I want to create a selector module to allow me to chose which table to pull data from.

我发现我不能使用参数替换表名，如下所示，所以我正在寻找一些替代方法来实现这一点.

I have found out that I can't use parameter substitution for a table name as shown bellow, so I'm looking for some alternative methods to accomplish this.

c.execute("SELECT * FROM ? ", DB)

有什么想法吗?

推荐答案

我不知道这是否是 python3 的东西，但这样做似乎最简单:

I don't know if this is a python3 thing but it seems easiest to just do this:

c.execute("SELECT * FROM %s "% tablename)

这篇关于如何解决无法将表名解析为 python sqlite 查询的问题?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！