在 SQLite 上连接表时如何进行更新?

How do I make an UPDATE while joining tables on SQLite?(在 SQLite 上连接表时如何进行更新?)
本文介绍了在 SQLite 上连接表时如何进行更新?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我试过了:

UPDATE closure JOIN item ON ( item_id = id ) 
SET checked = 0 
WHERE ancestor_id = 1

还有:

UPDATE closure, item 
SET checked = 0 
WHERE ancestor_id = 1 AND item_id = id

两者都适用于 MySQL,但在 SQLite 中会出现语法错误.

Both works with MySQL, but those give me a syntax error in SQLite.

如何使这个 UPDATE/JOIN 与 SQLite 版本 3.5.9 一起工作?

How can I make this UPDATE / JOIN works with SQLite version 3.5.9 ?

推荐答案

你不能.SQLite 不支持 UPDATE 语句中的 JOIN.

You can't. SQLite doesn't support JOINs in UPDATE statements.

但是,您可以使用子查询来代替:

But, you can probably do this with a subquery instead:

UPDATE closure SET checked = 0 
WHERE item_id IN (SELECT id FROM item WHERE ancestor_id = 1);

或者类似的东西;目前尚不清楚您的架构到底是什么.

Or something like that; it's not clear exactly what your schema is.

这篇关于在 SQLite 上连接表时如何进行更新?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编程学习网!

本站部分内容来源互联网,如果有图片或者内容侵犯您的权益请联系我们删除!

相关文档推荐

FastAPI + Tortoise ORM + FastAPI Users (Python) - Relationship - Many To Many(FastAPI+Tortoise ORM+FastAPI用户(Python)-关系-多对多)
Window functions not working in pd.read_sql; Its shows error(窗口函数在pd.read_sql中不起作用;它显示错误)
(Closed) Leaflet.js: How I can Do Editing Geometry On Specific Object I Select Only?((已关闭)Leaflet.js:如何仅在我选择的特定对象上编辑几何图形?)
Explicit join syntax(显式连接语法)
LEFT JOIN doesnamp;#39;t behave as expected as gives NULLs in MySQL(左联接的行为与MySQL中的空值不同)
in sqlite update trigger with multiple if/Case Conditions(在具有多个IF/CASE条件的SQLite UPDATE触发器中)