“SELECT ... IN（SELECT ...）”在CodeIgniter中查询

本文介绍了“SELECT ... IN（SELECT ...）”在CodeIgniter中查询的处理方法，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

我有类似这样的查询：

SELECT username FROM users WHERE locationid IN (SELECT locationid FROM locations WHERE countryid='$')

如何在CodeIgniter中运行此查询？我无法在CodeIgnite的用户指南中找到解决方案。

How could I run this query in CodeIgniter? I can't find a solution in CodeIgnite's user guide.

非常感谢您的回答！

尊敬的！

推荐答案

查看这里。

基本上你必须做绑定参数：

Basically you have to do bind params:

$sql = "SELECT username FROM users WHERE locationid IN (SELECT locationid FROM locations WHERE countryid=?)"; $this->db->query($sql, '__COUNTRY_NAME__');

但是，和E先生说的一样，使用连接：

But, like Mr.E said, use joins:

$sql = "select username from users inner join locations on users.locationid = locations.locationid where countryid = ?"; $this->db->query($sql, '__COUNTRY_NAME__');