Question

尝试运行一些没有运气的SQL，尝试从条件为真的两个表中选择数据。

表格是driver_details和位置。两个表都有user_id列，我想根据部分之间匹配的user_id从两者中获取数据。（该选择语句有效并返回ID）;

SELECT driver_details.firstName,
                    locations.lat,
                    locations.lng
                    FROM driver_details
                    INNER JOIN locations
                    WHERE user_id = 
                    (SELECT user_id FROM locations WHERE 
                    (lat BETWEEN 0 AND 5) AND 
                    (lng BETWEEN 0 AND 5))

我收到错误：致命错误：在第33行的C：\ xampp \ htdocs \ vector \ www \ scripts \ getDriversInRange.php中的非对象上调用成员函数bind_param（）

Answer 1

您需要使用两个表中常见的SOMETHING来关联正在联接的2个表，这些表用于连接它们...您提到user_id

...
FROM driver_details
INNER JOIN locations
ON driver_details.user_id = locations.user_id
WHERE ...

Answer 2

SELECT d.firstName,
    l.lat,
    l.lng
FROM driver_details d
    INNER JOIN locations l ON d.user_id = l.user_id
WHERE l.lat BETWEEN 0 AND 5
    AND l.lng BETWEEN 0 AND 5

这基本上是BWS发布的更完整的例子。

从条件为真的多个表中进行选择

2 个答案: