Question

我正在努力做到以下。

从名为hospital的表中选择id = hid的医院。为它添加一个值“overall_rating”并获取所有评级并从另一个名为hrating的表中获取平均值

这是我的查询

 $statement = $conn->prepare('SELECT hospital.*,(SELECT AVG(hrating.rating_h) FROM hrating WHERE hid = hospital.hid) as overall_rating WHERE hid=:hid LIMIT 1');

获取此错误

{"error":"SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE hid='44' LIMIT 1' at line 1"}

我错在哪里。

Answer 1

您的查询中似乎没有“FROM hospital”位？

SELECT hospital.*,(SELECT AVG(hrating.rating_h) 
                   FROM hrating 
                   WHERE hid = hospital.hid) as overall_rating 
FROM hospital -- this line seems to be missing ??
WHERE hid=:hid LIMIT 1

Answer 2

试试这个：

选择医院。*，temp.overall_rating 来自医院 LEFT JOIN（选择隐藏AVG（rating_h）作为overall_rating 来自hrating的hrating组）临时
在hid = hospital.hid 在hospital.hid =：hid LIMIT 1

获取另一个表php中特定行的一个表的平均值

2 个答案: