從另一個表中的特定行獲取一個表的平均值php

從名爲醫院的表中選擇id = hid的醫院。從命名爲hrating

這裏另一個表中添加值「overall_rating」給它，並得到所有的收視率，並利用它的魅力是我的查詢

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

收到此錯誤

{"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"}

我在哪裏錯了。

來源

2014-02-08 Muhammad Umar

看起來你的查詢中沒有「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

來源

2014-02-08 06:39:06 jwwishart

試試這個：

SELECT *醫院，temp.overall_rating 從醫院 LEFT JOIN（SELECT藏AVG（rating_h）由HID overall_rating FROM hrating組）對HID溫度
= hospital.hid WHERE hospital.hid =：hid LIMIT 1

來源

2014-02-08 06:45:42 shola

從另一個表中的特定行獲取一個表的平均值php

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