Question

我有以下mysql查询，

SELECT COUNT(*) as pg_total 
FROM products p , prod_sub s 
LEFT JOIN 
    (SELECT io.uni_id, count(io.uni_id) AS popular 
    FROM items_ord io GROUP BY io.uni_id ) as z 
ON z.uni_id = p.uni_id 
WHERE p_stock != 9 
AND p.cat_id = 'ct_54a2e9fed3689' 
AND p.uni_id = s.uni_id AND s.sub_id IN ('sb_54a2ea814a64d') 
GROUP BY p.uni_id 
ORDER BY z.popular DESC

但是这个查询工作正常，

SELECT COUNT(*) as pg_total 
FROM products p 
LEFT JOIN 
    (SELECT io.uni_id, count(io.uni_id) AS popular 
    FROM items_ord io GROUP BY io.uni_id ) z 
ON z.uni_id = p.uni_id
WHERE p_stock != 9 
AND p.cat_id = 'ct_54a2e9fed3689' 
ORDER BY z.popular DESC

此查询抛出Unknown column 'p.uni_id' in 'on clause'。我找不到问题。

Answer 1

您错过了as

中的FROM items_ord io GROUP BY io.uni_id ) z

SELECT COUNT(*) as pg_total 
FROM products p 
LEFT JOIN 
    (SELECT io.uni_id, count(io.uni_id) AS popular 
    FROM items_ord io GROUP BY io.uni_id ) as z 
ON z.uni_id = p.uni_id
WHERE p_stock != 9 
AND p.cat_id = 'ct_54a2e9fed3689' 
ORDER BY z.popular DESC

Answer 2

运行此代码并复制输出并在sql中过去并运行，您将获得哪个列未知

$sql = "SELECT COUNT(*) as pg_total 
FROM products p , prod_sub s 
LEFT JOIN 
    (SELECT io.uni_id, count(io.uni_id) AS popular 
    FROM items_ord io GROUP BY io.uni_id ) as z 
ON z.uni_id = p.uni_id 
WHERE p_stock != 9 
AND p.cat_id = 'ct_54a2e9fed3689' 
AND p.uni_id = s.uni_id AND s.sub_id IN ('sb_54a2ea814a64d') 
GROUP BY p.uni_id 
ORDER BY z.popular DESC";
echo $sql; exit;

mysql查询未知库仑错误

2 个答案: