[英]mysql subquery with inner join and limit
我有兩個帶有列的表recipes_sa
:
recipes_id recipes_name recipes_chef
---------- ------------ ------------
和帶有列的chefs_sa
:
chefs_id chefs_name
-------- ----------
而且我想使用INNER JOIN
和LIMIT
獲得有限數量的食譜及其廚師詳細信息
我做了以下功能:
function getLimitJoinData($data, $tbls, $ids, $abr, $type, $limit) {
$dataToSelect = implode($data, ',');
$q = "SELECT $dataToSelect";
$q.= " FROM (SELECT * FROM $tbls[0] LIMIT $limit) $abr";
for ($i=1; $i < count($tbls); $i++) {
$q .= " ".$type." JOIN ". $tbls[$i] ." ON " . $abr.'.recipes_chef' .' = '. $ids[$i-1][0];
}
}
和查詢是這樣的
SELECT chefs_sa.chefs_name,
recipes_sa.recipes_name
FROM (SELECT * FROM recipes_sa LIMIT 8) rec
INNER JOIN chefs_sa ON rec.recipes_chef = chefs_sa.chefs_id
但是,當我運行查詢時,出現以下警告:
警告:PDO :: query():SQLSTATE [42S22]:找不到列:1054未知列'recipes_sa.recipes_name'我不明白為什么
我在recipes_sa
recipes_name
中有“ recipes_name
recipes_sa
”列, recipes_sa
我了解,數據庫首先運行“ 內部查詢 ”(有限制的查詢 ),然后如何找不到“配方名稱”列!
另一種執行方法是按配方排序,然后限制為最新的8,而不是使用子查詢:
SELECT cs.chefs_name, rs.recipes_name
FROM recipes_sa rs
INNER JOIN chefs_sa cs ON rs.recipes_chef = cs.chefs_id
ORDER BY rs.recipes_name ASC LIMIT 8
你已經化名recipes_sa
AS rec
。 使用以下內容:
SELECT chefs_sa.chefs_name,
rec.recipes_name
FROM (SELECT * FROM recipes_sa LIMIT 8) rec
INNER JOIN chefs_sa ON rec.recipes_chef = chefs_sa.chefs_id
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