这些查询可以浓缩成2,甚至更好的1查询吗?
查询1
$stmt = $conn->
prepare("
SELECT providers.Provider_ID, providers.Description, reviews.Date, reviews.F_Name, reviews.S_Name, reviews.Website, reviews.Rating, reviews.Message, reviews.Email_Address
FROM providers
INNER JOIN reviews ON reviews.Provider=providers.Provider_ID
WHERE providers.Provider_ID LIKE CONCAT('%',:provider,'%') LIMIT $set_limit
");
$stmt->bindParam(':provider', $provider, PDO::PARAM_STR);
$stmt->execute();
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);查询2
$stmt = $conn->
prepare("
SELECT ROUND(AVG(rating),1) as AVG_Rating
FROM reviews
WHERE Provider LIKE CONCAT('%',:provider,'%')
");
$stmt->bindParam(':provider', $provider, PDO::PARAM_STR);
$stmt->execute();
$get_average = $stmt->fetchColumn(); 查询3
$stmt = $conn->
prepare("
SELECT COUNT(*)
FROM reviews
WHERE Provider LIKE CONCAT('%',:provider,'%')
");
$stmt->bindParam(':provider', $provider, PDO::PARAM_STR);
$stmt->execute();
$total_rows = $stmt->fetchColumn();发布于 2014-03-31 21:35:25
您正在将不同粒度级别的结果结合在一起。一个在提供者级别,另一个在审查级别。因此,您需要带有摘要信息的评论。您可以通过加入聚合结果来获得这个结果:
SELECT p.Provider_ID, p.Description, r.Date, r.F_Name, r.S_Name,
r.Website, r.Rating, r.Message, r.Email_Address,
tsum.AVG_Rating, tsum.cnt
FROM providers p INNER JOIN
reviews r
ON r.Provider = p.Provider_ID CROSS JOIN
(SELECT Provider, ROUND(AVG(rating),1) as AVG_Rating, COUNT(*) as cnt
FROM reviews
WHERE Provider LIKE CONCAT('%',:provider,'%')
) tsum
WHERE p.Provider_ID LIKE CONCAT('%',:provider,'%')
LIMIT $set_limit发布于 2014-03-31 21:40:04
查询2和查询3可以很容易地组合成一个查询:
SELECT ROUND(AVG(rating),1) as AVG_Rating
, COUNT(*) AS count_
FROM reviews
WHERE Provider LIKE CONCAT('%',:provider,'%')但是您将无法使用fetchColumn(),您需要获取行并访问每个列。
此查询(如查询2和查询3)返回一行。查询1有可能返回0、1或更多行,因此组合该查询会有一些问题。可以在每一行返回AVG_Rating和count_,但是处理结果集的代码就不那么简单了。
https://stackoverflow.com/questions/22772541
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