所以我得到了这个查询,是否有可能从avgscore中获取mysql结果?Mysql查询,获取平均结果
SELECT * FROM dog_clinic
LEFT JOIN (SELECT AVG(score) AS avgscore, input_id
FROM dog_clinic_score GROUP BY input_id) s ON s.input_id = dog_clinic.id
ORDER BY avgscore DESC, visited DESC LIMIT 0,10
在此先感谢
全声明:
if ($stmt_dog_clinic = $mysqli->prepare("SELECT dog_clinic.id, dog_clinic.name, dog_clinic.content, dog_clinic.city, dog_clinic.street, dog_clinic.picture, dog_clinic.visited, s.avgscore FROM dog_clinic LEFT JOIN (SELECT AVG(score) AS avgscore, input_id FROM dog_clinic_score GROUP BY input_id) s ON s.input_id = dog_clinic.id ORDER BY s.avgscore DESC, dog_clinic.visited DESC LIMIT 0,10"))
{
$stmt_dog_clinic->bind_result($id, $name, $content, $city, $street, $picture, $visited, $avgscore);
$stmt_dog_clinic->execute();
while ($stmt_dog_clinic->fetch())
{ echo $avgscore; }
$ avgscore dsn't产生的任何数据。
不应该是's.avgscore DESC,dog_clinic.visited DESC'吗? – Rahul
声明似乎是好的,我使用mysqli类,我不知道如何绑定avgscore的结果,我需要它作为php代码的结果。 – Cameleon
你得到的错误信息是什么? – stefgosselin