Mysql查询，获取avg结果

如何解决Mysql查询，获取avg结果

| 所以我得到了这个查询，是否可以从avgscore获取MysqL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

提前致谢 整个声明：

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不产生任何数据。     

解决方法

        好的，让我们再进一步...通过左连接，这意味着并非所有的狗诊所都具有“得分”等级...如果是这样，则该值为NULL并可能导致其窒息。

SELECT
      dc.*,COALESCE( s.avgscore,0 ) FinalAvg
   FROM 
      dog_clinic dc
         LEFT JOIN ( SELECT dcs.input_id,AVG(dcs.score) AS avgscore
                        FROM 
                           dog_clinic_score dcs
                        GROUP BY 
                           dcs.input_id) s 
            ON dc.id = s.input_id
   ORDER BY 
      COALESCE( s.avgscore,0 ) DESC,dc.visited DESC 
   LIMIT 
      0,10;

    ,        您的意思是：

SELECT dog_clinic.*,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;

我认为Rahul是正确的，它隐含为

s.avgscore DESC,dog_clinic.visited DESC

，我也将其集成到查询中。