SELECT AVG(列)的MySQL PHP返回资源ID 3
我想得到我的评论表的评分栏的平均值。 这就是迄今为止所做的事情SELECT AVG(列)的MySQL PHP返回资源ID 3,php,mysql,html,Php,Mysql,Html,我想得到我的评论表的评分栏的平均值。 这就是迄今为止所做的事情 $averagereviewquery = "SELECT AVG(review_rating) from reviews where product_id=".$primarykey; $averagereviewresult=mysql_query($averagereviewquery); if($insert_review){ //successful inse
$averagereviewquery = "SELECT AVG(review_rating) from reviews where product_id=".$primarykey;
$averagereviewresult=mysql_query($averagereviewquery);
if($insert_review){ //successful insert
echo $averagereviewresult;
}
发生的情况是averagereviewresult返回“资源Id 3”。我该如何更改它以显示平均值?mysql\u查询会返回一个资源。为了得到结果,您需要从查询中快速读取值。一种方法是使用:
您没有获取实际数据 $result=mysql\u result($averagereviewquery,0)
您需要先将结果提取到数组中
$averagereviewquery = "SELECT AVG(review_rating) from reviews where product_id=".$primarykey;
$averagereviewresult=mysql_query($averagereviewquery);
if($insert_review){ //successful insert
$row = mysql_fetch_array($averagereviewresult);
echo $row[AVG(review_rating)];
}
SELECT AVG(review_rating) AS review_rating_avg ...
$averagereviewquery = "SELECT AVG(review_rating) from reviews where product_id=".$primarykey;
$averagereviewresult=mysql_query($averagereviewquery);
if($insert_review){ //successful insert
$row = mysql_fetch_array($averagereviewresult);
echo $row[AVG(review_rating)];
}