我需要一个评级系统,它显示 php 中的平均评级。我完成了评级过程(保存和更新过程)。我只需要在 php 中显示平均评分(使用 php-Ajax 评分系统)。
当我从数据库中检索数据时,我遇到了错误。代码是这样的:
<?php
$con = mysql_connect("localhost","root","");
if(!$con){
echo "Connection to the Database Console was Unsuccessful";
}
$select = mysql_select_db("oilandgas13",$con);
if(!$select){
echo "Connection to the Database was Unsuccessful";
}
$add_coun= "SELECT sum(rating) sum, count(id) count from comments WHERE item_id = $itemID AND status=1";
$result = mysql_query($add_coun,$con);
if(!$result)
{
echo "query was not successfully";
}
$result = mysql_fetch_object($result);
$sum = $result->sum;
$count = $result->count;
$rating = $sum / $count;
echo $rating;
?>
我收到这样的错误:
警告:mysql_fetch_object():提供的参数不是第 19 行 C:\wamp\www\final work_apr51\final work_apr51\calculation.php 中的有效 MySQL 结果资源
警告:第 23 行 C:\wamp\www\final work_apr51\final work_apr51\calculation.php 中除以零