Question

我试图为我的网站创建一个评论系统，我想从我最近的5条评论中得到一个平均数，第一部分工作正常，如果我删除了AVG我没有得到任何错误只是最后一个数字的值我进入了数据库。一旦我添加了AVG，它就会给我未定义的索引：评级第23行？

任何帮助都将非常感谢继承人代码：

// collect 5 lastest reviews for viewing
$reviews = "";
    $sqli = mysqli_query($conn, "SELECT review, reviewer_name, review_date FROM reviews WHERE review_id='$id' ORDER BY review_date DESC LIMIT 5");
    $reviewCount = mysqli_num_rows($sqli); // count the output amount
    if($reviewCount > 0) {
        //get all the product details
        while($row = mysqli_fetch_array($sqli)){
        $review = $row["review"];
        $reviewer_name = $row['reviewer_name'];
        $review_date = $row['review_date'];
        $reviews .= ''.$review.''.$reviewer_name.''.$review_date.'<br />';
        }
}
$rating = "";
$test = mysqli_query($conn, "SELECT AVG(rating) FROM reviews WHERE review_id='$id' ORDER BY review_date DESC LIMIT 5");
$ratingCount = mysqli_num_rows($test); // count the output amount
    if($ratingCount > 0) {
        while($row = mysqli_fetch_array($test)){
            $rating = $row["rating"];
        }
}

Answer 1

输出中没有名为rating的列。使用列别名：

SELECT AVG(rating) as rating
FROM reviews
WHERE review_id = '$id';

请注意，order by和limit没用，因为没有group by的聚合查询只返回一行。

无法让MYSQL AVG工作

1 个答案: