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php - MySQL 和 PHP - 不是唯一的表/别名

转载 作者:IT老高 更新时间:2023-10-28 23:49:36 25 4
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我收到下面列出的以下错误,想知道如何解决此问题。

Not unique table/alias: 'grades'

这是我认为给我带来问题的代码。

function getRating(){
$dbc = mysqli_connect ("localhost", "root", "", "sitename");

$page = '3';

$sql1 = "SELECT COUNT(*)
FROM articles_grades
WHERE users_articles_id = '$page'";

$result = mysqli_query($dbc,$sql1);

if (!mysqli_query($dbc, $sql1)) {
print mysqli_error($dbc);
return;
}

$total_ratings = mysqli_fetch_array($result);

$sql2 = "SELECT COUNT(*)
FROM grades
JOIN grades ON grades.id = articles_grades.grade_id
WHERE articles_grades.users_articles_id = '$page'";

$result = mysqli_query($dbc,$sql2);

if (!mysqli_query($dbc, $sql2)) {
print mysqli_error($dbc);
return;
}

$total_rating_points = mysqli_fetch_array($result);
if(!empty($total_rating_points) && !empty($total_ratings)){
// set the width of star for the star rating
$rating = (round($total_rating_points / $total_ratings,1)) * 10;
echo $rating;
} else {
$rating = 100;
echo $rating;
}
}

最佳答案

问题似乎出在这里:

SELECT COUNT(*) 
FROM grades
JOIN grades ON grades.id = articles_grades.grade_id
WHERE articles_grades.users_articles_id = '$page'"

您正在尝试将成绩表连接到自身。您可能打算加入 articles_grades。

关于php - MySQL 和 PHP - 不是唯一的表/别名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2077355/

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