我收到下面列出的錯誤,想知道我該如何解決這個問題。MySQL&PHP - 非唯一表/別名
Not unique table/alias: 'grades'
這是我認爲給我的問題的代碼。
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;
}
}
ON條件表明它是一個錯字 – VolkerK 2010-01-16 12:45:41
這似乎是一個很舊的答案,但我被卡在查詢我認爲'JOIN等級'應該''JOIN articles_grades' – 2016-01-19 07:06:09