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Home/ Questions/Q 562643
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T12:34:34+00:00 2026-05-13T12:34:34+00:00

I get the following error listed below and was wondering how do I fix

  • 0

I get the following error listed below and was wondering how do I fix this problem.

Not unique table/alias: 'grades'

Here is the code I think is giving me the problem.

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;
}
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T12:34:35+00:00Added an answer on May 13, 2026 at 12:34 pm

    The problem seems to be here:

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

    You are trying to join the table grades to itself. You probably meant to join with articles_grades.

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