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php – warning problem: expects parameter 1 to be mysqli_result

Posted by: admin July 12, 2020 Leave a comment


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

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given on line 65

The code is around this section of PHP code listed below. I can list the full code if needed.

// function to retrieve average and votes
function getRatingText(){
    $dbc = mysqli_connect ("localhost", "root", "", "sitename");
    $sql1 = "SELECT COUNT(*) 
             FROM articles_grades 
             WHERE users_articles_id = '$page'";

    $result = mysqli_query($dbc,$sql1);
    $total_ratings = mysqli_fetch_array($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);
    $total_rating_points = mysqli_fetch_array($result);
    if (!empty($total_rating_points) && !empty($total_ratings)){
        $avg = (round($total_rating_points / $total_ratings,1));
        $votes = $total_ratings;
        echo $avg . "/10  (" . $votes . " votes cast)";
    } else {
        echo '(no votes cast)';
How to&Answers:

mysqli_query() returns FALSE if there was an error in the query. So you should test for it…

/* Select queries return a resultset */
if ($result = mysqli_query($dbc, "SELECT Name FROM City LIMIT 10")) {
    printf("Select returned %d rows.\n", $result->num_rows);

    /* free result set */

See this link for the mysqli_query reference


Waterfall is probably right. Revise your code as follows:

$result = mysqli_query($dbc,$sql1) or die(mysqli_error($dbc));
// and
$result = mysqli_query($dbc,$sql2) or die(mysqli_error($dbc));

PS: Just wondering what exactly is $page? Did you forget to do a:

global $page;