我收到下面列出的以下警告，我想知道如何修复它
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given on line 65
该代码围绕下面列出的这部分 PHP 代码。如果需要，我可以列出完整的代码。
// 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 (" . $votes . " votes cast)";
} else {
echo '(no votes cast)';
}
}
mysqli_query()回报FALSE如果查询中有错误。所以你应该测试一下...
/* Select queries return a resultset */
if ($result = mysqli_query($dbc, "SELECT Name FROM City LIMIT 10")) {
printf("Select returned %d rows.\n", $result->num_rows);
/* free result set */
$result->close();
}
请参阅此链接了解mysqli_query参考http://php.net/manual/en/mysqli.query.php
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