警告问题：期望参数1为mysqli_result [英] warning problem: expects parameter 1 to be mysqli_result

问题描述

可能重复：

mysql_fetch_array（）期望参数1为资源，在选择中给出布尔值

我收到以下列出的警告，我想知道如何解决它

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

代码是下面列出的这一部分PHP代码。如果需要，我可以列出完整的代码。

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)';
    }
}

推荐答案

code> mysqli_query（）如果查询中有错误，则返回 FALSE 。所以你应该测试一下...

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 */
    $result->close();
}

查看 mysqli_query 参考
http://php.net/manual/en/mysqli .query.php

这篇关于警告问题：期望参数1为mysqli_result的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！