警告问题：期望参数 1 为 mysqli_result [重复]

Posted 2023-02-24

【中文标题】警告问题：期望参数 1 为 mysqli_result [重复]

【英文标题】：warning problem: expects parameter 1 to be mysqli_result [duplicate]

【发布时间】：2011-01-05 20:14:09

【问题描述】：

我收到下面列出的以下警告，我想知道如何解决它

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

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)';
    

【参考方案1】：

mysqli_query() 返回 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();

请参阅此链接以获取 mysqli_query 参考 http://php.net/manual/en/mysqli.query.php

【参考方案2】：

瀑布可能是对的。修改你的代码如下：

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

PS：只是想知道$page 到底是什么？你忘了做一个：

global $page;