php 警告问题：期望参数 1 为 mysqli_result
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2077263/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
warning problem: expects parameter 1 to be mysqli_result
提问by tEcHnUt
I get the following warning listed below and I was wondering how do I fix it
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
The code is around this section of PHP code listed below. I can list the full code if needed.
该代码围绕下面列出的 PHP 代码部分。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 (" . $votes . " votes cast)";
} else {
echo '(no votes cast)';
}
}
回答by Michael Waterfall
mysqli_query()returns FALSEif there was an error in the query. So you should test for it...
mysqli_query()FALSE如果查询中有错误则返回。所以你应该测试一下...
/* Select queries return a resultset */
if ($result = mysqli_query($dbc, "SELECT Name FROM City LIMIT 10")) {
printf("Select returned %d rows.\n", $result->num_rows);
/* free result set */
$result->close();
}
See this link for the mysqli_queryreference
http://php.net/manual/en/mysqli.query.php
请参阅此链接以获取mysqli_query参考
http://php.net/manual/en/mysqli.query.php
回答by Salman A
Waterfall is probably right. Revise your code as follows:
瀑布可能是对的。修改您的代码如下：
$result = mysqli_query($dbc,$sql1) or die(mysqli_error($dbc));
// and
$result = mysqli_query($dbc,$sql2) or die(mysqli_error($dbc));
PS: Just wondering what exactly is $page? Did you forget to do a:
PS：只是想知道究竟是$page什么？您是否忘记执行以下操作：
global $page;
