Py学习  »  DATABASE

我不断得到这个…警告:mysqli_fetch_assoc()期望参数1是mysqli_result[duplicate]

Alan • 5 年前 • 1996 次点击  

我正在尝试集成HTML净化器 http://htmlpurifier.org/ 过滤我的用户提交的数据,但我得到以下错误。我在想我该怎么解决这个问题?

我得到以下错误。

on line 22: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given 

第22行是。

if (mysqli_num_rows($dbc) == 0) {

这是php代码。

if (isset($_POST['submitted'])) { // Handle the form.

    require_once '../../htmlpurifier/library/HTMLPurifier.auto.php';

    $config = HTMLPurifier_Config::createDefault();
    $config->set('Core.Encoding', 'UTF-8'); // replace with your encoding
    $config->set('HTML.Doctype', 'XHTML 1.0 Strict'); // replace with your doctype
    $purifier = new HTMLPurifier($config);


    $mysqli = mysqli_connect("localhost", "root", "", "sitename");
    $dbc = mysqli_query($mysqli,"SELECT users.*, profile.*
                                 FROM users 
                                 INNER JOIN contact_info ON contact_info.user_id = users.user_id 
                                 WHERE users.user_id=3");

    $about_me = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['about_me']));
    $interests = mysqli_real_escape_string($mysqli, $purifier->purify($_POST['interests']));



if (mysqli_num_rows($dbc) == 0) {
        $mysqli = mysqli_connect("localhost", "root", "", "sitename");
        $dbc = mysqli_query($mysqli,"INSERT INTO profile (user_id, about_me, interests) 
                                     VALUES ('$user_id', '$about_me', '$interests')");
}



if ($dbc == TRUE) {
        $dbc = mysqli_query($mysqli,"UPDATE profile 
                                     SET about_me = '$about_me', interests = '$interests' 
                                     WHERE user_id = '$user_id'");

        echo '<p class="changes-saved">Your changes have been saved!</p>';
}


if (!$dbc) {
        // There was an error...do something about it here...
        print mysqli_error($mysqli);
        return;
}

}
Python社区是高质量的Python/Django开发社区
本文地址:http://www.python88.com/topic/42944
 
1996 次点击  
文章 [ 2 ]  |  最新文章 5 年前
raveren
Reply   •   1 楼
raveren    15 年前

查询要么没有返回行,要么是erroneus,因此 FALSE 被退回。把它改成

if (!$dbc || mysqli_num_rows($dbc) == 0)

mysqli_num_rows :

返回值

成功时返回true或失败时返回false 失败。用于选择、显示、描述或 explain mysqli_query()将返回 结果对象。

Sean Vieira
Reply   •   2 楼
Sean Vieira    15 年前

$dbc 正在返回False。查询中有错误:

SELECT users.*, profile.* --You do not join with profile anywhere.
                                 FROM users 
                                 INNER JOIN contact_info 
                                 ON contact_info.user_id = users.user_id 
                                 WHERE users.user_id=3");

一般来说,raveren已经描述了解决这个问题的方法。