PHP＆MySQL：mysqli_num_rows（）期望参数1是mysqli_result，布尔给定

我想集成HTML净化器http://htmlpurifier.org/来筛选我的用户提交的数据，但我在下面得到以下错误。我想知道如何解决这个问题？PHP＆MySQL：mysqli_num_rows（）期望参数1是mysqli_result，布尔给定

我收到以下错误消息。

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) {

这里是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; 
} 

}

来源

2010-03-30 TaG

正如我确信你会从给出的答案中看到，这个错误与htmlpurifier无关。我将从您的问题中删除该标记并更新问题标题以反映此问题。 – 2010-03-30 15:18:08

$dbc返回false。您的查询中有错误：

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");

此问题的解决方法一般由Raveren描述。

来源

2010-03-30 15:17:10

短语“您的查询中有错误”为我完成了这项工作。谢谢！ – 2017-02-01 06:15:20

查询不返回行或错误，因此返回FALSE。将其更改为

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

mysqli_num_rows：

返回值

成功返回TRUE或FALSE的失败。对于SELECT，SHOW，DESCRIBE或 EXPLAIN mysqli_query（）将返回一个结果对象。

来源

2010-03-30 15:15:00 raveren

如果没有行返回，查询将会成功（0条记录不会使mysqli_query或mysqli_num_rows返回false）。 – 2017-12-27 15:08:04

PHP＆MySQL：mysqli_num_rows（）期望参数1是mysqli_result，布尔给定

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