PHP＆MySQL：mysqli_num_rows（）期望參數1是mysqli_result，布爾給定

我想集成HTML淨化器http://htmlpurifier.org/來篩選我的用戶提交的數據，但我在下面得到以下錯誤。我想知道如何解決這個問題？PHP＆MySQL：mysqli_num_rows（）期望參數1是mysqli_result，布爾給定

我收到以下錯誤消息。

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) {

這裏是php代碼。

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; 
} 

}

來源

2010-03-30 TaG

正如我確信你會從給出的答案中看到，這個錯誤與htmlpurifier無關。我將從您的問題中刪除該標記並更新問題標題以反映此問題。 – 2010-03-30 15:18:08

$dbc返回false。您的查詢中有錯誤：

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");

此問題的解決方法一般由Raveren描述。

來源

2010-03-30 15:17:10

短語「您的查詢中有錯誤」爲我完成了這項工作。謝謝！ – 2017-02-01 06:15:20

查詢不返回行或錯誤，因此返回FALSE。將其更改爲

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

mysqli_num_rows：

返回值

成功返回TRUE或FALSE的失敗。對於SELECT，SHOW，DESCRIBE或 EXPLAIN mysqli_query（）將返回一個結果對象。

來源

2010-03-30 15:15:00 raveren

如果沒有行返回，查詢將會成功（0條記錄不會使mysqli_query或mysqli_num_rows返回false）。 – 2017-12-27 15:08:04

PHP＆MySQL：mysqli_num_rows（）期望參數1是mysqli_result，布爾給定

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