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PHP和MySQL：mysqli_num_rows（）期望参数1是mysqli_result，布尔给定[重复]

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36

这个问题在这里已有答案：

我正在尝试集成HTML Purifier http://htmlpurifier.org/来过滤我的用户提交的数据，但我在下面收到以下错误 . 我想知道如何解决这个问题？

我收到以下错误 .

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;
}

}
php mysql

2 回答

  • 32

    查询不返回任何行或是错误的，因此返回 FALSE . 将其更改为

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

    mysqli_num_rows ：

    返回值成功时返回TRUE，失败时返回FALSE . 对于SELECT，SHOW，DESCRIBE或EXPLAIN，mysqli_query（）将返回一个结果对象 .

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  • 30

    $dbc 正在返回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");

    Raveren已经描述了对此的修复 .

    回复于

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