Question

我试图让分页在我的页面上工作，但遇到错误：

mysql_result() expects parameter 1 to be resource, object given in

我想我需要使用mysqli命令，但似乎无法弄明白。这是我的代码

$connect = mysqli_connect('localhost', 'root', 'password', 'vdb');
$per_page = 6;
$pages_query = mysqli_query($connect, "SELECT COUNT(id) FROM customers");

$pages = ceil(mysql_result($pages_query, 0) / $per_page);

$page = (isset($_GET['page'])) ? (int)$_GET['page'] : 1;
$start = ($page - 1) * $per_page;
$query = mysqli_query($connect, "SELECT cid, fname, lname,address, score  FROM customers");


while ($query_row = mysqli_fetch_assoc($query)) {
    $array[] = $query_row['fname'] . '<br />';
}
if($pages >= 1)
{
    for($x=1; $x <= $pages; $x++)
    {
        echo '<a href="?page='.$x.'">' .$x.'</a>';
    }

}

提前致谢！

Answer 1

你在mysql和mysqli

之间喋喋不休

$pages_query = mysqli_query($connect, "SELECT COUNT(id) FROM customers"); // MySQLi

$pages = ceil(mysql_result($pages_query, 0) / $per_page); // MySQL

Answer 2

应该有mysqli_result而不是mysql_result。切勿混淆它们，只使用mysqli_*个功能。

// line 3 of code above
$pages = ceil(mysqli_result($pages_query, 0) / $per_page);
                   ^^

Mysql_result（）错误

2 个答案: