下一张和上一张图片

Question

我有两个链接，下一个和上一个，代码相同，除了下一个链接的方向相反时，符号大于或小于。

仍在为此挣扎，有人可以帮忙吗？

这是错误

您的SQL语法有误； 检查与您的MySQL服务器版本相对应的手册以获取正确的语法，以在“ ORDER BY photo_id DESC LIMIT 1）附近使用UNION（在第1行中选择photo_id FROM userphotos WHERE pho）

$id=$_SESSION['id'];
//Now we'll get the list of the specified users photos
$sql = "SELECT id FROM albums WHERE user_id='$id' ORDER BY name ASC LIMIT 1 ";
$query = mysqli_query($mysqli,$sql)or die(mysqli_error($mysqli));

while($album = mysqli_fetch_array($query)){ ?>

<?php
    $var = $_GET['pid'] ;
    $photo_sql = "(SELECT photo_id FROM userphotos WHERE photo_id < ".$var." AND photo_ownerid = ".$user['id']." AND album_id=".$album['id']." ORDER BY photo_id DESC LIMIT 1)";
    $photo_sql.= " UNION (SELECT photo_id FROM userphotos WHERE photo_id > ".$var." AND photo_ownerid = ".$user['id']." AND album_id=".$album['id']." ORDER BY photo_id DESC LIMIT 1)";
    $photo_query = mysqli_query($mysqli,$photo_sql)or die(mysqli_error($mysqli));
    $photo_prev=mysqli_fetch_array($photo_query);

            echo " <a href='photo.php?pid=".$photo_prev['photo_id']."'>Previous</a> | ";

Answer 1

从顶部开始：

<?php //this is the very first line
$mysqli = new mysqli($this->DBIP, $this->UName, $this->pw, $this->DB); //set mysqli
$sql = "SELECT id FROM albums WHERE user_id='$id' ORDER BY name ASC LIMIT 1 ";
//can't query $mysqli unless you set it to a connection.
$query = mysqli_query($mysqli,$sql)or die(mysqli_error($mysqli));
while($album = mysqli_fetch_array($query)){ // remove this --> "?>"

$ photo_prev引用$ photo_query，引用$ mysqli。 所以死在那里