我创建了一个效果很好的next和prev图像链接。但我希望他们都能回到开头或结尾。所以说我有image101,图像202和image243,一旦我点击image243,它将返回到image101。
我想和以前一样,当我点击Image101时,我想要链接回到image243。 (自我解释)。
我该怎么做呢?
<?php $sql = "SELECT id FROM albums WHERE user_id=".$user['id']." ORDER BY id ASC LIMIT 1"; $query = mysql_query($sql)or die(mysql_error());
while($album = mysql_fetch_array($query)){ ?>
<?php
$photo_sql = "SELECT photo_id FROM userphotos WHERE photo_id > ".$_GET['pid']." AND photo_ownerid = ".$user['id']." AND album_id=".$album['id']." ORDER BY photo_id ASC LIMIT 1 ";
$photo_query = mysql_query($photo_sql)or die(mysql_error());
$photo_next=mysql_fetch_array($photo_query);
echo "<a href='photo.php?pid=".$photo_next['photo_id']."'>Next</a>";
$photo_sql = "SELECT photo_id FROM userphotos WHERE photo_id < ".$_GET['pid']." AND photo_ownerid = ".$user['id']." AND album_id=".$album['id']." ORDER BY photo_id DESC LIMIT 1 ";
$photo_query = mysql_query($photo_sql)or die(mysql_error());
$photo_prev=mysql_fetch_array($photo_query);
echo " | <a href='photo.php?pid=".$photo_prev['photo_id']."'>Prev</a>";
}
?>
答案 0 :(得分:1)
为什么不从PHP那样做?您可以获取照片ID,而不是拥有所有这些多余的MySQL查询,并使用PHP:
$photo_sql = "SELECT photo_id FROM userphotos WHERE photo_ownerid = ".$user['id']." AND album_id=".$album['id']." ORDER BY photo_id ASC"
$photo_result = mysql_query($photo_sql) or die(mysql_error());
while($row=mysql_fetch_array($photo_result)) {
$photos[]=$row[0];
}
$total = mysql_num_rows($photo_result);
$current = 2; // whatever your position is in the photo array
echo '<img src="'.$photos[$current].'" alt="my image!" />'; // display current photo
$next = ($current+1) % $total; // modulo
$prev = (($current-1) < 0) ? $total : $current -1;
echo "<a href='photo.php?pid=".$next."'>Next</a>";
echo " | <a href='photo.php?pid=".$prev."'>Prev</a>";
为了解释,我们使用modulo来获取正确的下一页。获取上一页不言自明:如果我们低于零,请使用$ total照片总数。
使用它这种方式在数据库方面比所有这些大于小于查询的效率更高效。
答案 1 :(得分:0)
您可以添加第一张图片的链接,如下所示:
$photo_sql = "SELECT MIN(photo_id) FROM userphotos WHERE photo_ownerid = ".$user['id']
$photo_query = mysql_query($photo_sql)or die(mysql_error());
$photo_first =mysql_fetch_array($photo_query);
echo "<a href='photo.php?pid=".$photo_first['photo_id']."'>First</a>";
当然,您可以使用MAX来获取最后一张照片。
答案 2 :(得分:0)
这将为您提供下一个条目,也是第一个条目,这意味着如果没有NEXT,它将给出第一个条目。
$photo_sql = "SELECT photo_id FROM userphotos WHERE photo_id > ".$_GET['pid']." AND photo_ownerid = ".$user['id']." ORDER BY photo_id ASC LIMIT 1 ";
$photo_sql. = "UNION SELECT photo_id FROM userphotos WHERE photo_ownerid = ".$user['id']." ORDER BY photo_id ASC LIMIT 1 ";
然后在之前使用ORDER BY photo_id DESC进行UNION