从当前ID获取上一行和下一行

Question

如何获取表格中的下一行？

`image_id` int(11) NOT NULL auto_increment
`image_title` varchar(255) NOT NULL
`image_text` mediumtext NOT NULL
`image_date` datetime NOT NULL
`image_filename` varchar(255) NOT NULL

如果当前图像为3，而下一个图像为7等，则无效：

$query = mysql_query("SELECT * FROM images WHERE image_id = ".intval($_GET['id']));
echo $_GET['id']+1;

我该怎么办？

感谢

Answer 1

SELECT * FROM images WHERE image_id < 3 ORDER BY image_id DESC LIMIT 1 -- Previous
SELECT * FROM images WHERE image_id > 3 ORDER BY image_id LIMIT 1 -- Next

Answer 2

如果您想圈出记录，可以使用：

-- previous or last, if there is no previous
SELECT *
FROM images
WHERE image_id < 12345 OR image_id = MAX(image_id)
ORDER BY image_id DESC
LIMIT 1

-- next or first, if there is no next
SELECT *
FROM images
WHERE image_id > 12345 OR image_id = MIN(image_id)
ORDER BY image_id ASC
LIMIT 1

---

与UNION相同，可能更有效率：

-- previous or last, if there is no previous
(SELECT * FROM images WHERE image_id < 12345 ORDER BY image_id DESC LIMIT 1)
UNION (SELECT * FROM images WHERE image_id = (SELECT MAX(image_id) FROM images))
LIMIT 1

-- next or first, if there is no next
(SELECT * FROM images WHERE image_id > 12345 ORDER BY image_id ASC LIMIT 1)
UNION (SELECT * FROM images WHERE image_id = (SELECT MIN(image_id) FROM images))
LIMIT 1

Answer 3

你非常接近。试试这个：

$query = mysql_query("SELECT * FROM images
                     WHERE image_id > ".intval($_GET['id'])." ORDER BY image_id LIMIT 1");

    <?php echo $_GET['id'] ?>

Answer 4

为了热爱互联网，请不要自己构建SQL查询。使用PDO。