Малко помощ за Next/Prev бутони

[vinsbg]

Здравейте,
дали ще може някой да помогне как точно ще стане това..
Нещо съвсем се омотах. Опитвам се да сложа бутоните за следваща/предишна снимка, но нещо не успявам. Снимките се пазят в папка, а в базата се записва 'id', 'caption' и 'filename'.

<?php
if (isset($_GET["id"])) {
    $id = $_GET["id"];
} else {
    $id = 1;

    $res = mysql_query("select * from images order by id ASC LIMIT $id, 1") or die (mysql_error());
    $prevSQL = mysql_query("SELECT * FROM images WHERE id < $id ORDER BY id  LIMIT  1") or die (mysql_error());
    $nextSQL = mysql_query("SELECT * FROM images WHERE id > $id ORDER BY id  LIMIT  1") or die (mysql_error());


   // $row = mysql_fetch_array($res);
    while($row = mysql_fetch_array($res)){

         echo "<div id=\"picture\">";
         echo "<img src=\"upload/" . $row['filename']  . "\" alt=\"\" /><br />";
         echo $row['caption'] . "<br />";
         echo "</p>";
         echo "<a href='show.php?id=".$prevSQL."'>Prev</a>";
         echo " | ";
         echo "<a href='show.php?id=".$nextSQL."'>Next</a>";
         echo "</div>";
    }
}
?>

 

[SVTerziev]

<?php
$id = isset($_GET['id']) ? (int)$_GET['id'] : "1";
$res = mysql_query("select * from images where id='$id'") or die (mysql_error());
$prevSQL = mysql_query("SELECT id FROM images WHERE id < $id ORDER BY id DESC LIMIT 1") or die (mysql_error());
$nextSQL = mysql_query("SELECT id FROM images WHERE id > $id ORDER BY id ASC LIMIT 1") or die (mysql_error());
$prevobj=mysql_fetch_object($prevSQL);
$nextobj=mysql_fetch_object($nextSQL);
$pc = mysql_fetch_object(mysql_query("SELECT COUNT(id) as pid FROM images WHERE id<$id ORDER BY id DESC")) or die (mysql_error());
$nc = mysql_fetch_object(mysql_query("SELECT COUNT(id) as nid FROM images WHERE id>$id ORDER BY id ASC")) or die (mysql_error());
$prev=$pc->pid>0 ? '<a href="show.php?id='.$prevobj->id.'">Prev</a> |' : '';
$next=$nc->nid>0 ? '<a href="show.php?id='.$nextobj->id.'">Next</a>' : '';
$row = mysql_fetch_array($res);
echo "<div id=\"picture\">";
echo "<img src=\"upload/" . $row['filename']  . "\" alt=\"\" /><br />";
echo $row['caption'] . "<br />";
echo "</p>";
echo $prev;
echo $next;
echo "</div>";
?>

 

[vinsbg]

Аз тъкмо спретнах нещо, което сработи...
Благодаря ти! :?:

<?php
include 'db.php';
$query = 'select * from images';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());

$line = mysql_fetch_array($result, MYSQL_BOTH);
if (!$line) die('Empty table');
$previd = -1;
$currid = $line[0];
if (isset($_GET['id'])) {
    do {
        $currid = $line[0];
        if ($currid == $_GET['id']) break;
        $previd = $currid;
        $line = mysql_fetch_array($result, MYSQL_BOTH);
    } while ($line);
}

if ($line) {
    echo "<div id=\"picture\">";
    echo "<img src=\"upload/" . $line['filename']  . "\" alt=\"\" /><br />";
    echo $line['caption'] . "<br />";
    echo "</p>";
    echo "</div>";


}
else echo "Няма запис.\n";

if ($previd > -1) echo "<a href=\"show.php?id=".$previd."\">Prev | </a>\n";

$line = mysql_fetch_array($result, MYSQL_BOTH);
if ($line) echo "<a href=\"show.php?id=".$line[0]."\">Next </a>\n";

?>