PHP Show next image in the table - php

Question

I am showing an image on my page from the db table like this:
<?php
if ($db_found) {
$SQL = "SELECT * FROM myTable where id='$posted_id'";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
echo '<img src="images/'.$db_field['image'].'" alt="" />';
}
mysql_close($db_handle);
}
?>
Next
How can I do so that if $posted_id is for example 1 ... when I click the "Next" link the image id = 2 appears and so on.

Answer 1

for that you need to either refresh page or use ajax.
you can pass variable posted_id in url like this.
Next
this way you can pass next id from database .. if your id falls in sequence.
you also need to programatically handle issue like what to do if record of next id does't exist in database..

Answer 2

You should work with the MySQL LIMIT and ORDER filters.
<?php
if (isset($_GET['current'])) {
$current = $_GET['current'];
} else {
$current = 0;
}
$request = "SELECT * FROM myTable ORDER BY id ASC LIMIT " . $current . ",1";
?>
And then, just to catch the next item, you can do something like that:
<?php
// make the last item point to the first one
$loop = true;
$count = "SELECT COUNT(*) FROM myTable";
if ($current < $count) {
$next = $current + 1;
} else if ($loop) {
$next = 0;
// no loop, then just stay at the end
} else {
$next = $current;
}
?>