Re: [PHP] << Previous / Next Buttons >>

[Chris Lee]

I just use LIMIT its alot better, alot less CPU intensive.

if (!isset($pos))
    $pos = 0

SELECT * FROM product LIMIT 0, $pos

$pos++
if ( !((SELECT count(*) FROM product) > $pos) )
    <a href='$PHP_SELF?pos=$pos'>Next</a>

you get the idea.

  Chris Lee
  [EMAIL PROTECTED]

"Jeff Oien" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > Hi, what's the simplest way to provide previous/next buttons that
navigate
> > through individual records of a database?  This is for a photo gallery,
so I
> > want to do one record at a time until I'm out of records, then I want
the
> > "next" button to disappear.  I've been fiddling with it for a while now,
but
> > haven't been very successful.  Anyone have any snippets, or ideas for
me?
> >
> > Thanks in advance.
>
> I'll show you what I did but realize I'm an advanced beginner/intermediate
> programmer. This may not be the most efficient way to do it but seems to
> be the simplest code compared to other ways of doing it that I've seen.
Not
> sure if I will explain this well enough either. And this is for a fairly
small
> database. With a larger one it may be too inefficient.
>
> First count how many records are in the database for the album and put
> it in $count.
>
> Then loop through all the records in the album, place the id for each
record
> in an array and find out where in the array the current record is.
$photo_id
> is fed to this script which tells it which picture to display:
>
> $x = 0;
> while ($row = mysql_fetch_array($result)) {
> $photo_id1 = $row['photoid'];
> $array[$x] =  $photo_id1;
> if ($photo_id1 == $photo_id) {
> $position = $x;
> }
> $x++;
> }
>
> In another sql query for retrieving the specific image (have to compensate
for
> array[] starting at zero but record count starting at 1):
>
> $row = mysql_fetch_array($result);
> $photo_id = $row['photoid'];
> if ($position == 0) {
> $photo_pos = $position + 1;
> echo "<p>Showing $photo_pos of $count &lt;&lt; Previous"; //no link
> }
> else {
> $photo_pos = $position + 1;
> $previous_pos = $photo_pos - 2;
> $previous = $array[$previous_pos];
> echo "<p>Showing $photo_pos of $count <a
> href=\"display.php?photo_id=$previous\">&lt;&lt; Previous</a>";
> }
> if ($count == $photo_pos) {
> echo " | Next &gt;&gt;"; //no link
> }
> elseif ($photo_pos < $count) {
> $next_pos = $photo_pos;
> $next = $array[$next_pos];
> echo " | <a
> href=\"display.php?photo_id=$next\">Next &gt;&gt;</a>";
> }
>
> Comments from more experienced programmers welcome. I don't want to lead
> astray.
> Jeff Oien



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