Hello,
I'm trying to put together a little photo album section for a website. I
have this query which returns the expected info from the command line.
SELECT * FROM photo_album as p INNER JOIN images AS i1 ON p.img_1=i1.img_id
INNER JOIN images AS i2 ON p.img_2=i2.img_id
INNER JOIN images AS i3 ON p.img_3=i3.img_id
INNER JOIN images AS i4 ON p.img_4=i4.img_id;
The problem is when i try to display this on a page. Here is what i have
<?php
include "includes/connect.php";
$result = mysql_query("SELECT * FROM photo_album as p INNER JOIN images
AS i1 ON p.img_1=i1.img_id
INNER JOIN images AS i2 ON p.img_2=i2.img_id
INNER JOIN images AS i3 ON p.img_3=i3.img_id
INNER JOIN images AS i4 ON p.img_4=i4.img_id",$db)or
die(mysql_error());
if ($myrow = mysql_fetch_array($result)) {
do {
print "<table>
<tr>
<td<img src=$myrow[img_path]></td>
</tr>
</table>";
} while ($myrow = mysql_fetch_array($result));
}
?>
This displays only the last inner join image not four like i was
expecting. How can i pull all the images from the array? I've tried lots
of different ways like <img src=$myrow[i3.img_path]> with no luck.
Any help would be greatly appreciated.
Thanks in advance,
Mike
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- [PHP-DB] Re: help w/Multiple Joins Michael Baerwolf
- [PHP-DB] Re: help w/Multiple Joins Bastian Vogt