how about like this:
echo table\ntr;
$result = @mysql_query(SELECT image FROM specials);
while($row = mysql_fetch_row($result)) {
echo tdimg src= . $row[0] . //td\n;
}
echo /tr\n/table\n;
hope it helps!
matt
On Jan 5, 2008 4:51 PM, Thomas [EMAIL PROTECTED] wrote:
I'm attempting to make a table with one row and 3 columns holding three
different images. I want each image URL to be called from my database.
Everything is set-up in my database. When I use the following code, it
places the same picture across the three columns and does this three times
(creating three rows.) I want a different picture to be placed across the
three columns and to have only one row. What can I do?
Here is the code:
$result = @mysql_query('SELECT image FROM specials');
$num=mysql_numrows($result);
$i = 0;
while ($i $num) {
$image=mysql_result($result,$i,image);
?
table
tr
td
img src=? echo $image; ?
/td
td
img src=? echo $image; ?
/td
td
img src=? echo $image; ?
/td
/tr
?
$i++;
}
echo /table;
?
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