[SOLVED] Need a loop

[blogit]

Hello, i have a photo gallery installed and working i'm able to connect to the db and select data from tables to display on any other page of my site but I want to be able to display say 4 pictures on a page if I just copy/paste my code it ends up grabbing and displaying the same image 4 times over. Below is the code i'm using and the image tag. I connect fine and it displays images, I need to create a loop so that each time the pid (picture id) comes up it increments by 1, or maybe each instance of pid in the image tag is a unique number from 1-1000. There was a while loop included with the code but it wasn't working nor was much else of the code so i changed it around and now im stuck, if anyone can give me some advice id really appreciate it

 

thanks

 

$result = mysql_query("SELECT filename, filepath, pid FROM cpg14x_pictures WHERE pwidth > pheight ORDER BY RAND() LIMIT 4");

 

$obj = mysql_fetch_object($result);

 

<img src="/Gallery/albums/<?php echo $obj->filepath; ?>thumb_<?php echo $obj->filename; ?>" id="thephoto_<?php echo $obj->pid; ?>" onLoad="initImage('thephoto_<?php echo $obj->pid; ?>')" alt="The Gallery" />

 

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The old loop that didn't work was to create a row after the 2nd and 4th picture for a small table, i know its missing a closing bracket for the while loop but it still wasn't working and i dont know what $i was.

 

while($obj = mysql_fetch_object($result)){

$i = $i + 1;

if ($i == 3 || $i == 5)

}

[~n[EO]n~]

Have a look at this example

http://www.phpfreaks.com/forums/index.php/topic,95426.0.html

[blogit]

Thanks I tested that code by having it output the picture id (pid) which works but i'm not sure how to insert the entire img tag into that echo/string, below is how it looks and its obviously not going to work what needs changing?

 

echo "<td><img src="/Gallery/albums/<?php echo $obj->filepath; ?>thumb_<?php echo $obj->filename; ?>" id="thephoto_<?php echo $obj->pid; ?>" onLoad="initImage('thephoto_<?php echo $obj->pid; ?>')" alt="The Gallery" /></td>";

[~n[EO]n~]

You can simply try this, I don't think <?php...?> inside a echo will work 

<td><img src="/Gallery/albums/<?php echo $obj->filepath; ?>thumb_<?php echo $obj->filename; ?>" id="thephoto_<?php echo $obj->pid; ?>" onLoad="initImage('thephoto_<?php echo $obj->pid; ?>')" alt="The Gallery" /></td>

[blogit]

No it does't any other suggestions on inserting a random number where pid should be?

[blogit]

Got this to work with the following code instead of echoing print worked out....Thanks for the link to that other code

 

print '<td>'.'<div class="img-shadow">'.'<a href="/Gallery/displayimage.php?pos=-'.$pid.'">'.'<img src="/Gallery/albums/' . $filepath . 'thumb_' . $filename . '">'.'</a>'.'</div>'.'</td>';