Another SQL Query Help thread :)

tsinc80697 · 29 May 2007 at 13:50

I need to output some data from a database, at the moment it is just 1 query that goes through everything and outputs them and styles them with CSS.

Everything that comes out is output into a div with an ID = package.

This div ID has a background-image set to Image_1.jpg

But what I need is the query to output 2 entries at a time into different divs so I can style them differently.

ie

Instead of everything being output to id=package, I need it to output to id=package then the next one to id=package2 then back to id=package etc.

Will upload a picture to show what I mean incase Im not clear!

pictures:

This one is how it looks at the moment, without the data in each DIV:
http://www.tomsinclair.co.uk/random/all_right.jpg

This is how I want it to look :
http://www.tomsinclair.co.uk/random/all_ok.jpg

is there a simple way or just outputting 2 entries in 1 query and within that I can style them differently?

tsinc80697 · 29 May 2007 at 14:19

AH thanks a lot, ill try that when I get back from lunch, think it might work!

edit: misread it at first!

tsinc80697 · 29 May 2007 at 19:18

think im a bit naff at php !

Heres the query part of the page:

PHP:

			<?php
			$result = mysql_query("SELECT DISTINCT thingy FROM thingy WHERE thingy = thingy");
			while($row = mysql_fetch_array($result))
			{
			echo "<div id='p_pair'>";
			echo "<div id='p_1'>";
			echo "<a href=package.php?id=" . $row['package_ID'] . ">" . $row['name'] . "</a>";
			echo "<br/>";
			echo "<h1>&pound;" . $row['price'] . "</h1>";
			echo "</div>";
			echo "<div id='p_2'>";
			echo content_truncate($row['description']);
			echo "</div><br/><br/><br/>";
			echo "<div id='p_3'>";
			$imageID = $row['package_ID'];
			$picture_q = "SELECT thingy FROM thingy WHERE thingy = '$imageID' LIMIT 1";
			$picture_r = mysql_query($picture_q) or die('Error, query failed');
			while($row = mysql_fetch_array($picture_r))
			{
			echo "<a href=package.php?id=" . $row['package_ID'] . "><img alt =\"image\" src=\"includes/getimage.php?id=" . $row['picture_ID'] . "\"/></a>";
			echo "<br/><br/>";
			}
			echo "</div></div>";		
			}
			?>

Cant get it to work, it just needs to output everything in the database with alternating "p_pair" to "p_pair_b" for each item

tsinc80697 · 13 Jun 2007 at 16:14

Forgot about this problem

Can anyone help me out with this, still have it and its annoying me now that I have realised it is still there!

tsinc80697 · 13 Jun 2007 at 18:01

ah, thanks a lot =] will try it later!

tsinc80697 · 13 Jun 2007 at 20:36

psyr33n said:
That echoing is pretty nasty, might want to exit out of PHP to show large chunks of HTML.

Yeah its changed now , just needed that PHP for my simple mind

Have to show people the site once its done, nearly finished and then the end of my first paid php job!