images showing in list view but not in page view

4 posts by 2 authors in: Forums > CMS Builder
Last Post: February 12, 2008 (RSS)

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By Chris_t - February 12, 2008

Here is the problem, I have to the left of the main area of content a box that will contain 2 images. The logos of the 2 teams playing. I set up a simple custom category called vs that has a title and 2 upload boxes, that was the easy part. I have it set not to create a thumbnail because the images are so small to start with it would make one about 10 pixels big.

I placed it in the area I wanted and it looks great, but when I click on a players name and it goes to display his page I get a "No record found!error" in that box.

http://www.ragefootball.com/test/playerList.php
Try clicking on any "read more" and you get this error.
I have on that page just a bit under the vs box on that has upcoming events and that displays great in any view.

Thanks for any help
Chris

Re: [ChrisTitchenal] images showing in list view but not in page view

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By Dave - February 12, 2008 - edited: February 12, 2008

It's reading the number off the end of the url and trying to load that record.

If it's a page viewer try hardcoding the record number you want it to load, like this:

$options['recordNum'] = '1';

Let me know if that works for you.

Dave Edis - Senior Developer
interactivetools.com

Re: [Dave] images showing in list view but not in page view

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By Chris_t - February 12, 2008

Thanks Dave,

I tried that and now get this error " getuploads: no 'recordNum' value specified!" on the part of the page that would normally come up with the page information.

thanks for your fast reply and hope this gets figured out by you or me

Chris

Re: [ChrisTitchenal] images showing in list view but not in page view

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By Dave - February 12, 2008

Sorry, you'll need to hardcode the record number for the getUploads call as well. Change this:

<?php foreach (getUploads($options['tableName'], 'uploads', $record['num']) as $upload): ?>

To this:

<?php foreach (getUploads($options['tableName'], 'uploads', 1) as $upload): ?>

The basic idea is that we're tell it exactly what record number to load rather than getting that information from the url.

Let me know if that works for you.

Dave Edis - Senior Developer
interactivetools.com