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page.cgi = a Parser?

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page.cgi = a Parser?
Sorry if asked before. Is it possible to call dynamically a page that's not part of the links static template system? Y ou can call correctly the Cool page by using:

page.cgi?g=Cool/&d=1

But if I try calling my own page the following occurs:

page.cgi?g=News/&d=1

ERROR: We don't seem to have a category by the name 'News'

Does not page.cgi act as a parser to anything you throw at it?
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Re: [Canoon] page.cgi = a Parser? In reply to
I believe if you use the t parameter, not the g parameter, you can call template files that are not part of the "static template system" using the full path and file name/extension of the file.
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Buh Bye!

Cheers,
Me
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Re: [Canoon] page.cgi = a Parser? In reply to
Close, but it's p=template to display template.html. (p == page, if you pass in t=string, it will use a different template set).

Cheers,

Alex
--
Gossamer Threads Inc.
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Re: [Alex] page.cgi = a Parser? In reply to
Thanks for that.

So can I do what I need? Where is my page supposed to reside for this to work?
I have it in the html_public tree as: /news/index.html

but I get errors:

A fatal error has occured:

Invalid Site HTML method: site_html_/news/index.html (/news/index.html). Most likely template does not exist in simple. at /cgi-bin/linksql/admin/Links/SiteHTML.pm line 208.

Please enable debugging in setup for more details.

Frown

Last edited by:

Canoon: Feb 11, 2002, 11:08 AM
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Re: [Canoon] page.cgi = a Parser? In reply to
Hi,

The template must be in your templates/default directory. It will get parsed and printed like any other template.

Cheers,

Alex
--
Gossamer Threads Inc.
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Re: [Alex] page.cgi = a Parser? In reply to
Thanks very much - it works great!