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View Full Version : How To: Search the Forums Using Google



captainkman
February 16th, 2013, 04:23 AM
Hi everyone. :)

I've seen the following method detailed in other threads, but I'm not sure many know about it, so I started a thread devoted to it.

Most of us know that the search box here has its flaws, like not displaying 2011 posts/threads. And so we use Google, but get results that aren't from here, which is frustrating if you're searching for an old topic you KNOW is here.

Here's how to get results from these forums only:

1. Go to http://www.google.com/ (http://www.google.com/)
2. In the search bar, type site:forums.auran.com/trainz
3. Leave a space
4. Type your search query
5. Press Enter, as clicking the search button is too mainstream.
6. Look! All your results are from the Forums, no other junk to cycle through!
7. Be amazed that something I told you actually works. :eek:

Happy searching. :)

Kieran.

trainz2010vinny
February 16th, 2013, 04:39 AM
Thanks Its really amazng. You are jerking technologies

captainkman
February 16th, 2013, 05:53 PM
I'm glad it works for you.

I must give credit to those who started showing this method. I just can't remember who, so I won't say anyone's name.

Any other great success stories?

mjolnir
February 17th, 2013, 12:06 AM
Another way to accomplish the same thing: type your query into the Google search bar, but before you do the search, click the settings button in the upper right hand corner of the Google window, and select "Advanced search". In the window that opens, type "forums.auran.com/trainz" into the space, about 3/4 of the way down, which is labeled "site or domain".

Note that this method also works for many third party sites, so that if you are looking for a missing dependency that you think was from a given site, you can search for the dependency, and then type the name of the site into the search bar, and search that site.

ns

captainkman
February 17th, 2013, 03:49 AM
Ah - thank you for that addition. Hopefully users will be able to get full benefit out of both methods.