Reputation: 241
I'm not sure how to word this in simple form to return the appropriate search results, so I've created a Stack.
I have two domains.
Domain1.com
Domain2.com
I want all instances of Domain2.com to redirect to Domain1.com automatically.
If a slug path is provided for Domain2.com I want that slug path carried over to be detected by Domain1.com.
Then, I want Domain1.com to detect anything coming from Domain2.com and to redirect accordingly to the matching slug path on Domain1.com.
Example:
Domain2.com/this-slug
automatically redirects to
Domain1.com/this-slug
... because Domain1.com understands what to do exactly with Domain2.com and any slug if passed along with it.
I would prefer (I think) this to be done via .htaccess but I am open to other methods. In the next year I expect to move to NGINX so I'm not sure how that would work which is why I'm open to other methods to carry over at that time.
I think the difference from search results I did find is that I want this done automatically and not to enter manually the slug path for each one needed. I'm essentially using Domain2.com as Share URL to Domain1.com.
Upvotes: 1
Views: 3155
Reputation: 3389
If you want to redirect domain2.com/this-slug to domain1.com/this-slug, add these rewrite rules to domain2.com's htaccess file:
RewriteEngine On
# First condition, requested resource is not an existing file in domain2.com (this condition can be removed if needed)
RewriteCond %{REQUEST_FILENAME} !-f
# Second condition, requested resource is not an existing directory file in domain2.com (this condition can be removed if needed)
RewriteCond %{REQUEST_FILENAME} !-d
# Third condition, slug must exist
RewriteCond %{REQUEST_URI} !^/?$
# Do the redirection
RewriteRule ^(.*)$ http://domain1.com/$1 [R=301,QSA,L]
Please note that with this rewrite rule domain2.com/something/this-slug would redirect to domain1.com/something/this-slug, if you don't want to redirect if there are intermediate slashes within the request URI, use this as third condition instead:
RewriteCond %{REQUEST_URI} ^/?[^/]+$
Upvotes: 1