Drupal 7.x

Hello

I'm very interested in how to extract only the configuration data from my database. This because at some point during development I'm only interested in the configuration (and not the test data) and export it to other developers or prepare something for the live site. But I'm actually also interested in only extracting the content and not the configurations. Any suggestions how to do this with drush? Or should I use a different tool ?

Hello,

I'm new in drupal.
I was installing drupal 7.7 and faced a lot of errors. I've fixed some errors by searching in google as well as drupal.org forums. but an error is still available and I can not find any fix for it:

Notice: Use of undefined constant WATCHDOG_ERR - assumed 'WATCHDOG_ERR' in file_unmanaged_copy() (line 909 of /home/username/public_html/includes/file.inc).

Hello,

I am developing a module for drupal integration with a certain API. I have done the part where I encode some data into a uri, make the http request and receive a response and all works swell.
To take it a step forward I need to have a way to receive some data at any time through the use of uri arguments. So the remote server would make an http request to say http://www.mysite.com/api.php?id=2&name=John.

I've made this change on the autoassignrole.module

// Sort the roles
      if (variable_get('autoassignrole_user_sort', 'SORT_ASC') == 'SORT_ASC') {
        asort($roles);
      } else if (variable_get('autoassignrole_user_sort', 'SORT_DESC') == 'SORT_DESC') {
        arsort($roles);
      } else {
		  drupal_sort_weight($roles, $roles);
	  }

I've added one more IF and the ELSE

then got to autoassignrole.admin.inc and changed variable

I have 2 content type that has several fields with advanced validation rules,
after build these, requirements was changed and I had to make a user after saving any instance of these content type.
how can I convert these content types to profile types in optimum way? (there is no data, just structure)

so I'm trying to debug some css error with Firebug and it lists the offending line as being in burlapdesign.com #6, burlapdesign.com is the domain name of site. When I look at my server I can't see that file. My n00b question is how do I find that file so as to make changes?