_____/ On Tue 27 Dec 2005 03:22:20 GMT, [Aaron Brazell] wrote : \_____
On 12/26/05, Scott Merrill <skippy@xxxxxxxxxx> wrote:
I'd like to send an email to all the users with the Administrator role.
Since the roles are stored as a serialized array in the usermeta table,
I need to iterate over the entire user table to find all the
administrators. I _could_ cheat for the time being and use the
deprecated wp_user_level value, but that's just avoiding the issue.
In many circumstances, iterating over the user table won't be a big
deal, because the number of registered users is so small. This is,
historically, because there is very little value for readers to become
registered users. The newest version of my subscribe2 plugin is likely
to change that, and I suspect that some of my users are going to see
some reasonably large user lists. Iterating over these to parse each
array of roles seems like a real waste of an RDBMS.
Short of re-working the roles and capabilities system, does anyone have
any tricks up their sleeves that might help?
Thanks,
Scott
Hmmmm... without iterating over the user table? Maybe create an item in
the options table with an array of user ids updated whenever a user is
added to the admin group? Would probably require a one time user table
scan when the plugin is installed to get all current admins and could be
added to after that.
I almost think iterating over the entire table would be more worthwhile,
but I don't know.
As Robert pointed out, hanging bits off the core is rarely the
desirable option.
I don't know particular database systems, but the correct thing is to do a
table join of all users and then apply a mask. In this case, it can even be
bitwise. I don't know how/if you can do this in MySQL. Here is a very
simplified example:
|user-id|role|email|
|0000000|0011|10000|
|0000001|0011|01011|
|0000010|0010|01101|
|0000011|0100|01010|
|0000100|0011|01000|
|0000101|0010|01111|
Assuming administrator corresponds to level 8, apply XOR mask to role:
0100
If 0, you found an administrator, so get fetch from the table.
Roy
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