Folks-

I have a table that is chock full of errors..(PLAN shows Table Access full)
so I know that it was created wrong
Is there an easy way to setup a query to locate all views that use this table?


Thanks,

Martin Gainty
(cell) 617-852-7822





From: Bill MacAllister <[EMAIL PROTECTED]>
To: Paul DuBois <[EMAIL PROTECTED]>, [EMAIL PROTECTED]
Subject: Re: mysqld_safe in 4.1.2
Date: Thu, 01 Jul 2004 09:21:24 -0700
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--On Wednesday, June 30, 2004 07:36:23 PM -0500 Paul DuBois <[EMAIL PROTECTED]> wrote:

At 13:58 -0700 6/30/04, Bill MacAllister wrote:
Hello,

I am a bit confused by processing of user= specifications in the
mysqld_safe script.  Here is what I see:

* If user= is specificed in the [mysqld] or [server] sections then it is
 ignored and a flag is set that prevents the specification of a user
 in any subsequent options parsing.

* Otherwise the user will be extracted and used from either the
[mysqld_safe], [safe_mysqld], or command line arguements.

Not quite. For security reasons, only the *first* encountered instance of the user option is used, and any others are ignored.

http://dev.mysql.com/doc/mysql/en/Server_options.html

See the descripion of --user in the option list.

Thanks for the pointer to the documentation. I just missed it.

But, my reading of the code in mysqld_safe (4.1.2) does not match the documentation. The first pass through the option sections [mysqld] and [server] have the SET_USER script variable set to 2. With this value set any user specification is ignored, but specifying a user disables any subsequent user specifications because SET_USER is set to 1.

Also, I don't see any where that would generate warnings if there are multiple user specifications. It took me quite a while to figure out why my mysqld_safe was trying to run the mysql server as mysql instead of the user that I specified. Still looks like a bug to me.

Bill


This is counter intuitive if nothing else, and I really don't see why user should be parsed any differently than any other options. Am I missing something or is this a bug?

It's by intent.

--
Paul DuBois, MySQL Documentation Team
Madison, Wisconsin, USA
MySQL AB, www.mysql.com



+--------------------------------------------------- | Bill MacAllister | 14219 Auburn Road | Grass Valley, CA 95949 | 530-272-8555

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