NYCPHP Meetup

[Analysis & Solutions]

Hi Nestor:

On Fri, Oct 17, 2003 at 12:07:09PM -0700, Nestor Florez wrote:
> 
> BUT WHEN I put in variables in the hash array it fails:
> $cfg['Servers'][$i]['user']             = $user;      // MySQL user
> $cfg['Servers'][$i]['password']      = $pwd; // MySQL password
> 
> YES, the variable do have the information that I want to pass.

So, what happens if you go to a command line and try to log in with the 
information you provided in the variables $user and $pwd.   If you
don't get in, you need to issue the "flush privileges" command to the 
MySQL server from the command line interface.

Also, in another email, you mention the fact that you're using an older 
version of the program.  Not good.  There are security vulnerabilities in 
them.  Upgrade now.  You'll solve your problem of trying to hack your own 
login page AND tighten security.

--Dan

... what's so hard about deleting unnecessary text? ...

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