Re: How to debug: password authentication failed for user

Adrian Klaver <adrian(dot)klaver(at)aklaver(dot)com> writes:
> My guess it has to do with:
> -c 'ALTER USER timeshift_user PASSWORD '"'"'timeshift_pass'"'"';'
> I am still trying to work out what that quoting is doing?

That makes my head hurt, too. Using log_statement to see what's
actually getting sent to the server, I can see that as given
it looks to be fine --- but if you put a shell variable in for
the password as per the original intention, it's not expanded.
I think what you need is

-c 'ALTER USER timeshift_user PASSWORD '"'$PASSWORD'"';'

Note this will fall over with potential for SQL injection if there's a
single quote in the password, so better not use it with untrusted
input. On the whole I'd say "find some other way to do that".

regards, tom lane