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From: Tom Lane <[email protected]>
To: Adrian Klaver <[email protected]>
Cc: Alexander Farber <[email protected]>
Cc: pgsql-general <[email protected]>
Subject: Re: How to debug: password authentication failed for user
Date: Thu, 27 Feb 2025 13:32:38 -0500
Message-ID: <[email protected]> (raw)
In-Reply-To: <[email protected]>
References: <CAADeyWjQNmfiQxaiRF0tkZ-kZyApPJT00xbmcMtj+oToSPGoWQ@mail.gmail.com>
	<[email protected]>
	<CAADeyWjAUV8q_gdmEA=G4H5b66gaZopp42bTch27okCxPTS=hQ@mail.gmail.com>
	<[email protected]>

Adrian Klaver <[email protected]> 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

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