From: Tom Lane <tgl@sss.pgh.pa.us>
To: Adrian Klaver <adrian.klaver@aklaver.com>
cc: Alexander Farber <alexander.farber@gmail.com>,
        pgsql-general <pgsql-general@postgresql.org>
Subject: Re: How to debug: password authentication failed for user
In-reply-to: <d9cb5aab-417d-40b3-80f1-6c26acc7c839@aklaver.com>
References: <CAADeyWjQNmfiQxaiRF0tkZ-kZyApPJT00xbmcMtj+oToSPGoWQ@mail.gmail.com> <056ebaa8-94bb-461d-905c-4035eab9ceac@aklaver.com> <CAADeyWjAUV8q_gdmEA=G4H5b66gaZopp42bTch27okCxPTS=hQ@mail.gmail.com> <d9cb5aab-417d-40b3-80f1-6c26acc7c839@aklaver.com>
Comments: In-reply-to Adrian Klaver <adrian.klaver@aklaver.com>
	message dated "Thu, 27 Feb 2025 10:21:04 -0800"
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Date: Thu, 27 Feb 2025 13:32:38 -0500
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Adrian Klaver <adrian.klaver@aklaver.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