Re: Command Line option misunderstanding

On Mon, Dec 2, 2024 at 4:50 PM wrote: The problem seems to be, as alluded to by others attempting to help me that the problem only exists when using -c on the same line a

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On Mon, Dec 2, 2024 at 4:50 PM <punch-hassle-guise@duck.com> wrote:

The problem seems to be, as alluded to by others attempting to help me

that the problem only exists when using -c on the same line as -v.

The word "line" here belies further misunderstanding of how shell-executed commands work (the following "two line" command is still just one actual multiple-option command invocation).

psql -v a=1 \

-c 'select :a'

It is best to just say "using -c and -v together".

It is correct that we haven't pointed out, probably because for experienced people it seems obvious, that using -v and -c (or putting \set in -c) is a pointless thing to do.  But psql doesn't go about trying to analyze intent here so, yes, you either get useless successful output in response or a confused server.

That said...

psql -v a=1 -c '\echo :a' postgres
1

So it truly is just this specific SQL-related usage that is pointless, not combining -v and -c generally (I'm sure a useful backslash command can be substituted for \echo)

Related Question:

Documentation says:

command must be either a command string that is completely parsable by the server (i.e., it contains no psql-specific features), or a single backslash command.

$psql -h anna -d GT7   -c "\set  a '11117' \\ select evt_id from events where sport_mode_evt_id=:a"

Really not caring that you are turning on autocommit...

Anyway, what I believe you managed to accomplish here is to set the named variable "a" to the value <single-quote 11117 single-quote blah-blah-blah equals colon a>

Then proceeded to do nothing with that variable since the -c command was done being evaluated in the "single backslash command" mode.

David J.

Thanks for your time and explanations.

Ken