(PHP/Postgresql) Call a function which returns void?

Question

I'm getting crazy trying to call, inside PHP code, a function that returns void, with pg_prepare/pg__execute...

$result = pg_prepare($dbconn, "",'SELECT * FROM dungeonasdb.crea_personaggio( "$nome", "$descrizione", "$email", "$password")');
$result = pg_execute($dbconn, "", NULL ) or die('Query creazione fallita');

Is there any settings of pg_prepare/pg_execute that I do wrong? Thanks

dungeonasdb.crea_personaggio is your table name ?

Hamza Abdaoui
– Hamza Abdaoui

2017-07-01 11:20:45 +00:00
Commented Jul 1, 2017 at 11:20 — Hamza Abdaoui
– Hamza Abdaoui, Commented Jul 1, 2017 at 11:20

joanolo · Accepted Answer · 2017-07-02 09:48:46Z

1

You probably want to do something different:

pg_prepare: don't put specific values, but placeholders. You will bind them to values later on. I prefer to always give statements a name.

If dungeonasdb.crea_personaggio is the function that returns void, you can call it by means of:

SELECT dungeonasdb.crea_personaggio(...)

You cannot call your function by means of SELECT * FROM dungeonasdb.crea_personaggio(...) because this would imply that dungeonasdb.crea_personaggio is a set returning function, which is not.

So you probably want your statements to be:

$result = pg_prepare($dbconn, "crea_personnagio", 
   'SELECT dungeonasdb.crea_personaggio($1, $2, $3, $4)'); 
$result = pg_execute($dbconn, "crea_personnagio", 
    ARRAY($nome, $descrizione, $email, $password)) 
    or die('Query execuzione fallita');

Your second statment is a pg_execute statement that binds parameters to the placedholders ($1, ...), and executes.

edited Jul 2, 2017 at 9:48

answered Jul 1, 2017 at 14:16

joanolo

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4 Comments

Simo S Over a year ago

ok I solved the problem, it was the function in the database, I wrote all the functions using only the tables name and not like 'dbname.tablename' . Also I think your answer it's not correct, beacause PERFORM works only inside postgresql functions , you can't use it from outside. But thanks for your time :)

joanolo Over a year ago

You're right about PERFORM. I've corrected it. In PostgreSQL, what you're actually writing is schema_name.function_name(...) or schema_name.table_name (or, in general schema_name.object_name). You don't access two databases from the same connection (except through FDW). In PostgreSQL you have database -> schema -> object. In MySQL (for instance), you have only database (=schema) -> object.

joanolo Over a year ago

Did the answer solve your problem? If so, it is customary to mark it as accepted, so that people can concentrate on the unanswered ones.

Simo S Over a year ago

I haven't used your method but it works the same, now I accept it, thanks again!

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