Re: problem with variable

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От Michael Wood
Тема Re: problem with variable
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Msg-id AANLkTimGKvYg9sm6ZPh18Gc_7hcQwK3ycVlM6qb4xbaS@mail.gmail.com
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Ответ на Re: problem with variable  (Michael Wood <esiotrot@gmail.com>)
Список pgsql-novice
On 9 June 2010 16:51, Michael Wood <esiotrot@gmail.com> wrote:
> Hi
>
> On 9 June 2010 16:24, coviolo@libero.it <coviolo@libero.it> wrote:
>> something like this:
>>
>> IF (TG_OP = 'UPDATE') THEN
>> EXECUTE 'CREATE TABLE '||NEW.nome_tabella||' (ordinativo serial PRIMARY KEY
>> CHECK (nome_tabella = '''||NEW.nome_tabella||'''::text))
>> INHERITS (database_t);
>>
>> 3 quotes first and 3 quotes after the second variable?
>
> Just a guess, but I think this is what you want:
>
> IF (TG_OP = 'UPDATE') THEN
> EXECUTE 'CREATE TABLE ' || NEW.nome_tabella || ' (ordinativo serial PRIMARY KEY
> CHECK (nome_tabella = "' || NEW.nome_tabella || '"::text))
> INHERITS (database_t);'
>
> i.e. you want:
>
> CREATE TABLE table_name (x serial PRIMARY KEY
> CHECK (column_name = "table_name"::text))
> INHERITS (database_t);

Sorry, I was talking nonsense.  You want column_name = 'table_name'
but because the ' will be inside a quoted string, you need to double
it.  So you were right.  Use ...nome_tabella = ''' || NEW.nome_tabella
|| '''::text...

--
Michael Wood <esiotrot@gmail.com>

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