Re: Users functions library

Hello

safe it to pgfoundry http://pgfoundry.org/
regards
Pavel Stehule

2008/7/13 Alejandro D. Burne <alejandro(dot)dburne(at)gmail(dot)com>:
> 2008/7/12 Alejandro D. Burne <alejandro(dot)dburne(at)gmail(dot)com>:
>> Hi, I need to write a function that gives me a password string, no
>> just a numbers-characters string; something like people wrote in php,
>> its based on determined syllables and numbers.
>> I think it be useful to other people, is there a site where one can
>> post it and share with other postgres users?
>>
>> Alejandro
>>
>
> Sorry, the function:
>
> CREATE OR REPLACE FUNCTION gen_password(plenght smallint)
> RETURNS bpchar AS
> $BODY$
>
> DECLARE
> lValid_Consonant bpchar DEFAULT 'BCDFGHJKMNPRSTV';
> lValid_Vowel bpchar DEFAULT 'AEIOUY';
> lValid_Numbers bpchar DEFAULT '23456789';
>
> lConsonant_Length smallint DEFAULT char_length(lValid_Consonant);
> lVowel_Length smallint DEFAULT char_length(lValid_Vowel);
> lNumbers_Length smallint DEFAULT char_length(lValid_Numbers);
>
> lPassword bpchar DEFAULT '';
>
> BEGIN
> LOOP
> IF ROUND(RANDOM()*3)<>1 THEN
> lPassword:=lPassword||SUBSTRING(lValid_Consonant FROM
> (ROUND(RANDOM()*(lConsonant_Length-1))+1)::integer FOR 1)||
> SUBSTRING(lValid_Vowel FROM
> (ROUND(RANDOM()*(lVowel_Length-1))+1)::integer FOR 1);
> IF ROUND(RANDOM()*2)<>1 THEN
> lPassword:=lPassword||SUBSTRING(lValid_Consonant FROM
> (ROUND(RANDOM()*(lConsonant_Length-1))+1)::integer FOR 1);
> END IF;
> ELSE
> lPassword:=lPassword||SUBSTRING(lValid_Numbers FROM
> (ROUND(RANDOM()*(lNumbers_Length-1))+1)::integer FOR 1);
> END IF;
> IF char_length(lPassword) >= plenght THEN
> EXIT;
> END IF;
> END LOOP;
>
> RETURN SUBSTRING(lPassword FROM 1 FOR plenght);
> END;
>
> $BODY$
> LANGUAGE 'plpgsql' VOLATILE;
>
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