Re: [SQL] simple "select / if found" isn't

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Re: [SQL] simple "select / if found" isn't

От:
Gary Stainburn <gary.stainburn@ringways.co.uk>
Дата:
On Friday 16 December 2016 11:34:48 Jan Otto wrote:
>
> return exists(select 1 from user_previous_passwords
>    where u_id=ID and crypt(PASS,u_previous_password) =
> u_previous_password);
>
> > END;
> > $$ LANGUAGE plpgsql;
>
> regards, jan

What a muppet am I???????

It had to be something that simple, but I was going code blind.  The ironic 
bit was that the where clause was the only thing that remained the same with 
ever different idea I tried.

Thanks Jan

Re: [SQL] simple "select / if found" isn't

От:
Jan Otto <asche@me.com>
Дата:
hi gary,

> On 16 Dec 2016, at 12:02, Gary Stainburn  wrote:
> 
> I'm creating a simple function that must have been done millions of times 
> before, but I can't get it to work.  In this case, I'm checking a user ID and 
> password against previously used passwords:
> 
> All I want to do is return 'found' based on the select but I can't get it to 
> work.  
> 
> If I run 
> 
> select 1 from user_previous_passwords 
> 	where u_id=25 and 
> 	crypt('MyPaSSword',u_previous_password) = u_previous_password;
> 
> then it returns the matching row(s)
> 
> If I run my function
> 
> create or replace function check_previous_passwords (ID int4, PASS varchar) 
> returns boolean as $$
> DECLARE
>  UID int4;
> BEGIN
>  return exists(select 1 from user_previous_passwords 
>    where u_id=ID and crypt(PASS,u_previous_password) = PASS);

return exists(select 1 from user_previous_passwords   where u_id=ID and crypt(PASS,u_previous_password) = u_previous_password);

> END;
> $$ LANGUAGE plpgsql;

regards, jan
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