Re: ('dog$house' = quote_ident('dog$house')) is surprisingly FALSE

On 10/5/22 17:16, Bryn Llewellyn wrote:
> The doc for "quote_ident()" says this:
> 
> «
> https://www.postgresql.org/docs/14/functions-string.html
> Returns the given string suitably quoted to be used as an identifier in an SQL statement string. Quotes are added
onlyif necessary (i.e., if the string contains non-identifier characters or would be case-folded). Embedded quotes are
properlydoubled.
 
> »
> 
> B.t.w, the value of "quote_ident()" rests on the distinction between a name (what you provide with the function's
actualargument) and an identifier (what it returns). Some of you flatly reject (borrowing a phrase from Tom) the
distinctionbetween these two terms of art. Oh well…
 

What it returns is text, quoted if needed:

create table "$dog"(n int);

select  pg_typeof(quote_ident('$dog')), quote_ident('$dog');
  pg_typeof | quote_ident
-----------+-------------
  text      | "$dog"

The way I see is if it where an actual identifier then this:

select * from quote_ident('$dog');
  quote_ident
-------------
  "$dog"

would be equal to this:

select * from "$dog";
  n
---



-- 
Adrian Klaver
adrian.klaver@aklaver.com