Thread: converting from bytea to integers

converting from bytea to integers

From
John DeSoi
Date:
I'd like to convert some bytea data to an array of four byte integers
(and vice versa). I'm probably missing something obvious, but I don't
see an efficient way to generate a 4 byte integer from a bytea string
(could be big endian or little endian).  Converting back to bytea
seems easy enough using to_hex.

Thanks for any suggestions,



John DeSoi, Ph.D.





Re: converting from bytea to integers

From
"Daniel Verite"
Date:
    John DeSoi wrote:

> I'd like to convert some bytea data to an array of four byte integers

> (and vice versa). I'm probably missing something obvious, but I don't

> see an efficient way to generate a 4 byte integer from a bytea string

> (could be big endian or little endian).

get_byte()?

mailtest=> \set e '\'\12\15\107\20\'::bytea'

mailtest=> select
get_byte(:e,0),get_byte(:e,1),get_byte(:e,2),get_byte(:e,3);
 get_byte | get_byte | get_byte | get_byte
----------+----------+----------+----------
       10 |      13 |         71 |    16
(1 row)

Best regards,
--
 Daniel
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Re: converting from bytea to integers

From
John DeSoi
Date:
On Apr 20, 2009, at 5:23 PM, Daniel Verite wrote:

> get_byte()?
>
> mailtest=> \set e '\'\12\15\107\20\'::bytea'
>
> mailtest=> select get_byte(:e,0),get_byte(:e,1),get_byte(:e,
> 2),get_byte(:e,3);
> get_byte | get_byte | get_byte | get_byte ----------+----------
> +----------+----------
>      10 |      13 |         71 |    16

That's what I ended up with. My first attempts at it were unsuccessful
because I did not notice that get_byte uses zero indexing. Earlier in
the routine I extracted bytes using substring and just assumed they
used the same indexing. They don't. It might be worthy of a
documentation note -- it seems easy to miss if you have not used the
binary functions before.

I generated the integer from the bytes using something like this:

b1 = get_byte(p_array, i+3);
b2 = get_byte(p_array, i+2);
b3 = get_byte(p_array, i+1);
b4 = get_byte(p_array, i);
val = (b1 << 24) + (b2 << 16) + (b3 << 8) + b4;


Thanks,


John DeSoi, Ph.D.