"Ioseph Kim" <[EMAIL PROTECTED]> writes:
> when @-@ operator used at path type,
> below query maybe returns 1.
> because this path is just line.
No, because it's a closed path, so it's a loop from (0,0) to (1,0)
and back again. If you don't want to count the return segment,
use an open path.
I misunderstood. :)
path '((0,0),(1,0))' is 'closed' path.
in this case, it's maybe operator calculated return length too.
- Original Message -
From: "Ioseph Kim" <[EMAIL PROTECTED]>
To:
Sent: Saturday, February 03, 2007 6:00 AM
Subject: Re
when @-@ operator used at path type,
below query maybe returns 1.
because this path is just line.
- Original Message -
From: "Tom Lane" <[EMAIL PROTECTED]>
To: "Ioseph Kim" <[EMAIL PROTECTED]>
Cc:
Sent: Saturday, February 03, 2007 5:36 AM
Subject: Re
"Ioseph Kim" <[EMAIL PROTECTED]> writes:
> => select @-@ lseg '((0,0),(1,0))';
> ?column?
> --
> 1
> (1 row)
> => select @-@ path '((0,0),(1,0))';
> ?column?
> --
> 2
> (1 row)
> It's maybe bug in v8.2.1
What do you think is wrong with those answers?
=> select @-@ lseg '((0,0),(1,0))';
?column?
--
1
(1 row)
=> select @-@ path '((0,0),(1,0))';
?column?
--
2
(1 row)
--
It's maybe bug in v8.2.1
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