Re: pg_type.typname of array types.
От | Dmitriy Igrishin |
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Тема | Re: pg_type.typname of array types. |
Дата | |
Msg-id | AANLkTikQ-62xM49qZR8qK8M4KU_PfJB7Ga3Dnm8kcgKw@mail.gmail.com обсуждение исходный текст |
Ответ на | Re: pg_type.typname of array types. (Andrew Dunstan <andrew@dunslane.net>) |
Список | pgsql-hackers |
Hey Andrew,<br /><br />Finally convinced. Thanks!<br /><br /><div class="gmail_quote">2010/12/8 Andrew Dunstan <span dir="ltr"><<ahref="mailto:andrew@dunslane.net">andrew@dunslane.net</a>></span><br /><blockquote class="gmail_quote"style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;"><divclass="im"><br /><br /> On 12/08/2010 05:35 AM, Dmitriy Igrishin wrote:<br /><blockquote class="gmail_quote" style="margin:0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;"> Hey hackers@,<br /><br />Is it guaranteed that name of array types in pg_type system<br /> catalog will always be prefixed by underscore or thisconvention<br /> can be changed in future ?<br /><br /><br /></blockquote><br /><br /></div> It is not guaranteed today,let alone in the future, that the array type for x will be _x for any x.<br /><br /> Consider:<br /><br /> andrew=#create type _foo as (x int); create type foo as (y<br /> text);select typname from pg_type where oid = (selecttyparray from<br /> pg_type where typname = 'foo');<br /> CREATE TYPE<br /> CREATE TYPE<br /> typname<br/> ---------<br /> ___foo<br /> (1 row)<br /><br /><br /> cheers<br /><font color="#888888"><br /> andrew<br/><br /></font></blockquote></div><br /><br clear="all" /><br />-- <br />// Dmitriy.<br /><br /><br />
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