获取lobject在PostgreSQL的大小 [英] Obtaining the size of lobject in PostgresSQL

问题描述

我不知道为什么没有函数直接获得PostgreSQL大对象的大小。我相信人能求（）对象的末尾，然后告诉（）的位置，而不是它太贵了？我找不到关于它的谷歌的任何信息？那么，什么是获得lobject例如大小的正确方法如果你想以填补内容的大小 http头？

I wonder why is there no function to directly obtain the size of large object in PostgreSQL. I believe one can seek() the end of object and then tell() the position, but isn't it too expensive? I can't find any info about it in google? So what is the proper way to obtain the size of lobject e.g. if you want to fill the Content-Size http header?

推荐答案

本功能已经足以让我有效的，您可能希望与您的数据来试试吧：

This function has been efficient enough for me, you may want to try it with you data:

CREATE OR REPLACE FUNCTION lo_size(oid) RETURNS integer 
AS $$ 
declare 
 fd integer; 
 sz integer; 
begin 
 fd = lo_open($1, 262144);
 if (fd<0) then
   raise exception 'Failed to open large object %', $1;
 end if;
 sz=lo_lseek(fd,0,2);
 if (lo_close(fd)!=0) then
   raise exception 'Failed to close large object %', $1;
 end if;
 return sz;
end; 
$$ LANGUAGE 'plpgsql'; 

另一个选择是从pg_largeobject选择总和（长（数据）），其中二倍体= the_oid ，但它需要读取访问pg_largeobject，我认为已经燮pressed在PG 9.0+非超级用户

Another option is select sum(length(data)) from pg_largeobject where loid=the_oid but it requires read access to pg_largeobject which I think has been suppressed in pg 9.0+ for non-superusers

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