Postgres的9.0.13更新值

Question

基础上，文档https://www.postgresql.org/docs/9.1/static/sql-update.html这是不可能什么，我试图做（在该页面的最下方）

======== ========

UPDATE accounts SET (contact_last_name, contact_first_name) = 
    (SELECT last_name, first_name FROM salesmen 
    WHERE salesmen.id = accounts.sales_id); 

================

是否有其他办法做到这一点？

db=> update "Template" set ("MsgCategoryName","FromAddress","FromName","ToAddress","ToName","BccAddress","EmailSubject","EmailHTML","EmailPlanText") = (Select "MsgCategoryName","FromAddress","FromName","ToAddress","ToName","BccAddress","EmailSubject","EmailHTML","EmailPlanText" from "Template" where "AccountID" = 1016020479 and "LocaleID" = 'nl' and "TmplName" = 'Invoice/Nobrand/PDF') where "AccountID" = 1017069459 and "TmplName" = 'Invoice/Nobrand/PDF' and "LocaleID" = 'nl'; 
ERROR: syntax error at or near "Select" 
LINE 1: ...s","EmailSubject","EmailHTML","EmailPlanText") = (Select "Ms... 

db=> select version(); 
                 version              
------------------------------------------------------------------------------------------------------------------- 
PostgreSQL 9.0.13 on x86_64-unknown-linux-gnu, compiled by GCC gcc (GCC) 4.4.6 20110731 (Red Hat 4.4.6-3), 64-bit 
(1 row) 
                 ^

Answer 1

我没有旧版本检查，但我相信，你不能使用CTE的更新(what you can for sure with 9.1），在这种情况下，你可以plpgsql它：

do 
$$ 
declare _r record; 
begin 
    for _r in (SELECT id,last_name, first_name FROM salesmen JOIN accounts on salesmen.id = accounts.sales_id) loop 
    UPDATE accounts SET contact_last_name = _r.last_name, contact_first_name = _r.first_name where sales_id = _r.id; 
    end loop; 
end; 
$$ 
;