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如何从以下查询中获得预期的结果，postgres 目前不支持在此处输入链接描述，（参见关于兼容性的最后一节）

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

实际上，我正在向表中添加一个新的空列，并用另一个表/视图中相同大小和类型的列的内容填充它

do $$DECLARE
    k record;
    _current_view text := '';
    _current_var text := '';
    _new_column_sql text := '';
    _insert_sql text := '';
    _select_sql text := '';

    begin
        for k in SELECT column_name from information_schema.columns where table_name='train'
        and column_name <> 'action' 
        loop
            _current_view := 'av_' || k.column_name;
            _current_var := k.column_name;

            _new_column_sql := 'alter table train_average_imputed add column ' || _current_var || ' float';

            _select_sql := 'select prob as ' || _current_var || ' from ' || _current_view || ', train 
                where train.' || _current_var || '=' || _current_view || '.' || _current_var ;

            _insert_sql := 'insert into train_average_imputed(' || _current_var || ') ' || _select_sql;

            raise notice '%', _select_sql;
            --EXECUTE _test_sql;
            EXECUTE _new_column_sql;
            EXECUTE _insert_sql;
            exit;
        end loop;
        end$$;

谢谢。

更新

实际上我需要沿着行号更新，我尝试使用这个

drop table play2;
create table play2 as (select action from play);
alter table play2 add column color text;

with trans_table as(select color, row_number() over() as rn from play2)
update trans_table set color = a.color from (
    select color, row_number() over() as rn from play) as a
where trans_table.rn =a.rn;

但我收到以下错误

DROP TABLE
SELECT 100
ALTER TABLE
ERROR:  relation "trans_table" does not exist
LINE 2: update trans_table set color = a.color from (
           ^

我看不出我的代码与 postgreswith页面在此处输入链接描述有何不同。请注意，如果我在它返回预期结果之后使用 aselect而不是update查询。with

更新 2

正如伊戈尔建议的那样，只需要复制表格并从那里添加列。

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1 回答 1

4 
UPDATE accounts 
SET (contact_last_name, contact_first_name) = (SELECT last_name, first_name FROM salesmen WHERE salesmen.id = accounts.sales_id);

在 Postgres 中可以写成：

UPDATE accounts 
SET contact_last_name = salesmen.last_name,
    contact_first_name = salesmen.first_name 
FROM salesmen 
WHERE salesmen.id = accounts.sales_id;
于 2013-06-17T09:36:24.563 回答