我在PostgreSQL 9.3中有两个不同列的表:
CREATE TABLE person1(
NAME TEXT NOT NULL,
AGE INT NOT NULL
);
CREATE TABLE person2(
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);
INSERT INTO person2 (Name, Age, ADDRESS, SALARY)
VALUES ('Piotr', 20, 'London', 80);
我想将记录从person2
复制到person1
,但是列名可以在程序中更改,因此我想在程序中选择联合列名。所以我创建了一个包含列名称交集的数组。接下来我使用函数insert into .... select
,但是当我按名称将数组变量传递给函数时,我收到错误。像这样:
select column_name into name1 from information_schema.columns where table_name = 'person1';
select column_name into name2 from information_schema.columns where table_name = 'person2';
select * into cols from ( select * from name1 intersect select * from name2) as tmp;
-- Create array with name of columns
select array (select column_name::text from cols) into cols2;
CREATE OR REPLACE FUNCTION f_insert_these_columns(VARIADIC _cols text[])
RETURNS void AS
$func$
BEGIN
EXECUTE (
SELECT 'INSERT INTO person1 SELECT '
|| string_agg(quote_ident(col), ', ')
|| ' FROM person2'
FROM unnest(_cols) col
);
END
$func$ LANGUAGE plpgsql;
select * from cols2;
array
------------
{name,age}
(1 row)
SELECT f_insert_these_columns(VARIADIC cols2);
ERROR: column "cols2" does not exist
这里有什么问题?
答案 0 :(得分:2)
您似乎假设SQL中的SELECT INTO
会分配变量。但事实并非如此。
它创建了一个新的表,并且在Postgres中不鼓励使用它。请改用上级CREATE TABLE AS
。尤其是,因为plpgsql中SELECT INTO
的含义是不同:
我之前已经提出了关于SQL变量的相关问题:
因此你不能这样调用这个函数:
SELECT f_insert_these_columns(VARIADIC cols2);
这样可行:
SELECT f_insert_these_columns(VARIADIC (TABLE cols2 LIMIT 1));
关于短TABLE
语法:
要复制两个表之间具有相同名称的列的所有行:
CREATE OR REPLACE FUNCTION f_copy_rows_with_shared_cols(
IN _tbl1 regclass
, IN _tbl2 regclass
, OUT rows int
, OUT columns text) AS
$func$
BEGIN
SELECT INTO columns -- proper use of SELECT INTO!
string_agg(quote_ident(attname), ', ')
FROM (
SELECT attname
FROM pg_attribute
WHERE attrelid IN (_tbl1, _tbl2)
AND NOT attisdropped -- no dropped (dead) columns
AND attnum > 0 -- no system columns
GROUP BY 1
HAVING count(*) = 2
) sub;
EXECUTE format('INSERT INTO %1$s(%2$s) SELECT %2$s FROM %3$s'
, _tbl1, columns, _tbl2);
GET DIAGNOSTICS rows = ROW_COUNT; -- return number of rows copied
END
$func$ LANGUAGE plpgsql;
呼叫:
SELECT * FROM f_copy_rows_with_shared_cols('public.person2', 'public.person1');
结果:
rows | columns
-----+---------
3 | name, age
请注意在plpgsql中正确使用SELECT INTO
进行分配。
请注意使用数据类型regclass
。这允许使用模式限定的表名(可选)并防止SQL注入尝试:
关于GET DIAGNOSTICS
:
关于OUT
参数: