Question

我有下表：

示例：

create table test
( 
 id int,
 name varchar(10),
 city varchar(10)
);

我想在函数中将ID值从表分配给变量。

功能：

create or replace function testing(ids int,names varchar(10),citys varchar(10)
returns void as
$body$
Declare
       ident int;
BEGIN
       ident := SELECT ID FROM test;
       raise info '%',ident;
END;
$body$
Language plpgsql;

错误详情：

ERROR:  syntax error at or near "SELECT"
LINE 12:  ident := SELECT ID from test;

Answer 1

使用select ... into

create or replace function testing(ids int,names varchar(10),citys varchar(10)
returns void as
$body$
Declare
       ident int;
       foo   text;
BEGIN
       SELECT ID, some_col  
          into ident, foo 
       FROM test;
       raise info '%',ident;
END;
$body$
Language plpgsql;

手册中有更多细节和示例：
http://www.postgresql.org/docs/current/static/plpgsql-statements.html#PLPGSQL-STATEMENTS-SQL-ONEROW

将选择值分配给PostgreSQL 9.3中的变量

1 个答案: