关于postgreSQL的search_path参数功能及效果

如下是对search_path的一些测试：

查询当前search_path：
highgo=# show search_path;
  search_path   
----------------
 "$user",public
(1 row)


创建一张表：
highgo=# create table one (int int);
CREATE TABLE
查询该表所属模式：
highgo=# \dt
       List of relations
 Schema | Name | Type  | Owner  
--------+------+-------+--------
 public | one  | table | highgo
(1 row)
该表被创建到了public模式下。


那为什么没有被创建到"$user"下呢？
查询当前数据库中的模式：
highgo=# select catalog_name,schema_name,schema_owner from information_schema.schemata;
 catalog_name |    schema_name     | schema_owner 
--------------+--------------------+--------------
 highgo       | pg_toast           | highgo
 highgo       | pg_temp_1          | highgo
 highgo       | pg_toast_temp_1    | highgo
 highgo       | pg_catalog         | highgo
 highgo       | public             | highgo
 highgo       | hgdb_catalog       | highgo
 highgo       | oracle_catalog     | highgo
 highgo       | information_schema | highgo
原来并没有与"$user"同名的highgo模式。
再创建一个模式：
highgo=# create schema first;
CREATE SCHEMA
创建表：
highgo=# create table two (int int);
CREATE TABLE
highgo=# \dt
       List of relations
 Schema | Name | Type  | Owner  
--------+------+-------+--------
 public | one  | table | highgo
 public | two  | table | highgo
(2 rows)


highgo=# show search_path;
  search_path   
----------------
 "$user",public
(1 row)
修改search_path，将public放到前边：
highgo=# set search_path=public,"$user";
SET
highgo=# create table three (int int);
CREATE TABLE
检查：
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 public | one   | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(3 rows)
看上去好像没什么影响。


再创建与用户同名模式：
highgo=# create schema highgo;
CREATE SCHEMA
修改search_path，将"$user"放到前边：
highgo=# set search_path="$user",public;
SET
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 public | one   | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(3 rows)
创建表：
highgo=# create table four (int int);
CREATE TABLE
查询：
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 highgo | four  | table | highgo
 public | one   | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(4 rows)
发现模式发生了改变。
再次修改search_path，修改第一个为不存在的模式。
highgo=# set search_path=asp,"$user",public;
SET
创建表：
highgo=# create table five (int int);
CREATE TABLE
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 highgo | five  | table | highgo
 highgo | four  | table | highgo
 public | one   | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(5 rows)
发现它跳过了不存在的模式，该表被创建到了第一个存在的模式。
创建asp模式。
highgo=# create schema asp;
CREATE SCHEMA
创建表：
highgo=# create table six (int int);
CREATE TABLE
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 asp    | six   | table | highgo
 highgo | five  | table | highgo
 highgo | four  | table | highgo
 public | one   | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(6 rows)
预料之中。
显式指定模式名。
highgo=# create table public.seven(int int);
CREATE TABLE
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 asp    | six   | table | highgo
 highgo | five  | table | highgo
 highgo | four  | table | highgo
 public | one   | table | highgo
 public | seven | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(7 rows)
与预期的一样，创建到了指定的模式下。
修改search_path，去掉asp：
highgo=# set search_path="$user",public;
SET
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 highgo | five  | table | highgo
 highgo | four  | table | highgo
 public | one   | table | highgo
 public | seven | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(6 rows)
此时已经看不到asp模式下的表。
尝试查询：
highgo=# select * from six;
ERROR:  relation "six" does not exist at character 15
STATEMENT:  select * from six;
ERROR:  relation "six" does not exist
LINE 1: select * from six;
                      ^
再次创建表six：
highgo=# create table six (int int);
CREATE TABLE
highgo=# \dt
        List of relations
 Schema | Name  | Type  | Owner  
--------+-------+-------+--------
 highgo | five  | table | highgo
 highgo | four  | table | highgo
 highgo | six   | table | highgo
 public | one   | table | highgo
 public | seven | table | highgo
 public | three | table | highgo
 public | two   | table | highgo
(7 rows)
此时存在six表在highgo模式下。
highgo=# insert into six values(1);
INSERT 0 1
highgo=# select * from six;
 int 
-----
   1
(1 row)
修改search_path：
highgo=# set search_path=asp,"$user",public;
SET
highgo=# select * from six;
 int 
-----
(0 rows)
发现并没有数据。
原因是此时查询的是asp模式下的six表。
验证当前的schema：
highgo=# select current_schema();
 current_schema 
----------------
 asp
(1 row)
查询数据库配置文件：
[highgo@db1 data]$ cat postgresql.conf|grep search_path
#search_path = '"$user",public'         # schema names
默认是注释掉的。
现在尝试修改它。
search_path = 'php,"$user",public'              # schema names
重载：
[highgo@db1 data]$ pg_ctl reload
在之前的会话查询：
highgo=# show search_path;
     search_path      
----------------------
 asp, "$user", public
(1 row)
新开会话查询：
highgo=# show search_path;
    search_path     
--------------------
 php,"$user",public
(1 row)
说明该参数在数据库层面设置后，不会影响现有会话。
实验过程确认在会话中设置该参数，仅会对此会话造成影响。
也可以对用户层面进行设置:
highgo=# alter user highgo set search_path=asp,php,"$user",public;
ALTER ROLE
在现有会话查询：
highgo=# show search_path;
    search_path     
--------------------
 php,"$user",public
(1 row)
新开会话查询：
highgo=# show search_path;
        search_path        
---------------------------
 asp, php, "$user", public
(1 row)
说明该参数在用户层面设置后，不会影响现有会话，并且用户层面做的设置的优先级高于数据库层面。

By 徐云鹤