psql（9.6.1，server 9.5.5）
员工
Column | Type | Modifiers | Storage | Stats target | Description
----------------+-----------------------------+-----------------------------------------------------------------+----------+--------------+---- ---------
employee_id | integer | not null default nextval('employees_employee_id_seq'::regclass) | plain | |
first_name | character varying(20) | | extended | |
last_name | character varying(25) | not null | extended | |
email | character varying(25) | not null | extended | |
phone_number | character varying(20) | | extended | |
hire_date | timestamp without time zone | not null | plain | |
job_id | character varying(10) | not null | extended | |
salary | numeric(8,2) | | main | |
commission_pct | numeric(2,2) | | main | |
manager_id | integer | | plain | |
department_id | integer
对于自学，我想使用变量。
此请求的结果适合我：
hr=> select last_name, char_length(last_name) as Length from employees where substring(last_name from 1 for 1) = 'H' order by last_name;
last_name | length
-----------+--------
Hartstein | 9
Higgins | 7
Hunold | 6
(3 rows)
但是对于自我教育，我想使用变量：
\set chosen_letter 'H'
hr=> select last_name, char_length(last_name) as Length from employees where substring(last_name from 1 for 1) = :chosen_letter order by last_name;
ERROR: column "h" does not exist
LINE 1: ...ployees where substring(last_name from 1 for 1) = H order by...
^
那些撇号似乎毁了一切。我无法解决这个问题。
您能否帮我理解如何使用变量获取上述结果？
答案 0 :(得分：2)
尝试使用：
\set chosen_letter '''H'''