create table ithelp171214 (
  id serial primary key
, shop char(1) not null
, buy_date date not null
, buy_time time not null
, item text not null
);

insert into ithelp171214 (shop, buy_date, buy_time, item) values
('A', date '2017-01-01', time '16:00', '保護貼'),
('A', date '2017-01-01', time '16:00', '手機'),
('A', date '2017-01-01', time '16:00', '充電器'),
('D', date '2017-02-01', time '18:17', '手機'),
('D', date '2017-02-01', time '18:17', 'USB線'),
('A', date '2017-03-01', time '20:21', '刮鬍刀'),
('A', date '2017-03-01', time '20:21', 'USB線');
--
select * 
  from ithelp171214
 order by shop, buy_date, buy_time;

 id | shop |  buy_date  | buy_time |  item  
----+------+------------+----------+--------
  3 | A    | 2017-01-01 | 16:00:00 | 充電器
  1 | A    | 2017-01-01 | 16:00:00 | 保護貼
  2 | A    | 2017-01-01 | 16:00:00 | 手機
  6 | A    | 2017-03-01 | 20:21:00 | 刮鬍刀
  7 | A    | 2017-03-01 | 20:21:00 | USB線
  4 | D    | 2017-02-01 | 18:17:00 | 手機
  5 | D    | 2017-02-01 | 18:17:00 | USB線
(7 筆資料列)

--
select i.*
  from (select shop
             , buy_date
             , buy_time
         from ithelp171214
        where item = '手機'
       ) a
     , ithelp171214 i
 where a.shop = i.shop
   and a.buy_date = i.buy_date
   and a.buy_time = i.buy_time
 order by shop, buy_date, buy_time; 

 id | shop |  buy_date  | buy_time |  item  
----+------+------------+----------+--------
  1 | A    | 2017-01-01 | 16:00:00 | 保護貼
  2 | A    | 2017-01-01 | 16:00:00 | 手機
  3 | A    | 2017-01-01 | 16:00:00 | 充電器
  4 | D    | 2017-02-01 | 18:17:00 | 手機
  5 | D    | 2017-02-01 | 18:17:00 | USB線
(5 筆資料列)

這次的問題其實不難,只是你不太願意寫subquery而已.
本來我是不想回答的.但是想想還是覺得回答一下好了.
資料型態轉換是有成本的.
而且那樣轉換合併,不利於使用index.
這部份就不再多說了.自己多想想囉.

幸運叔祝大家愉快~~~

基本上版主的SAMPLE中少了"交易單號"之類的資料.(一家店每天應該不只買一支手機)
就假設SHOP 就是"交易單號"吧 !

SELECT *
FROM TABLE
WHERE SHOP IN (SELECT SHOP FROM TABLE WHERE cno='手機')

觀念很簡單: 先將含有手機的交易單號找出,成一個LIST,再將這個LIST 中所有的交易資料找出來.