Question

我正在使用SQL语言。我想结合查询的结果。这是我的代码和预期输出。

 create table customer(name varchar,dpt varchar);

insert into customer (name,dpt)VALUES ('albe','cse');
insert into customer (name,dpt)VALUES ('bine','cse');
insert into customer (name,dpt)VALUES ('alfred','ece');
insert into customer (name,dpt)VALUES ('booshan','ece');
insert into customer (name,dpt)VALUES ('antony','eee');
insert into customer (name,dpt)VALUES ('job','ece');

查询1 ：

select
dpt,
count(name)
  from customer
  where dpt='ece' or dpt='cse'
    group by dpt;

结果：

| dpt | count |
|-----|-------|
| cse |   2   |
| ece |   3   |

我需要结果为：

| cse/ece | 5 |

怎么可能？ sqlfiddle在这里：

http://sqlfiddle.com/#!12/a1c88/13

Answer 1

好像你正在使用PostgreSQL（可以从sqlfiddle链接中推断出来。）

如果是这种情况，那么您可以在外部查询中对查询应用array_agg和sum：

   SELECT array_agg(dpt), sum(cnt)
   FROM (
      SELECT dpt, count(name) as cnt
      FROM customer
      WHERE dpt='ece' OR dpt='cse'
      GROUP BY dpt) t;

SQL Fiddle Demo

Answer 2

您也可以使用GROUP_CONCAT

SELECT GROUP_CONCAT(dpt), SUM(cnt)
FROM(select
     dpt,
     count(name) cnt
     from customer
     where dpt='ece' or dpt='cse'
     group by dpt) t;

SQL Fiddle

在sql中组合查询的结果

2 个答案: