sql - 显示工作，emp表中每个工作下的员工总数

Question

我想要像这样的输出：

----------

 job          count          ename
----------
salesman      4            name1
                           name2
                           name3
                           name4

clerk         4            name1
                           name2
                           name3
                           name4
manager       3            name1
                           name2
                           name3
analyst       2            name1
                           name2
president     1            name

.......等等。

它不应该重复职位并计算该职位中的每个名称。我得到了重复的答案。

Answer 1

SQL小提琴

Oracle 11g R2 模式设置：

CREATE TABLE employee
    ("job" varchar2(9), "ename" varchar2(5))
;

INSERT ALL 
    INTO employee ("job", "ename")
         VALUES ('salesman', 'name1')
    INTO employee ("job", "ename")
         VALUES ('salesman', 'name4')
    INTO employee ("job", "ename")
         VALUES ('clerk', 'name1')
    INTO employee ("job", "ename")
         VALUES ('clerk', 'name2')
    INTO employee ("job", "ename")
         VALUES ('manager', 'name1')
    INTO employee ("job", "ename")
         VALUES ('manager', 'name2')
    INTO employee ("job", "ename")
         VALUES ('manager', 'name3')
    INTO employee ("job", "ename")
         VALUES ('salesman', 'name2')
    INTO employee ("job", "ename")
         VALUES ('salesman', 'name3')
    INTO employee ("job", "ename")
         VALUES ('clerk', 'name3')
    INTO employee ("job", "ename")
         VALUES ('president', 'name')
    INTO employee ("job", "ename")
         VALUES ('clerk', 'name4')
    INTO employee ("job", "ename")
         VALUES ('analyst', 'name1')
    INTO employee ("job", "ename")
         VALUES ('analyst', 'name2')
SELECT * FROM dual
;

查询 1：

SELECT case when j."ename" = firstname then j."job" else ' ' end as job, 
       case when j."ename" = firstname then cast(c.cnt as varchar(5)) else ' ' end as "count", 
       j."ename"
FROM employee j
INNER JOIN (SELECT "job",  min("ename") as firstname, count(*) as cnt FROM employee GROUP BY "job") c ON c."job" = j."job"
ORDER BY cnt desc, j."job", j."ename"

结果：

|       JOB | COUNT | ENAME |
|-----------|-------|-------|
|     clerk |     4 | name1 |
|           |       | name2 |
|           |       | name3 |
|           |       | name4 |
|  salesman |     4 | name1 |
|           |       | name2 |
|           |       | name3 |
|           |       | name4 |
|   manager |     3 | name1 |
|           |       | name2 |
|           |       | name3 |
|   analyst |     2 | name1 |
|           |       | name2 |
| president |     1 |  name |