+----------+--------+
| emp_name | rating |
+----------+--------+
| Sameer | 4 |
| Sameer | 9.8 |
| Sameer | 9 |
| Sameer | 7 |
| Sameer | 8.2 |
| Sameer | 9.5 |
| Sameer | 10 |
| Ashwath | 9 |
| Ashwath | 4 |
| Ashwath | 9 |
+----------+--------+
我刚刚开始学习SQL并且我写了一个查询并得到了上面的输出,但是我想显示Sameer和Ashwath的评级,而不是评级,我该怎么办?
查询:
SELECT
emp_name,
bus.rating
FROM employees
JOIN drives
ON employees.emp_id = drives.emp_id
JOIN bus
ON bus.bus_no = drives.bus_no
WHERE
drives.emp_id IN (select emp_id from drives group by emp_id having count(bus_no) > 2);
答案 0 :(得分:2)
只按员工汇总并取平均评分:
SELECT
emp_name,
AVG(bus.rating) AS avg_rating
FROM employees
INNER JOIN drives
ON employees.emp_id = drives.emp_id
INNER JOIN bus
ON bus.bus_no = drives.bus_no
WHERE
drives.emp_id IN (SELECT emp_id FROM drives
GROUP BY emp_id HAVING COUNT(bus_no) > 2)
GROUP BY
emp_name;
答案 1 :(得分:0)
在学习SQL时,最好做正确的事情:
尝试始终使用INNER JOIN或LEFT JOIN。使用IN比执行INNER JOIN或LEFT JOIN更“昂贵”。 最后,当您使用聚合函数(COUNT,SUM,AVG)时,您需要使用GROUP BY 如果我在这里重写你的查询就是我要做的事情:
SELECT
emp_name,
AVG(bus.rating)
FROM employees
INNER JOIN drives on employees.emp_id=drives.emp_id
INNER JOIN
(select emp_id, count(bus_no) from drives
group by emp_id having count(bus_no) > 2 ) AS d
ON drives.emp_id = d.emp_id
INNER JOIN bus
ON bus.bus_no = drives.bus_no
GROUP BY emp_name
答案 2 :(得分:-1)
SELECT emp_name,avg(rating)FROM table GROUP BY emp_name
答案 3 :(得分:-1)
试试这个 -
$sql = "SELECT emp_name, AVG(rating) as avg_rating FROM table_name GROUP BY emp_name";