如何在MySQL中对相似的条目组执行平均

Question

+----------+--------+
| 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 |
+----------+--------+

I just started learning SQL and I wrote a query and got the above output but i want to display the averege rating of Sameer and Ashwath instead of rating, how can i do it?

Query:

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);

Answer 1

Just aggregate by employee and take the average rating:

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;

Answer 2

As you are learning SQL, it's better to do the right things :

Try to always use INNER JOIN OR LEFT JOIN. Using IN is more "expensive" than doing an INNER JOIN or LEFT JOIN. Last, when you use an aggregate function (COUNT, SUM, AVG) you need to use a GROUP BY If I rewrite your query here is what I would do :

 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

Here is a famous image to explain all the joins

Answer 3

从表GROUP BY emp_name中选择emp_name，avg（rating）

Answer 4

尝试这个 -

$sql = "SELECT emp_name, AVG(rating) as avg_rating FROM table_name GROUP BY emp_name";