“分组依据”表达

Question

我在尝试打印department_id，department_name和每个部门的工资总额时遇到了一些问题，但我无法弄清楚原因。我收到错误：00979 - ＆＃34;不是GROUP BY表达式＆＃34;

SELECT d.department_id, d.department_name, SUM(e.salary)
FROM departments d, employees e
WHERE d.department_id = e.department_id
GROUP BY d.department_id;

Answer 1

只需将name添加到group by表达式即可。在此过程中，还要修复查询以使用显式join语法：

SELECT d.department_id, d.department_name, SUM(e.salary)
FROM departments d JOIN
     employees e
     ON d.department_id = e.department_id
GROUP BY d.department_id, d.department_name;

虽然您没有问，但我会指出您的查询版本确实有意义且符合ANSI标准（尽管大多数数据库不支持此功能）。您正在通过主键进行聚合，因此允许引入其他列 - 尽管Oracle不支持此功能。

Answer 2

SELECT d.department_id, d.department_name, SUM(e.salary)
FROM departments d, employees e
WHERE d.department_id = e.department_id
GROUP BY d.department_id, d.department_name;

只需在group by子句中添加department_name列。

Answer 3

或者，您也可以使用MIN()或MAX()汇总部门名称：

SELECT d.department_id, MAX(d.department_name) AS department_name
     , SUM(e.salary) AS department_salary
  FROM departments d INNER JOIN employees e
    ON d.department_id = e.department_id
 GROUP BY d.department_id;

请注意，我将您的语法从旧的ANSI标准更新为更新的标准。如果您更喜欢较旧的语法（就像我一样），那么只需使用它：

SELECT d.department_id, MAX(d.department_name) AS department_name
     , SUM(e.salary) AS department_salary
  FROM departments d, employees e
 WHERE d.department_id = e.department_id
 GROUP BY d.department_id;