我认为错误信息非常清楚：你的子选择需要别名。

SELECT t.salary 
FROM (
      SELECT salary,
          DENSE_RANK() OVER (ORDER BY SALARY DESC) AS DENSE_RANK 
      FROM geosalary
      ) as t  --- this alias is missing
WHERE t.dense_rank = 2

错误信息非常明显：您需要为子查询提供别名。
备选查询：

SELECT DISTINCT salary
FROM   geosalary
ORDER  BY salary DESC NULLS LAST
OFFSET 1
LIMIT  1;

这将根据请求查找“第二高工资”（1行）。其他查询查找具有第二高工资的所有员工（1-n行）。
我添加了NULLS LAST，因为null值通常不应该排在第一位。参见：

• 按ASC列排序，但先按NULL值排序？

SELECT department_id, salary, RANK1 FROM (
    SELECT department_id,
           salary,
           DENSE_RANK ()
              OVER (PARTITION BY department_id ORDER BY SALARY DESC)
              AS rank1
    FROM employees) result  
WHERE rank1 = 3

上面的查询将为您提供单个部门中第三高的工资。如果你想不考虑部门，那么只删除PARTITION BY department_id

您的SQL引擎不知道您正在使用哪个表的“salary”列，这就是为什么您需要使用别名来区分这两个列的原因。试试这个：

SELECT salary
FROM (SELECT G.salary ,DENSE_RANK() OVER(ORDER BY G.SALARY) AS DENSE_RANK FROM geosalary G)
WHERE DENSE_RANK=2;

WITH salaries AS (SELECT salary, DENSE_RANK() OVER(ORDER BY SALARY) AS DENSE_RANK FROM geosalary)
SELECT * FROM salaries WHERE DENSE_RANK=2;

select  level, max(salary)
from   geosalary
where   level=2
connect  by
prior   salary>salary
group  by  level;

如果下面的工资列重复，查询将给出正确的结果：

WITH tmp_tbl AS
  (SELECT salary,
    DENSE_RANK() OVER (ORDER BY SALARY) AS DENSE_RANK
  FROM geosalary
  )
SELECT salary
FROM tmp_tbl
WHERE dense_rank =
  (SELECT MAX(dense_rank)-1 FROM tmp_tbl
  )
AND rownum=1;

这里是SQL标准

SELECT name, salary
FROM   geosalary 
ORDER BY salary desc
OFFSET 1 ROW
FETCH FIRST 1 ROW ONLY

计算第n个最高薪资变动冲抵值

SELECT MAX(salary) FROM geosalary WHERE salary < ( SELECT MAX(salary) FROM geosalary )