Question

我在查找SQL查询的解决方案时遇到问题。这可能是一个非常明显的初学者问题，但我似乎无法得到我之后的结果。我有一个类似于以下内容的表：

|Name |Station|Salary|
|Bob  |1      |2000  |
|Steve|2      |1750  |
|Mark |3      |2050  |
|Lisa |4      |2200  |
|Hans |5      |2000  |

我想选择此表中具有相同薪水的人员姓名。结果当然应该是 Bob 和 Hans 。

Answer 1

如果你在薪水上加入自己的表格，但名字是分开的，那么这应该给你任何匹配的受薪人员：

SELECT s1.Name, s1.Station, s1.Salary
FROM Staff s1
INNER JOIN Staff s2 ON s1.Salary = s2.Salary AND s1.Name <> s2.Name

Here's a SQLFiddle以显示其实际效果

Answer 2

SELECT Name
FROM table1 
WHERE Salary IN (
    SELECT Salary
    FROM table1
    GROUP BY Salary
    HAVING COUNT(*) > 1
)

Answer 3

如果确定拥有多名员工的工资范围，则可以将其作为派生表重新加入：

SELECT e.Name, e.Station 
FROM Employee e
INNER JOIN
(
   SELECT Salary
   FROM Employee
   GROUP BY Salary
   HAVING COUNT(*) > 1
) grp ON e.Salary = grp.Salary;

SqlFiddle here

Answer 4

大多数数据库都支持窗口功能。这样做的简单方法是计算具有给定薪水的人数。然后选择那些人。这是一个简单的方法：

select t.*
from (select t.*, count(*) over (partition by salary) as salarycnt
      from table t
     ) t
where salarycnt > 1
order by salary;

Answer 5

使用以下查询：

with datatab as
(
 select 'Bob' Name,   1 Station, 2000  Salary from dual union
 select 'Steve' Name,   2 Station, 1750  Salary from dual union
 select 'Mark' Name,   3 Station, 2050  Salary from dual union
 select 'Lisa' Name,   4 Station, 2200  Salary from dual union
 select 'Hans' Name,   5 Station, 2000  Salary from dual union
 select 'Test' Name,   6 Station, 1750  Salary from dual
)

SELECT NAME, sTATION, SALARY  FROM DATATAB
WHERE SALARY IN 
(
   SELECT Salary
   FROM datatab
   GROUP BY Salary
   HAVING COUNT(1) > 1
);

正如有人在Edits中建议的那样，The Query将是：

SELECT NAME, sTATION, SALARY  FROM TABLE_NAME
WHERE SALARY IN 
(
   SELECT Salary
   FROM TABLE_NAME
   GROUP BY Salary
   HAVING COUNT(1) > 1
)

Answer 6

尝试这个：

SELECT E1.Name, E1.Salary
FROM Employee E1, Employee E2
WHERE E1.Salary = E2.Salary
AND E1.Name <> E2.Name

http://sqlfiddle.com/#!2/1e34b

Answer 7

从中选择a.name，a.salary （选择计数（工资）作为cnt，工资组的工资由有工资的工资（工资）＆gt; 1

）作为X. 内部联接人员a a.Salary = x.salary

按顺序排序：

Answer 8

当您必须将同一个表放两次或更多次时，请使用别名：

select t1.Name,
       t2.Name,
       t1.Salary 
  from MyTable t1,
       MyTable t2
 where t1.Station > t2.Station and -- if Station is id 
       -- t1.Name > t2.Name -- if Name is in fact an id
       t1.Salary = t2.Salary

Answer 9

试试这个：

double start = omp_get_wtime();

            #pragma omp parallel for collapse(2)
            for(int y=0;y<hl;y++) 
                for(int x=0;x<wl;x++) 
                {
                    dI_l[x + y*wl][0] = 0.25f * (dI_lm[2*x   + 2*y*wlm1][0] +
                                                dI_lm[2*x+1 + 2*y*wlm1][0] +
                                                dI_lm[2*x   + 2*y*wlm1+wlm1][0] +
                                                dI_lm[2*x+1 + 2*y*wlm1+wlm1][0]);
                }

            double end = omp_get_wtime();
            std::cout << "   wall time : " << end - start << std::endl;

Answer 10

对于MySQL：

从工作人员中选择* 工资在哪里（从工人那里选择工资按薪水分组具有count（1）> 1）;

Answer 11

让表名称为：员工

选择唯一的e.name，例如雇员e，雇员e1的薪水，其中 e.salary = e1.salary和e.name！= e1.name;

Answer 12

您是否想要拥有相同工资的人数。

SELECT count(*) FROM table GROUP BY salary

选择所有具有相同薪水的人员

12 个答案: