Question

对不起标题，但有点难以在一个单行中解释这个主题..

我有一张这样的表格，我想知道（一年中的每个月）第一次获得奖金的员工人数。

EMPLOYEE_NAME   MONTH   BONUS_RECEIVED
AAA                 1                1
BBB                 1                1

CCC                 2                1
AAA                 2                1
DDD                 2                1

AAA                 3                1
BBB                 3                1
XXX                 3                1

所以，结果应该是

MONTH       TOTAL_BONUS
1                     2
2                     2
3                     1

第1个月，员工AAA和BBB获得奖金（结果为2）
第2个月，员工CCC和DD获得奖金（全年已收到AAA），结果为2
第3个月，只有员工XXX获得奖金，因为AAA和BBB已经全年收到奖金

Answer 1

双重聚合解决了您的问题：

select month, count(1) as total_bonus
from (
    select employee_name, min(month) as month
    from table_like_this
    where bonus_received = 1
    group by employee_name
)
group by month;

首先，对于每位员工，您会发现他/她获得奖金的第一个月。然后，您计算每个“第一个奖金 - 收到的月份”的员工数量。

Answer 2

你也可以使用RANK（）

SELECT MONTH
        ,COUNT(BONUS) AS BONUS
    FROM (
        SELECT EMPLOYEE
            ,MONTH
            ,BONUS
            ,RANK() OVER (
                PARTITION BY EMPLOYEE ORDER BY MONTH
                ) AS RN
        FROM TBTEST
        )
    WHERE RN = 1
    GROUP BY MONTH

Answer 3

您可以使用 ROW_NUMBER（）分析函数。

例如，

<强>设置

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>proxy-list</title>
</head>
	<body>
		<table>
			<tr>
				<td>192.168.1.10</td>     
				<td>HTTP1</td>
				<td>Vietnam</td>
				<td>8080</td>
				<td>a</td>
				<td>%d1</td>
			</tr>
			<tr>
				<td>10.25.100.10</td>
				<td>HTTPS2</td>
				<td>Campuchia</td>
				<td>3214</td>
				<td>b</td>
				<td>%d2</td>
			</tr>
			<tr>
				<td>203.25.10.110</td>
				<td>HTTP3</td>
				<td>ThaiLan</td>
				<td>123</td>
				<td>c</td>
				<td>%d3</td>
			</tr>
			<tr>
				<td>220.155.10.13</td>
				<td>HTTP4</td>
				<td>Itali</td>
				<td >1000</td>
				<td>d</td>
				<td>%d4</td>
			</tr>
			<tr>
				<td>220.155.10.113</td>
				<td>HTTP5</td>
				<td>Itali</td>
				<td >505</td>
				<td>d</td>
				<td>%d4</td>
			</tr>
			<tr>
				<td>220.155.10.115</td>
				<td>HTTPS6</td>
				<td>Itali</td>
				<td >321</td>
				<td>d</td>
				<td>%d4</td>
			</tr>
		</table>
	</body>
</html>

<强>查询

CREATE TABLE t
    (EMPLOYEE_NAME varchar2(3), MONTH number, BONUS_RECEIVED number);

INSERT ALL 
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('AAA', 1, 1)
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('BBB', 1, 1)
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('CCC', 2, 1)
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('AAA', 2, 1)
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('DDD', 2, 1)
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('AAA', 3, 1)
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('BBB', 3, 1)
    INTO t (EMPLOYEE_NAME, MONTH, BONUS_RECEIVED)
         VALUES ('XXX', 3, 1)
SELECT * FROM dual;

Oracle SQL - 独特值的总和属于年份，这是第一次

3 个答案: