Question

我在显示数据时遇到问题，这些数据在表中重复出现次数最多。例如，我有一个名为JOB，TASK和TASK_TYPE的表。

JOB包含job_id和job_price，并列出作业。每个作业都包含几个任务

TASK包含job_id和task_id，例如

****job_id**        **task_id**
J01                     T01
J01                     T03
J02                     T01
J02                     T02
J02                     T03
J03                     T01
J04                     T01

TASK_TYPE包含task_id和task_description。它列出了可供客户使用的不同任务。例如。

**task_id**    **task_description**
T01                   shopping
T02                   dog walking
T03                   car wash

我想显示最常用的任务类型，以及它用于的作业，在本例中为T01。到目前为止，我已经获得了此代码，但我不知道如何仅显示最重复的任务。

SELECT j.job_id, t.task_id, tt.type
    FROM job j
        JOIN task t ON j.job_id = t.job_id
            JOIN task_type tt ON t.task_id = tt.task_id GROUP BY tt.type, t.task_id, j.job_id ORDER BY type;

Answer 1

使用TOP 1逻辑作为@LONG提到：

SELECT A.TASK_ID, A.JOB_ID, B.TASK_DESCRIPTION
FROM
(SELECT TASK_ID, JOB_ID
FROM TASK
WHERE TASK_ID IN (SELECT TOP 1 TASK_ID FROM TASK GROUP BY TASK_ID ORDER BY COUNT(*) DESC)) A
INNER JOIN
TASK_TYPE B
ON A.TASK_ID = B.TASK_ID;

如果有多个TASK_ID s具有相同的最高出现频率，请尝试以下操作：

SELECT B.JOB_ID, Z.TASK_ID, C.TASK_DESCRIPTION
(SELECT A.TASK_ID, DENSE_RANK() OVER(ORDER BY A.FREQ DESC) AS MOST_FREQ
FROM
    (SELECT TASK_ID, COUNT(*) AS FREQ FROM TASK GROUP BY TASK_ID) A
) Z
INNER JOIN
TASK B
ON Z.TASK_ID = B.TASK_ID AND Z.MOST_FREQ = 1
INNER JOIN
TASK_TYPE C
ON B.TASK_ID = C.TASK_ID;

这会在不同的行中为您提供结果。根据您在帖子中尝试过的查询，这似乎是您所需要的：）

Answer 2

尝试以下查询，您可以忽略前面的CTE，仅用于样本数据。

WITH CTE
AS
(
select 'J01' as job_id,            'T01' as task_id
union all select 'J01',                    'T03'
union all select 'J02' ,                    'T01'
union all select 'J02' ,                    'T02'
union all select 'J02' ,                    'T03'
union all select 'J03'  ,                   'T01'
union all select 'J04' ,                    'T01'
),

TASK_TYPE
AS
(
SELECT 'T01' as task_id, 'shopping' as task_description
UNION ALL SELECT 'T02' , 'dog walking'
UNION ALL SELECT 'T03' , 'car wash'
)

SELECT TOP 1 B.TASK_ID,C.task_description,STUFF((SELECT ',' + A.job_id FROM CTE as A where A.task_id = B.task_id FOR XML PATH('')),1,1,'') as Related_Job
FROM CTE as B
INNER JOIN TASK_TYPE as C
ON C.task_id = B.task_id
GROUP BY B.TASK_ID,C.task_description
ORDER by COUNT(*) DESC

输出：

TASK_ID    task_description           Related_Job
T01        shopping                   J01,J02,J03,J04

Answer 3

我自己对SQL很陌生，因此我很抱歉我的查询混乱，但它在MySQL中适用于我。

SELECT j.job_id, t.task_id, tt.task_description 
FROM job j 
JOIN task t ON j.job_id = t.job_id 
JOIN task_type tt ON t.task_id = tt.task_id 
where t.task_id = (
select tx.task_id 
from (SELECT t.task_id, count(t.task_id) as accurance FROM job j JOIN task t ON j.job_id = t.job_id JOIN task_type tt ON t.task_id = tt.task_id group by t.task_id limit 1) as tx) 
GROUP BY tt.task_description, t.task_id, j.job_id order by task_description;

我希望这对你来说很容易理解，因为我们都是数据库新手。如果您有任何问题，请随时提出。

这是我的输出：

+--------+---------+------------------+
| job_id | task_id | task_description |
+--------+---------+------------------+
| J01    | T01     | shopping         |
| J02    | T01     | shopping         |
| J03    | T01     | shopping         |
| J04    | T01     | shopping         |
+--------+---------+------------------+

Answer 4

此解决方案使用WITH语法（使用Oracle 11g R2测试）来组织子查询以使查询易于阅读。它还将处理task_id计数绑定的情况（在提供的示例中，T03和T01绑定）。希望这对你有用！

查询

-- set up example tables
WITH TASK AS (
  SELECT 'J01' AS job_id, 'T01' AS task_id FROM dual
  UNION ALL
  SELECT 'J01' AS job_id, 'T03' AS task_id FROM dual
  UNION ALL
  SELECT 'J02' AS job_id, 'T03' AS task_id FROM dual
  UNION ALL
  SELECT 'J02' AS job_id, 'T02' AS task_id FROM dual
  UNION ALL
  SELECT 'J02' AS job_id, 'T03' AS task_id FROM dual
  UNION ALL
  SELECT 'J03' AS job_id, 'T01' AS task_id FROM dual
  UNION ALL
  SELECT 'J04' AS job_id, 'T01' AS task_id FROM dual),
TASK_TYPE AS (
  SELECT 'T01' AS task_id, 'shopping' AS task_description FROM dual
  UNION ALL
  SELECT 'T02' AS task_id, 'dog walking' AS task_description FROM dual
  UNION ALL
  SELECT 'T03' AS task_id, 'car wash' AS task_description FROM dual),
-- rank order by task_id count
TASK_RANK AS (
  SELECT task_id, COUNT(task_id) AS task_counts
  FROM TASK
  GROUP BY task_id),
-- get name of most frequent tasks (will handle tie)
TOP_TASK AS (
  SELECT task_id 
  FROM TASK_RANK
  WHERE task_counts IN (SELECT MAX(task_counts) FROM TASK_RANK))

-- simple query to get tasks
SELECT T.task_id, T.job_id, TT.task_description
FROM TASK T
INNER JOIN TASK_TYPE TT
ON T.task_id = TT.task_id
WHERE T.task_id IN (SELECT task_id FROM TOP_TASK)
ORDER BY task_id

输出

| TASK_ID | JOB_ID | TASK_DESCRIPTION |
|---------|--------|------------------|
|     T01 |    J04 |         shopping |
|     T01 |    J03 |         shopping |
|     T01 |    J01 |         shopping |
|     T03 |    J02 |         car wash |
|     T03 |    J02 |         car wash |
|     T03 |    J01 |         car wash |

SQL - 显示重复发生次数最多的数据

4 个答案:

查询

输出