我想在MySQL数据库中取出重复的记录。这可以用:
SELECT address, count(id) as cnt FROM list
GROUP BY address HAVING cnt > 1
结果是:
100 MAIN ST 2
我想要拖动它，以便它显示复制的每一行。喜欢的东西:
JIM JONES 100 MAIN ST
JOHN SMITH 100 MAIN ST
有什么想法可以做到吗?我试图避免做第一个，然后在代码中用第二个查询查找重复。
当前回答
最快重复项删除查询过程:
/* create temp table with one primary column id */
INSERT INTO temp(id) SELECT MIN(id) FROM list GROUP BY (isbn) HAVING COUNT(*)>1;
DELETE FROM list WHERE id IN (SELECT id FROM temp);
DELETE FROM temp;
其他回答
Find duplicate Records:
Suppose we have table : Student
student_id int
student_name varchar
Records:
+------------+---------------------+
| student_id | student_name |
+------------+---------------------+
| 101 | usman |
| 101 | usman |
| 101 | usman |
| 102 | usmanyaqoob |
| 103 | muhammadusmanyaqoob |
| 103 | muhammadusmanyaqoob |
+------------+---------------------+
Now we want to see duplicate records
Use this query:
select student_name,student_id ,count(*) c from student group by student_id,student_name having c>1;
+--------------------+------------+---+
| student_name | student_id | c |
+---------------------+------------+---+
| usman | 101 | 3 |
| muhammadusmanyaqoob | 103 | 2 |
+---------------------+------------+---+
这将在一次表传递中选择重复项，没有子查询。
SELECT *
FROM (
SELECT ao.*, (@r := @r + 1) AS rn
FROM (
SELECT @_address := 'N'
) vars,
(
SELECT *
FROM
list a
ORDER BY
address, id
) ao
WHERE CASE WHEN @_address <> address THEN @r := 0 ELSE 0 END IS NOT NULL
AND (@_address := address ) IS NOT NULL
) aoo
WHERE rn > 1
这个查询实际上模拟了Oracle和SQL Server中的ROW_NUMBER()
详见我博客上的文章:
分析函数:SUM, AVG, ROW_NUMBER -在MySQL模拟。
这样不是更简单吗?
SELECT *
FROM tc_tariff_groups
GROUP BY group_id
HAVING COUNT(group_id) >1
?
就我个人而言，这个问题解决了我的问题:
SELECT `SUB_ID`, COUNT(SRV_KW_ID) as subscriptions FROM `SUB_SUBSCR` group by SUB_ID, SRV_KW_ID HAVING subscriptions > 1;
这个脚本所做的是在表中显示所有存在过一次以上的订阅者ID，以及找到的重复的数量。
这是表的列:
| SUB_SUBSCR_ID | int(11) | NO | PRI | NULL | auto_increment |
| MSI_ALIAS | varchar(64) | YES | UNI | NULL | |
| SUB_ID | int(11) | NO | MUL | NULL | |
| SRV_KW_ID | int(11) | NO | MUL | NULL | |
希望对你也有帮助!
SELECT * FROM booking WHERE DATE(created_at) = '2022-01-11' 和代码在( 从预订中选择代码 按代码分组 have COUNT(code) > )由id DESC订购
