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
SELECT t.*,(select count(*) from city as tt where tt.name=t.name) as count FROM `city` as t where (select count(*) from city as tt where tt.name=t.name) > 1 order by count desc
用你的表格替换城市。 将name替换为字段名
其他回答
SELECT t.*,(select count(*) from city as tt where tt.name=t.name) as count FROM `city` as t where (select count(*) from city as tt where tt.name=t.name) > 1 order by count desc
用你的表格替换城市。 将name替换为字段名
寻找重复地址比看起来要复杂得多，特别是当你要求准确性时。在这种情况下，一个MySQL查询是不够的…
我在SmartyStreets工作，在那里我们解决验证和重复数据删除等问题，我看到过很多类似问题的不同挑战。
有一些第三方服务会在列表中为你标记重复项。仅使用MySQL子查询来执行此操作不会考虑地址格式和标准的差异。美国邮政总局(USPS)有一定的指导方针来制定这些标准，但只有少数供应商获得了执行此类操作的认证。
因此，我建议您最好的答案是，例如，将表导出到CSV文件中，并将其提交给有能力的列表处理程序。其中一个是LiveAddress，它会在几秒钟到几分钟内自动为你完成。它将用一个名为“duplicate”的新字段和一个Y值标记重复的行。
SELECT firstname, lastname, address FROM list
WHERE
Address in
(SELECT address FROM list
GROUP BY address
HAVING count(*) > 1)
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
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 |
+---------------------+------------+---+
