Question

我有一个包含两列的表格 - artist，release_id

我可以运行哪些查询来显示重复记录？

e.g。我的桌子是

ArtistX : 45677
ArtistY : 378798
ArtistX : 45677
ArtistZ : 123456
ArtistY : 888888
ArtistX : 2312
ArtistY: 378798

查询应显示

ArtistX : 45677
ArtistX : 45677
ArtistY : 378798
ArtistY : 378798

Answer 1

如果存在重复项，您可以在感兴趣的列中使用分组。

SELECT
    artist, release_id, count(*) no_of_records
FROM table
GROUP BY artist, release_id
HAVING count(*) > 1;

Answer 2

SELECT id,artist,COUNT(*) FROM myTable
GROUP BY artist, release_id HAVING COUNT(*) > 1

Answer 3

你可以尝试这样的事情

select artist, count(*) from mytable group by artist having count(*) > 1;

将输出

artist   count(*)
45677    2
378798   2

Answer 4

SELECT row, COUNT(row) AS num FROM mytable GROUP BY row HAVING (num > 1);

Answer 5

选择 artist，release_id，count（*）no_of_records，group_concat（id）从表 GROUP BY艺术家，release_id 有计数（*）＆gt; 1;

同时添加group_concat（id）可以获得重复项的所有ID。

Answer 6

您可以将此查询用于相同的结果。它对我有用

SELECT firstname，lastname，list.address FROM list INNER JOIN（选择地址FROM列表 GROUP BY地址HAVING count（id）＆gt; 1）dup ON list.address = dup.address

Answer 7

从艺术家IN的表格中选择*（从具有计数（ID）＆gt; 1的艺术家中选择表格组中的艺术家）和release_id IN（通过具有计数的release_id从表格组中选择release_id（release_id）＆gt; 1）;

将获取： ArtistX：45677 ArtistX：45677 ArtistY：378798 ArtistY：378798

Answer 8

SELECT id,artist,COUNT(id) as found FROM table GROUP by id HAVING found > 1

Answer 9

SELECT artist, count(*) 
FROM tableName 
GROUP BY artist 
HAVING count(*) > 1;

Answer 10

试试这个：

SELECT A.ARTIST,A.RELEASE_ID FROM ARTISTS A
WHERE EXISTS(
SELECT 'X' FROM ARTISTS B
WHERE B.ARTIST = A.ARTIST AND B.RELEASE_ID = A.RELEASE_ID
GROUP BY B.ARTIST,B.RELEASE_ID
HAVING COUNT(B.ARTIST)>1)
ORDER BY A.ARTIST;

Answer 11

这种方法对你来说可能不是很好，但是如果你想要摆脱重复并做到这一点，同时确保它们是重复的，你可以试试这个：

将您的table1复制到table2，例如：

CREATE TABLE table2 AS SELECT * FROM table1;
向table1添加新列，例如将其命名为kount
运行查询（这假定release_id应该是唯一的列）：

UPDATE table1 AS t1 SET t1.kount =（SELECT COUNT（*）FROM table2 AS t2 WHERE t1.release_id = t2.release_id）
drop table table2
使用table1。kount查找重复项并删除它们。最好是PHP / Python / Perl。例如，通过这种方式，您可以确保它们确实是重复的，并且只具有相同的release_id。同样的release_id可能是偶然发生的，标题，出版年限等可能会有所不同。所以只需将您的代码放在此处以过滤重复项（伪代码）：

foreach（sql（SELECT * FROM table1 WHERE kount＆gt; 1））do //做点什么

Answer 12

您还可以尝试以下操作：

SELECT W.artist, W.release_id FROM table W, table W1 
WHERE W.artist= W1.artist AND W.release_id = W1.release_id;

Answer 13

如果您在一行中有更多唯一列，则可以使用：

subsystem: com.apple.UIKit, category: HIDEventFiltered, enable_level: 0, persist_level: 0, default_ttl: 0, info_ttl: 0, debug_ttl: 0, generate_symptoms: 0, enable_oversize: 1, privacy_setting: 2, enable_private_data: 0
subsystem: com.apple.UIKit, category: HIDEventIncoming, enable_level: 0, persist_level: 0, default_ttl: 0, info_ttl: 0, debug_ttl: 0, generate_symptoms: 0, enable_oversize: 1, privacy_setting: 2, enable_private_data: 0
subsystem: com.apple.BaseBoard, category: MachPort, enable_level: 1, persist_level: 0, default_ttl: 0, info_ttl: 0, debug_ttl: 0, generate_symptoms: 0, enable_oversize: 0, privacy_setting: 0, enable_private_data: 0
subsystem: com.apple.UIKit, category: StatusBar, enable_level: 0, persist_level: 0, default_ttl: 0, info_ttl: 0, debug_ttl: 0, generate_symptoms: 0, enable_oversize: 1, privacy_setting: 2, enable_private_data: 0
subsystem: com.apple.libsqlite3, category: logging, enable_level: 0, persist_level: 0, default_ttl: 0, info_ttl: 0, debug_ttl: 0, generate_symptoms: 0, enable_oversize: 1, privacy_setting: 2, enable_private_data: 0
subsystem: com.apple.SystemConfiguration, category: SCPreferences, enable_level: 0, persist_level: 0, default_ttl: 0, info_ttl: 0, debug_ttl: 0, generate_symptoms: 0, enable_oversize: 0, privacy_setting: 2, enable_private_data: 0

在MySQL的同一个表中查找重复项

13 个答案: