我有一个表与以下字段:
id (Unique)
url (Unique)
title
company
site_id
现在，我需要删除具有相同标题、company和site_id的行。一种方法是使用下面的SQL和脚本(PHP):
SELECT title, site_id, location, id, count( * )
FROM jobs
GROUP BY site_id, company, title, location
HAVING count( * ) >1
运行此查询后，可以使用服务器端脚本删除重复项。
但是，我想知道这是否只能使用SQL查询。
当前回答
删除表中重复的记录。
delete from job s
where rowid < any
(select rowid from job k
where s.site_id = k.site_id and
s.title = k.title and
s.company = k.company);
or
delete from job s
where rowid not in
(select max(rowid) from job k
where s.site_id = k.site_id and
s.title = k.title and
s.company = k.company);
其他回答
我必须对文本字段执行此操作，并且遇到了索引上100字节的限制。
我通过添加一个列来解决这个问题，对字段进行md5哈希，并进行更改。
ALTER TABLE table ADD `merged` VARCHAR( 40 ) NOT NULL ;
UPDATE TABLE SET merged` = MD5(CONCAT(`col1`, `col2`, `col3`))
ALTER IGNORE TABLE table ADD UNIQUE INDEX idx_name (`merged`);
这个解决方案将把重复的数据移到一个表中，唯一的数据移到另一个表中。
-- speed up creating uniques table if dealing with many rows
CREATE INDEX temp_idx ON jobs(site_id, company, title, location);
-- create the table with unique rows
INSERT jobs_uniques SELECT * FROM
(
SELECT *
FROM jobs
GROUP BY site_id, company, title, location
HAVING count(1) > 1
UNION
SELECT *
FROM jobs
GROUP BY site_id, company, title, location
HAVING count(1) = 1
) x
-- create the table with duplicate rows
INSERT jobs_dupes
SELECT *
FROM jobs
WHERE id NOT IN
(SELECT id FROM jobs_uniques)
-- confirm the difference between uniques and dupes tables
SELECT COUNT(1)
AS jobs,
(SELECT COUNT(1) FROM jobs_dupes) + (SELECT COUNT(1) FROM jobs_uniques)
AS sum
FROM jobs
还有另一种解决方案:
DELETE t1 FROM my_table t1, my_table t2 WHERE t1.id < t2.id AND t1.my_field = t2.my_field AND t1.my_field_2 = t2.my_field_2 AND ...
我找到了一个简单的方法。(保持最新的)
DELETE t1 FROM table_name t1 INNER JOIN table_name t2
WHERE t1.primary_id < t2.primary_id
AND t1.check_duplicate_col_1 = t2.check_duplicate_col_1
AND t1.check_duplicate_col_2 = t2.check_duplicate_col_2
...
使用Delete JOIN语句删除重复的行:
DELETE t1 FROM table_name t1
JOIN table_name t2
WHERE
t1.id < t2.id AND
t1.title = t2.title AND t1.company = t2.company AND t1.site_id = t2.site_id;
