有没有办法在忽略给定字符串的同时找到重复数据?
例如,如果我有一个名称表,是否有一种方法可以连接名称为“Ann Smith”但忽略字符串“Dr.”的行。例如,包含“Ann Smith”和“Ann Smith博士”的行应连接成一行,其名称为“Ann Smith博士”。如果名称匹配(减去“dr。”字符串)并且两行的地址匹配,则连接电话号码。我想取两个名字中较大的一个,我认为这将涉及使用MAX语句。
目前我有一个名为t的表:
name | phone | address
ann smith | 1234567899 | 123 home address
dr. ann smith | 1234567890 | 123 home address
brian smith | 1235551234 | 789 city street
我想去:
name | phone | address
dr. ann smith | 1234567890, 1234567899 | 123 home address
brian smith | 1235551234 | 789 city street
要做你想做的事,你可能需要CTE(公用表格表达式)和LATERAL
查询。不幸的是,MySQL 5.x没有实现它们中的任何一个。
以下查询查找重复的名称:
select plain_name, count(*)
from (
select name, trim(replace(lower(name), lower('Dr.'), '')) as plain_name
from my_table
) x
group by plain_name
having count(*) > 1
这是朝着正确方向迈出的一步,但您需要进一步处理以获得所需的结果。
如果升级到MySQL 8,您将获得CTE,但仍然无法获得LATERAL查询。
编辑:我更进一步确定重复的名称。没有CTE,这个查询看起来越来越丑陋:
select z.*, y.times
from (
select name, trim(replace(lower(name), lower('Dr.'), '')) as plain_name
from my_table
) z,
(
select plain_name, count(*) as times
from (
select name, trim(replace(lower(name), lower('Dr.'), '')) as plain_name
from my_table
) x
group by plain_name
having count(*) > 1
) y
where z.plain_name = y.plain_name;
假设这些是完全嵌套的,你可以通过以下方式获得“长形式”:
select name,
(select t2.name
from t t2
where t2.name like concat('%', t.name, '%')
order by length(t2.name) desc
limit 1
) as long_form
from t;
然后,您可以在聚合中使用它。我会使用子查询:
select long_form, group_concat(distinct phone) as phones,
group_concat(distinct address) as addresses
from (select t.*,
(select t2.name
from t t2
where t2.name like concat('%', t.name, '%')
order by length(t2.name) desc
limit 1
) as long_form
from t
) tt
group by long_from;
我最终使用了上述答案的组合。首先,我创建了一个临时表,用于修剪和替换'博士'带有空字符串的字符串。
create temporary table if not exists temp_names AS (
select *,
case when name like lower('dr. %') then trim(replace(lower(name), lower('dr. %'), ''))
else name end as plain_name from t);
然后我使用select和group by来连接该表中具有相同plain_name值的值。
select max(name) as name, group_concat(distinct phone_number) as phone_number, address from temp_names
group by plain_name, address having count(*) >=1;
这给出了一个具有所需结果的表格:
name | phone_number | address
dr. ann smith | 1234567890, 1234567899 | 123 home address
brian smith | 1235551234 | 789 city street