Mysql:连接重复数据但忽略重复的字符串

Mysql: Concatenate Duplicate Data but ignore string in duplicates

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 table 个名称,是否有办法连接名称都为 "Ann Smith" 但忽略字符串 "Dr. " 的行。例如,包含 "Ann Smith" 和 "Dr. Ann Smith" 的行应该连接成一个名为 "Dr. Ann Smith" 的行。如果名称匹配(减去 "dr." 字符串)并且两行的地址匹配,则连接 phone 数字。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 table 叫 t:

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

要执行您想要的操作,您可能需要 CTE(通用 Table 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才能获得所需的结果。

如果您升级到 MySQL 8,您将获得 CTE,但仍然不会获得 LATERAL 查询。

编辑: 我进一步确定了重复的名称。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,您可以通过以下方式获得 "long form":

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 table 来修剪 'Dr. ' 字符串并将其替换为空字符串。

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 和分组依据将 table 中的值与相同的 plain_name 值连接起来。

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 table 所需的结果:

name          | phone_number           | address
dr. ann smith | 1234567890, 1234567899 | 123 home address
brian smith   | 1235551234             | 789 city street