Question

我有一张包含以下记录的表

id | fname | lname | skills 

22 | Jane |  Doe | php,mysql 
43 | Jane |  Doe | java,oracle,php
45 | Jane |  Doe | mongo,mysql

id是主键。我想执行一个查询，以便在表中看起来像这样

 id | fname | lname | skills 

45 | Jane |  Doe | php,mysql,java,oracle,mongo

技能将与最新的id和其他重复记录合并

任何帮助对我都非常有用。

Answer 1

试试这个：

UPDATE YourTable t
INNER JOIN(SELECT max(s.id) as max_id,s.fname,s.lname,GROUP_CONCAT(distinct(s.skills)) as skill_str
           FROM YourTable s
           GROUP BY s.fname,s.lname) t1
 ON (t.lname = t1.lname and t.fname = t1.fname AND t.id = t1.max_id)
SET t.skills = t1.skill_str

这会将表更新为所需的concat，然后删除：

DELETE FROM YourTable t
WHERE t.ID NOT IN(SELECT MAX(s.id) FROM YourTable s
                  GROUP BY s.lname,s.fname)

只有当ID是唯一的时，删除才有效！它将删除所有不是最大ID

的记录

编辑：试试这个：

DELETE t
FROM candidate t
LEFT JOIN(SELECT MAX(s.candidate_id) as max_id FROM candidate s
          GROUP BY s.fname,s.lname) t1
 ON (t.candidate_id = t1.max_id)
WHERE t1.max_id is null

Answer 2

试试这个，

SELECT 
        MAX(id) AS id, fname, lname, 
        GROUP_CONCAT(DISTINCT(skills)) AS skills
FROM
        < your_table >
GROUP BY
        fname, lname

参考文献：

http://www.w3resource.com/mysql/aggregate-functions-and-grouping/aggregate-functions-and-grouping-group_concat.php

http://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_group-concat

Answer 3

假设您每行有10个技能限制，您可以使用以下查询规范化skills列：

select distinct 
    s.fname, s.lname, 
    replace(
        substring(substring_index(s.skills, ',', p.pos),
        char_length(substring_index(s.skills, ',', p.pos -1)) + 1),
        ',', ''
    ) as skill
from
    skills s
cross join 
    (select 1 pos 
    union all select 2 
    union all select 3 
    union all select 4 
    union all select 5 
    union all select 6 
    union all select 7 
    union all select 8 
    union all select 9 
    union all select 10
) p on p.pos <= (char_length(s.skills) - char_length(replace(s.skills, ',', ''))) + 1

结果：

| fname | lname |  skill |
|-------|-------|--------|
|  Jane |   Doe |    php |
|  Jane |   Doe |   java |
|  Jane |   Doe |  mongo |
|  Jane |   Doe |  mysql |
|  Jane |   Doe | oracle |

sqlfiddle

(char_length(s.skills) - char_length(replace(s.skills, ',', ''))) + 1

将返回技能数量（计算逗号）。

    replace(
        substring(substring_index(s.skills, ',', p.pos),
        char_length(substring_index(s.skills, ',', p.pos -1)) + 1),
        ',', ''
    )

将在特定位置提取技能。

您可以使用具有不间断数字序列的任何现有表格，而不是创建内联表格。

要再次对结果进行非规范化，您可以使用GROUP_CONCAT(DISTINCT skill)：

select 
    s.fname, s.lname, 
    group_concat(distinct replace(
        substring(substring_index(s.skills, ',', p.pos),
        char_length(substring_index(s.skills, ',', p.pos -1)) + 1),
        ',', ''
    )) as skills
from skills s
cross join (
    select 1 pos 
    union all select 2 
    union all select 3 
    union all select 4 
    union all select 5 
    union all select 6 
    union all select 7 
    union all select 8 
    union all select 9 
    union all select 10
) p on p.pos <= (char_length(s.skills) - char_length(replace(s.skills, ',', ''))) + 1
group by s.fname, s.lname

结果：

| fname | lname |                      skills |
|-------|-------|-----------------------------|
|  Jane |   Doe | php,mysql,mongo,oracle,java |

sqlfiddle

这样，列表中没有两次技能出现。

更新和删除合并记录的查询

3 个答案: