如何使用MySQL比较两个逗号分隔的字符串列表 [英] How to compare two comma-separated string lists using MySQL

问题描述

我使用了一个名为"containsAll()"的Java方法来检查ArrayList是否具有共同的内容.

I used a Java method called 'containsAll()' to check if ArrayLists have common content.

比方说，我在MySQL表中有一个列表A(一行)和其他几个列表(在名称"列中，逐行). 所有列表均由逗号分隔的字符串(列表中至少一个字符串)组成-名称或其他名称.

Let's say I have a list A (one row), and several other lists in a MySQL table (in column 'name', row by row). All lists consist of comma-separated Strings (at least one String in a list) - names or whatever.

现在，我想检查列表A中的所有字符串是否都可以在名称"列中的任何行中找到. 结果集应显示名称"中匹配的所有行，包括行/列表的行必须具有列表A中的所有字符串，并且可以具有其他字符串.

Now, I want to check if all Strings in list A can be found in any of the rows in column 'name'. The result set should show all the rows in 'name' that match, that includes rows/lists must have all Strings in list A, and can have additional Strings.

示例I

A:"Mr.T"

____name_________________________________________
 'Hannibal'
 'Hannibal','Face','Murdock','Mr.T','Donald Duck'
 'Face','Donald Duck'
 'Superman','Chuck Norris','Mr.T'
_________________________________________________   

结果集:汉尼拔"，脸部"，默多克"，"T先生"，唐老鸭"-并且- '超人'，查克·诺里斯'，'T先生'

Result set: 'Hannibal','Face','Murdock','Mr.T','Donald Duck' -AND- 'Superman',Chuck Norris','Mr.T'

示例II

A:"Rocky"，"Mr.T"，"Apollo"

A: 'Rocky', 'Mr.T', 'Apollo'

 ______name__________________________________________________
 'Hannibal','Face','Murdock','Donald Duck','Superman','Mr.T'
 'Rocky','Apollo','Ivan'
 'Apollo', 'Superman','Hannibal','Rocky','Mr.T','Chuck Norris'
 'Rocky','Mr.T','Apollo','Chuck Norris'
_____________________________________________________________   

结果集:阿波罗"，超人"，汉尼拔"，洛克"，"T先生"，查克·诺里斯"-并且- 'Rocky'，'T先生'，'Apollo'，'Cuck Norris'

Result set: 'Apollo', 'Superman','Hannibal','Rocky','Mr.T','Chuck Norris' -AND- 'Rocky','Mr.T','Apollo','Cuck Norris'

我想知道是否可以使用MySQL查询来执行那些结果. 预先谢谢你.

I wonder if one can carry out those results using a MySQL query. Thank you in advance.

推荐答案

您似乎想做一个数组交集，除了您的数组是单列.可以做到，但是会很慢，难以调试，并且不会利用关系数据库的功能.更好的方法是将表架构更改为以下形式:

It appears you want to do an array intersection, except your array is a single column. It can be done, but it will be slow, difficult to debug and will not leverage the power of relational databases. A better way would be to change your table schema to something like this:

表组

group_id int unsigned not null auto_increment primary key,
character_list text

表members_in_group

Table members_in_group

group_id int unsigned not null,
group_member varchar(45) not null

然后您可以像这样查询:

Then you can query like this:

SELECT group_id, character_list
FROM groups g 
  JOIN members_in_groups m USING (group_id)
WHERE m.group_member IN ('Mr. T', ...);

groups表可能非常类似于您当前的表. members_in_groups表是相同的数据，分为易于搜索的部分.

The groups table is probably very like your current table. The members_in_groups table is the same data chopped up into easily searchable parts.

ETA给出您的评论，如果您可以保证每个character_list仅包含每个字符的一个实例，这应该有效.

ETA given your comment, this should work if you can guarantee that each character_list contains only one instance of each character:

SELECT group_id, 
       SUM(CASE m.group_member IN ('Mr. T', 'Apollo', 'Rocky') THEN 1 ELSE 0 END) AS tally, 
       character_list
FROM groups g
  JOIN members_in_groups m ON (g.group_id=m.group_id)
GROUP BY group_id
  HAVING SUM(CASE m.group_member IN ('Mr. T', 'Apollo', 'Rocky') THEN 1 ELSE 0 END) = 3;

在这种情况下，HAVING子句必须等于3，因为IN ('Mr. T', 'Apollo', 'Rocky')中有3个成员.

In this case the HAVING clause must equal 3 because there are 3 members in IN ('Mr. T', 'Apollo', 'Rocky').

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