show last comment which just 1 comment per user [d

问题:

This question already has an answer here:

• Select the 3 most recent records where the values of one column are distinct 8 answers

I have a normal comments table:

| id | comment | date | user | post | status |

I want to fetch my 10 last comments with just 1 comment per user, I mean something like this:

I have this data: | id | comment | date | user | post | | 1 | text1 | 2016-04-01| 1 | 1 | | 2 | text2 | 2016-04-02| 2 | 1 | | 3 | text3 | 2016-04-03| 1 | 2 | | 4 | text4 | 2016-04-04| 4 | 3 | | 5 | text5 | 2016-04-04| 3 | 5 | | 6 | text6 | 2016-04-05| 2 | 4 | | 7 | text7 | 2016-04-07| 5 | 3 | | 8 | text8 | 2016-04-10| 4 | 9 | | 9 | text9 | 2016-04-11| 3 | 7 | | 10 | text10 | 2016-04-12| 5 | 8 |

and I want get this : | 10 | text10 | 2016-04-12| 5 | 8 | | 9 | text9 | 2016-04-11| 3 | 7 | | 8 | text8 | 2016-04-10| 4 | 9 | | 6 | text6 | 2016-04-05| 2 | 4 | | 3 | text3 | 2016-04-03| 1 | 2 |

I use this sql statement:

SELECT  *
FROM  `comments`
GROUP BY  `user` 
ORDER BY MAX(`id`) DESC

This fetches comments with correct order for user but wrong order for comment's date: | 7 | text7 | 2016-04-07| 5 | 3 | | 9 | text9 | 2016-04-11| 3 | 7 | | 4 | text4 | 2016-04-04| 4 | 3 | | 6 | text6 | 2016-04-05| 2 | 4 | | 1 | text1 | 2016-04-01| 1 | 1 |

回答1:

Why it doesn't work with GROUP BY

SELECT * cannot be used with GROUP BY; it is invalid SQL. GROUP BY doesn't select table rows. It creates groups of rows using the provided expressions then, from each group, it generates a new record and computes each column of this new record using the values involved in the expression.

The columns that appear in the SELECT clause must satisfy one of the following rules:

• also appear in the GROUP BY clause;
• are used with GROUP BY aggregate functions;
• are functionally-dependent on the columns that appear in the GROUP BY clause.

While * is a shortcut for all the column names of the table(s) used by the query, for your query only the user column satisfy one of the requirements above.

Before version 5.7.5 MySQL didn't implement the third rule above. It used to accept queries that contain in the SELECT clause columns that do not follow any of the GROUP BY requirements. The value returned by the query for such columns was indeterminate.

Since version 5.7.5, MySQL rejects the GROUP BY queries that do satisfy the requirements.

The solution

Either way, the solution for your problem doesn't involve GROUP BY. It can be easily accomplished using a LEFT JOIN with the correct conditions:

SELECT lc.*
FROM comments lc               # 'lc' from 'last comment'
    LEFT JOIN comments nc      # 'nc' from 'newer comment'
        ON lc.user = nc.user   # both comments belong to the same user
        AND lc.id < nc.id      # 'nc' is newer than 'lc'
WHERE nc.id IS NULL            # there is no 'newer comment'
ORDER BY lc.id DESC
LIMIT 10

How it works

It joins the table comments, aliased as lc ("lc" from the "last comment" of a user) against itself, aliased as nc ("nc" from "newer comment"). The join clause matches each entry of lc with all entries of nc that belong to the same user (lc.user = nc.user) and are newer (lc.id < nc.id; I assumed the IDs are assigned sequentially and newer comments have larger values for id).

The use of LEFT JOIN ensures that every row of lc appears in the result of the join, even when no matching row is found in nc (because there is no newer comment of the same user). In this case, NULL is used instead of the fields of nc. The WHERE clause keeps in the final result set only the rows that have NULL in nc.id; this means in the lc part they contain the most recent comment of each user.

The SELECT clause contains all the fields of lc (those of nc are all NULL, anyway). The ORDER BY clause can be used to sort the result set. ORDER BY lc.id DESC puts the most recent comments first and the LIMIT clause keeps the result set to a decent size.

回答2:

You can try this:

SELECT u.*
FROM (
    SELECT id,comment,MAX(date) as date,`user`,post,status
    FROM comments 
    GROUP BY `user`) AS t
JOIN comments AS u USING (id)