问题
I need 2 id for each group.
SELECT `id`, `category`.`cat_name`
FROM `info`
LEFT JOIN `category` ON `info`.`cat_id` = `category`.`cat_id`
WHERE `category`.`cat_name` IS NOT NULL
GROUP BY `category`.`cat_name`
ORDER BY `category`.`cat_name` ASC
How to do this?
Sample Data:
id cat_name
1 Cat-1
2 Cat-1
3 Cat-2
4 Cat-1
5 Cat-2
6 Cat-1
7 Cat-2
Output Will be:
id cat_name
6 Cat-1
4 Cat-1
7 Cat-2
5 Cat-2
回答1:
If you need two arbitrary ids, then use min() and max():
SELECT c.`cat_name` , min(id), max(id)
FROM `info` i INNER JOIN
`category` c
ON i.`cat_id` = c.`cat_id`
WHERE c.`cat_name` IS NOT NULL
GROUP BY c`.`cat_name`
ORDER BY c.`cat_name` ASC ;
Note: You are using a LEFT JOIN and then aggregating by a column in the second table. This is usually not a good idea, because non-matches are all placed in a NULL group. Furthermore, your WHERE clause turns the LEFT JOIN to an INNER JOIN anyway, so I've fixed that. The WHERE clause may or may not be necessary, depending on whether or not cat_name is ever NULL.
If you want the two biggest or smallest -- and can bear to have them in the same column:
SELECT c.`cat_name`,
substring_index(group_concat id order by id), ',', 2) as ids_2
FROM `info` i INNER JOIN
`category` c
ON i.`cat_id` = c.`cat_id`
WHERE c.`cat_name` IS NOT NULL
GROUP BY c`.`cat_name`
ORDER BY c.`cat_name` ASC ;
回答2:
SELECT id, cat_name
FROM
( SELECT @prev := '', @n := 0 ) init
JOIN
( SELECT @n := if(c.cat_name != @prev, 1, @n + 1) AS n,
@prev := c.cat_name,
c.cat_name,
i.id
FROM `info`
LEFT JOIN `category` ON i.`cat_id` = c.`cat_id`
ORDER BY c.cat_name ASC, i.id DESC
) x
WHERE n <= 2
ORDER BY cat_name ASC, id DESC;
More discussion in Group-wise-max blog.
回答3:
In a database that supported window functions, you could enumerate the position of each record in each group (ROW_NUMBER() OVER (PARTITION BY cat_name ORDER BY id DESC)) and then select those records in relative position 1 or 2.
In MySQL, you can mimic this by a self-join which counts the number of records whose id is greater-than-or-equal-to a record of the same cat_name (PARTITION ... ORDER BY id DESC). Record #1 in a cat_name group has only one record of >= its id, and record #N has N such records.
This query
SELECT id, cat_name
FROM ( SELECT c.id, c.cat_name, COUNT(1) AS relative_position_in_group
FROM category c
LEFT JOIN category others
ON c.cat_name = others.cat_name
AND
c.id <= others.id
GROUP BY 1, 2) d
WHERE relative_position_in_group <= 2
ORDER BY cat_name, id DESC;
produces:
+----+----------+
| id | cat_name |
+----+----------+
| 6 | Cat-1 |
| 4 | Cat-1 |
| 7 | Cat-2 |
| 5 | Cat-2 |
+----+----------+
回答4:
Your query is like Select id, cat_name from mytable group by cat_name then update it to Select SELECT SUBSTRING_INDEX(group_concat(id), ',', 2), cat_name from mytable group by cat_name and you will get output like as follows
id cat_name
1,2 Cat-1
3,5 Cat-2
Does it helps?
回答5:
the only thing you need is adding limit option to the end of your query and ordering in descending order as shown below:
SELECT `id`, `category`.`cat_name`
FROM `info`
LEFT JOIN `category` ON `info`.`cat_id` = `category`.`cat_id`
WHERE `category`.`cat_name` IS NOT NULL
GROUP BY `category`.`cat_name`
ORDER BY `category`.`cat_name` DESC
LIMIT 2
回答6:
Very simple Group By ID. it is group duplicate data
回答7:
I have written query for you. I hope it will resolve your problem :
SELECT
id, cat_name
FROM
(SELECT
*,
@prevcat,
CASE
WHEN cat_name != @prevcat THEN @rownum:=0
END,
@rownum:=@rownum + 1 AS cnt,
@prevcat:=cat_name
FROM
category
CROSS JOIN (SELECT @rownum:=0, @prevcat:='') r
ORDER BY cat_name ASC , id DESC) AS t
WHERE
t.cnt <= 2;
回答8:
Better to use rank function the below sample query for your ouput will be helpful check it
select a.* from
(
select a, b ,rank() over(partition by b order by a desc) as rank
from a
group by b,a) a
where rank<=2
回答9:
please try this, it worked in the sample data given
SELECT `id`, `category`.`cat_name`
FROM `info`
LEFT JOIN `category` ON `info`.`cat_id` = `category`.`cat_id`
WHERE `category`.`cat_name` IS NOT NULL and (SELECT count(*)
FROM info t
WHERE t.id>=info.id and t.cat_id=category.cat_id )<3
GROUP BY `category`.`cat_name`,id
ORDER BY `category`.`cat_name` ASC
回答10:
Well, it's pretty ugly, but it looks like it works.
select
cat_name,
max(id) as maxid
from
table1
group by cat_name
union all
select
cat_name,
max(id) as maxid
from
table1
where not exists
(select
cat_name,
maxid
from
(select cat_name,max(id) as maxid from table1 group by cat_name) t
where t.cat_name = table1.cat_name and t.maxid = table1.id)
group by cat_name
order by cat_name
SQLFiddle
Basically, it takes the max for each cat_name, and then unions that to a second query that excludes the actual max id for each cat_name, allowing you to get the second largest id for each. Hopefully all that made sense...
回答11:
select id, cat_name from
(
select @rank:=if(@prev_cat=cat_name,@rank+1,1) as rank,
id,cat_name,@prev_cat:=cat_name
from Table1,(select @rank:=0, @prev_cat:="")t
order by cat_name, id desc
) temp
where temp.rank<=2
You may verify result at http://sqlfiddle.com/#!9/acd1b/7
来源：https://stackoverflow.com/questions/32093420/mysql-group-by-column-with-2-rows-for-each-group