I need 2 id for each group.
翻译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-2Output Will be:
id cat_name 6 Cat-1 4 Cat-1 7 Cat-2 5 Cat-2If 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 ;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; 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 | +----+----------+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-2Does it helps?
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 Very simple Group By ID. it is group duplicate data
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;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<=2please 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` ASCWell, 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_nameBasically, 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...
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<=2You may verify result at http://sqlfiddle.com/#!9/acd1b/7