Lets say I have the following mysql query
SELECT friends.id, sum(if(friends.online =1,1,0)) AS online_friends, ...,...
FROM some_table
GROUP BY id
ORDER BY sum(if(friends.online =1,1,0)) ASC;
What I want is to sort my resulting table bases on online_friends either ASC or DESC, the problem is that this query does not sort them by online_friends.
Can anyone help me fixed my query?
This is an example of the results
id group online_friends
-- ------- ----------------
1 west 5
2 east 3
3 north 7
4 south 2
I want to sort the results based on online_friends
Your query should work:
SELECT friends.id, sum(if(friends.online =1,1,0)) AS online_friends, ...,...
FROM friends
GROUP BY id
ORDER BY sum(if(friends.online =1,1,0)) ASC;
However, I would write it as:
SELECT friends.id, sum(friends.online = 1) AS online_friends, ...,...
FROM friends
GROUP BY id
ORDER BY sum(friends.online = 1) ASC;
You can also change the order by to:
ORDER BY online_friends
If you find that it still doesn't work, then please set up a SQL Fiddle so we can see what is happening.
You may try this:
SELECT friends.id, groups, sum(friends.online) AS online_friends
FROM some_table
where friends.online = 1
GROUP BY friends.id, groups
ORDER BY friends.online ASC;
I assumed that online and offline are having the value as 1 and 0 respectively.
Related
I have a table called deals, it has records like this for example
id deal_ref objectname status
1 1234 tom correct
2 1234 tom correct
3 1234 jerry wrong
4 1234 tom correct
I need to identify all latest deals where the status is "correct for example, but the last entry(row 4) must meet the following criteria, where the Max ID is equal to the deal_ref and the status is correct
I tried this
select distinct deal_ref, deal_status
from dealv1 d
where d.deal_ref = max(id)
and d.deal_status = 'Prospect'
and date_created between '2022-11-02 00:00:00' and '2022-11-04 00:00:00'
You use other names in your SQL than in the table (deal_status, date_created).
Nevertheless try do it the following:
SELECT *
FROM dealv1 d
WHERE status = 'correct'
ORDER BY ID DESC
LIMIT 1
i don´t get exactly what you are trying to do with the maxID. You just want the one row where deal_ref=max(id) and status is correct?
Then add
AND deal_ref = (SELECT MAX(id) from dealv1)
after 'correct' from the above statement.
Guys the query below worked, but its displaying multiple deal_refs that has the same deal_ref, for example 2 rows below each other with deal_ref 1234 twice.
SELECT *
FROM dealv1 d
WHERE status = 'correct'
ORDER BY ID DESC
LIMIT 1
I have a query which actually have a sorting using order by clause. i have a table like following...
user_id user_name user_age user_state user_points
1 Rakul 30 CA 56
2 Naydee 29 NY 144
3 Jeet 40 NJ 43
.....
i have following query...
select * from users where user_state = 'NY' order by user_points desc limit 50;
This gives me the list of 50 people with most points. I wanted to give least preference to few people who's id's were known. Incase if i do not have enough 50 records then those id's should come in the last in the list. I do not want the users 2 and 3 to come on top of the list even though they have higher points... those people should come on the last of the list from the query. Is there any way to push specific records to last on result set irrespective of query sorting ?
If you want to move specific records (like user_id = 2 and 3) down to the list; Then you can run below Query:
mysql> select *,IF(user_id=2 or user_id=3,0,1) as list_order from users where user_state = 'NY' order by list_order desc, user_points desc limit 50;
select * from (
select *
from users
where user_state = 'NY'
-- this order by ensures that 2 and 3 are included
order by case when user_id in (2,3) then 1 else 2 end, user_points desc
limit 50
) as top48plus2n3
-- this order by ensures that 2 and 3 are last
order by case when user_id in (2,3) then 2 else 1 end, user_points desc
Edit: changed id by user_id and corrected outside order by (sorry about that)
On the inner select:
By using this case calculation, what you do is ensuring that records with ids equal to 2 and 3 are "important" (firstly ordered in the order by). Those receive 1 while the others receive 2 as order value, only after that points are relevant.
On the outer select:
Records with ids 2 and 3 recieve 2 as order value, while the rest recieve 1. So they go last irrespective of its "default"
Here you have a reduced fiddle http://sqlfiddle.com/#!9/377c1/1
I'm fighting a bit with a query I'm building. Let's say I've got a DB table like this:
id | some_string
----------------
1 | 'lala'
2 | 'jeje'
3 | 'poopoo'
4 | 'wicked wicked'
I now want to get the last three records (2, 3, and 4) ordered ascending by key. I tried this:
SELECT * FROM tableName LIMIT 3 ORDER BY id ASC
This gets me the first three records, instead of the last three. I can of course also use the query below, which gets me the correct records, but then I don't get them in Ascending order:
SELECT * FROM tableName LIMIT 3 ORDER BY id DESC
Does anybody know how I can get the last three records in an ascending order? All tips are welcome!
select * from (
select * from table_name order by id desc limit 3
) last_3_rows
order by id
Sort on the resulting result set ie. do a select * from (<your query here>) order by id
This is a query inside another query. that reorders your query.(SQL - How to reorder a select query that uses the limit constraint)
select * FROM (SELECT * FROM tableName LIMIT 3 ORDER BY id DESC) AN_UNUSUAL_NAME ORDER BY id ASC
I have the following table (user_record) with millions of rows like this:
no uid s
================
1 a 999
2 b 899
3 c 1234
4 a 1322
5 b 933
-----------------
The uid can be duplicate .What I need is to show the top ten records(need inclued uid and s) with no duplicate uid order by s (desc). I can do this by two steps in the following SQL statements:
SELECT distinct(uid) FROM user_record ORDER BY s DESC LIMIT 10
SELECT uid,s FROM user_record WHERE uid IN(Just Results)
I just wana know is there a bit more efficient way in one statement?
Any help is greatly appreciated.
ps:I also have following the SQL statement:
select * from(select uid,s from user_record order by s desc) as tb group by tb.uid order by tb.s desc limit 10
but it's slow
The simpliest would be by using MAX() to get the highest s for every uid and sorted it based on the highest s.
SELECT uid, MAX(s) max_s
FROM TableName
GROUP BY uid
ORDER BY max_s DESC
LIMIT 10
SQLFiddle Demo
The disadvantage of the query above is that it doesn't handles duplicates if for instance there are multiple uid that have the same s and turn out to be the highest value. If you want to get the highest value s with duplicate, you can do by calculating it on the subquery and joining the result on the original table.
SELECT a.*
FROM tableName a
INNER JOIN
(
SELECT DISTINCT s
FROM TableName
ORDER BY s DESC
LIMIT 10
) b ON a.s = b.s
ORDER BY s DESC
I need to do GROUP BY after ORDER BY. I don't understand why MySQL doesn't support that.
This is my code:
SELECT
`pages`.`id`,
`contents`.`id_language`,
[...]
[...]
ORDER BY
FIND_IN_SET(`languages`.`id`, '3') DESC
[the GROUP BY]
The results will be something like this:
id | id_language | ...
1 3
1 1
2 3
2 5
2 1
I need to group by ID, I need only the first result and I need to save in a view. I can't use a SUBQUERY because of that.
The result need to be:
id | id_language | ...
1 3
2 3
Note: Don't get confused by id_language = 3, because it isn't a rule.
SELECT id, idl
FROM (SELECT
`pages`.`id` as id,
`contents`.`id_language` as idl,
[...]
[...]
ORDER BY
FIND_IN_SET(`languages`.`id`, '3') DESC
) d
GROUP BY d.id
Group By will group result sets, and is generally used for aggregation.
Order By is the way that results are sorted.
You may want an additional column in your original query that you GROUP BY, along with whatever you're currently grouping by. That column, when grouped, could then be used to order afterward. For instance:
SELECT
SUM(IF(`languages`.`id` = 3, 1, 0)) AS languageOrder,
`pages`.`id`,
`contents`.`id_language`,
[...]
[...]
[GROUP BY...]
ORDER BY languageOrder DESC
I would intend for languageOrder to be positive for groups that contain language #3, 0 otherwise. So groups that contain language 3 will be at the top.
Very amusing, try
select * from your_table
where id_language=3
order by id;
As far I can tell, the rule set is id_language=3,
which make no differences from using where