Hoping you will be able to help me with this MYSQL statement. I have a table like so:
|id |duration |start |
|1110460 |8.2 |20171211 |
|2221104 |8.9 |20171112 |
|1110460 |3.2 |20171113 |
|1110460 |4.4 |20171214 |
|3331938 |3.2 |20180115 |
|3331722 |5.4 |20171216 |
|1948212 |9.2 |20171217 |
|9219302 |3.2 |20171218 |
What I want to do is list the top 4 IDs by total duration for a given start month in descending order.
For example, for the top 4 IDs for 201712:
|id |duration |
|1110460 |12.6 |
|1948212 |9.2 |
|3331722 |5.4 |
|9219302 |3.2 |
Any help would be appreciated. This is what I have so far, but it has been returning the incorrect results:
SELECT id, sum(duration) FROM table WHERE
start LIKE '201712%' ORDER BY sum(duration) DESC LIMIT 4
You should use group by
SELECT id, sum(duration)
FROM table
WHERE start LIKE '201712%'
GROUP BY id
ORDER BY sum(duration) DESC LIMIT 4
To get 4 distinct maximum sums and their ids you could use following
SELECT a.id, a.sum_duration
FROM (SELECT id, sum(duration) sum_duration
FROM demo
WHERE `start` LIKE '201712%'
GROUP BY id
) a
JOIN (SELECT distinct sum(duration) max_durations
FROM demo
WHERE `start` LIKE '201712%'
GROUP BY id
ORDER BY sum(duration) DESC
LIMIT 4
) b on a.sum_duration = b.max_durations
The above version will return morethan 4 rows if 2 ids have same result of sum
DEMO
Related
I would like to count how many new unique users the database gets each day for all days recorded.
There will not be any duplicate ids per day, but there will be duplicates over multiple days.
If my table looks like this :
ID | DATE
---------
1 | 2022-05-21
1 | 2022-05-22
2 | 2022-05-22
1 | 2022-05-23
2 | 2022-05-23
1 | 2022-05-24
2 | 2022-05-24
3 | 2022-05-24
I would like the results to look like this :
DATE | NEW UNIQUE IDs
---------------------------
2022-05-21 | 1
2022-05-22 | 1
2022-05-23 | 0
2022-05-24 | 1
A query such as :
SELECT `date` , COUNT( DISTINCT id)
FROM tbl
GROUP BY DATE( `date` )
Will return the count per day and will not take into account previous days.
Any assistance would be appreciated.
Edit : Using MySQL 8
The user is new when the date is the least date for this user.
So you need in something like
SELECT date, COUNT(new_users.id)
FROM calendar
LEFT JOIN ( SELECT id, MIN(date) date
FROM test
GROUP BY id ) new_users USING (date)
GROUP BY date
calendar is either static or dynamically generated table with needed dates list. It can be even SELECT DISTINCT date FROM test subquery.
Start with a subquery showing the earliest date where each id appears.
SELECT MIN(`date`) `firstdate`, id
FROM tbl
GROUP BY id
Then do your count on that subquery. here.
SELECT firstdate, COUNT(*)
FROM (
SELECT MIN(`date`) `firstdate`, id
FROM tbl
GROUP BY id
) m
GROUP BY firstdate
That gives you what you want.
But it doesn't have rows for the dates where no new user ids first appeared.
Only count (and sum) the rows where the left join fails:
SELECT
m1.`DATE` ,
sum(CASE WHEN m2.id is null THEN 1 ELSE 0 END) as C
FROM mytable m1
LEFT JOIN mytable m2 ON m2.`DATE`<m1.`DATE` AND m2.ID=m1.ID
GROUP BY m1.`DATE`
see: DBFIDDLE
I have a table. It has the following structure
goods_receiving_items
id
item_id
quantity
created_at
I am trying to fetch rows against which have the following conditions
Has one item_id
When the sum of the quantity column equals a certain value
So for example I have the following data
+----+---------+----------+------------+
| id | item_id | quantity | created_at |
+----+---------+----------+------------+
| 1 | 2 | 11 | 2019-10-10 |
| 2 | 3 | 110 | 2019-10-11 |
| 3 | 2 | 20 | 2019-11-09 |
| 4 | 2 | 5 | 2019-11-10 |
| 5 | 2 | 1 | 2019-11-11 |
+----+---------+----------+------------+
I have tried the following query:
SET #sum:= 0;
SELECT item_id, created_at, (#sum:= #sum + quantity) AS SUM, quantity
FROM goods_receiving_items
WHERE item_id = 2 AND #sum<= 6
ORDER BY created_at DESC
If I don't use ORDER BY, then the query will give me ID '1'. But if I use ORDER BY it will return all the rows with item_id = 2.
What should be returned are IDs '5' and '4' exclusively in this order
I can't seem to resolve this and ORDER BY is essential to my task.
Any help would be appreciated
You should use the order by on the resulting set
you could do this using a subquery
SET #sum:= 0;
select t.*
from t (
SELECT item_id
, created_at
, (#sum:= #sum + quantity) as sum
, quantity
FROM goods_receiving_items
WHERE item_id = 2 AND #sum<= 6
) t
ORDER BY created_at DESC
You should try an INNER JOIN with SELECT min(created_at) or SELECT max(created_at)
From MYSQL docs:
...the selection of values from each group cannot be influenced by
adding an ORDER BY clause. Sorting of the result set occurs after
values have been chosen, and ORDER BY does not affect which values the
server chooses.
The answers on the following might help in more detail: MYSQL GROUP BY and ORDER BY not working together as expected
After searching around, I have made up the following query
SELECT
t.id, t.quantity, t.created_at, t.sum
FROM
( SELECT
*,
#bal := #bal + quantity AS sum,
IF(#bal >= $search_number, #doneHere := #doneHere + 1 , #doneHere) AS whereToStop
FROM goods_receiving_items
CROSS JOIN (SELECT #bal := 0.0 , #doneHere := 0) var
WHERE item_id = $item_id
ORDER BY created_at DESC) AS t
WHERE t.whereToStop <= 1
ORDER BY t.created_at ASC
In the above query, $search_number is a variable that holds the value that has to be reached. $item_id is the item we are searching against.
This will return all rows for which the sum of the column quantity makes up the required sum. The sum will be made with rows in descending order by created_at and then will be rearranged in ascending order.
I was using this query to calculate the cost when a certain amount of items are being used in an inventory management system; so this might help someone else do the same. I took most of the query from another question here on StackOverflow
I have an table like that:
id | name | v (lvl)
11 | Jane | 6
12 | John | 5
13 | Jane | 6
14 | John | 5
15 | Jane | 7
16 | Jane | 5
In my autocomplete form now id like to group the names but get the last value (value with biggest id). In the example above would be
Jane | 5
I tried with combinations like distinct, group by, order by. But im always get
Jane | 6
or grouped like this and reversed:
Jane | 6
Jane | 7
Jane | 5
I would need something like this:
SELECT name,lvl FROM
(
SELECT DISTINCT name, lvl FROM pora WHERE name LIKE 'Jane' ORDER BY lvl DESC
)
GROUP BY name
EDIT: I won't get the highest lvl, i want get the lvl of the highest id, grouped by name. Thats all. My example above would be the best explanation what i like to get.
In the inner query i change the order to DESC for all and in the outer i group it by names. But i get an error for this.
EDIT 2 I finally did at my own. The correct solution (i was already close):
SELECT a.name, a.lvl FROM
(
SELECT DISTINCT name, lvl FROM pora WHERE name LIKE 'Jane' ORDER BY id DESC
)as a
GROUP BY name
LIKE without % is just =
SELECT *
FROM yourTable
WHERE name = 'Jane'
ORDER BY id DESC
LIMIT 1
But because you mention autocomplete functionality you should use:
WHERE name LIKE 'Jane%'
To have the latest, you need to have a field dateAdded which stores the date you ran the insert command.
Following which, you use MAX(dateAdded) to get the latest ID (since, as you mentioned, it may decrease as well)
UPDATE:
if ID doesn't decrease, you can always use MAX(ID)
SELECT MAX(id), v from tablename where name = 'Jane'
UPDATE:
This has been tested:
SELECT ID, v from tableName where ID = (SELECT MAX(ID) as ID from tableName where name like '%Jane%')
Try the following query (h/t #lamak)
WITH CTE AS
(
SELECT *,
RN = ROW_NUMBER() OVER(PARTITION BY name
ORDER BY [id] DESC)
FROM poro
)
SELECT *
FROM CTE
WHERE RN = 1
my mysql table is like:
+---------+---------+------------+-----------------------+---------------------+
| visitId | userId | locationId | comments | time |
+---------+---------+------------+-----------------------+---------------------+
| 1 | 3 | 12 | It's a good day here! | 2012-12-12 11:50:12 |
+---------+---------+------------+-----------------------+---------------------+
| 2 | 3 | 23 | very beautiful | 2012-12-12 12:50:12 |
+---------+---------+------------+-----------------------+---------------------+
| 3 | 3 | 52 | nice | 2012-12-12 13:50:12 |
+---------+---------+------------+-----------------------+---------------------+
witch records visitors' trajectory and some comments on the places visited
I want to find visitors visited more than one place in a day, along with the specific day AND the places, Not only the count.
I tried the subquery:
mysql> SELECT userId, locationId, time FROM visits
WHERE (userId,DATE(time)) in (
SELECT userNum, Date(weiboTime) from visits GROUP BY userNum, Date(wei
boTime) Having COUNT(*)>1);
And the joint query:
mysql> select v2.userId, v1.loacationId, v1.time from visits as v1, visits as
v2 where v1.userId=v2.userId GROUP BY v2.userId, Date(v2.time) HAVING
COUNT(DISTINCT v2.locationId);
I am not sure whether it is correct for the second one. But both of them take too long time. Any suggestions for what should I do?
UPDATE
mysql> SELECT t.userId, locationId, t.time FROM (
SELECT userId, time
FROM visits GROUP BY userId,Date(time)
HAVING COUNT(*) > 1) AS t, visits
WHERE t.userId=visits.userId AND t.time=visits.time;
hope this will make myself more clear.
Your queries are including locationId, but your stated goal is to get user/date combos that had more than 1 visit in a day. Here's the sql to get that:
select userId, date(time), count(*)
from visits
group by userId, date(time)
having count(*) > 1;
Update:
To get all visits from user/day combo visits greater than 1:
select *
from visits
where (userId, date(time)) in (
select userId, date(time)
from visits
group by userId, date(time)
having count(*) > 1);
I think you'd be better suited using a GROUP BY in a subquery with your count. From MySQL's count documentation, you can do something like:
mysql> SELECT userId, locationId, time, visitCount
FROM
(SELECT COUNT(*) as visitCount
FROM visits
GROUP BY userId)
WHERE visitCount > 1;
I'd assume the slowness you're encountering comes from the HAVING and DISTINCT in your WHERE clauses.
My table looks like this (and I'm using MySQL):
m_id | v_id | timestamp
------------------------
6 | 1 | 1333635317
34 | 1 | 1333635323
34 | 1 | 1333635336
6 | 1 | 1333635343
6 | 1 | 1333635349
My target is to take each m_id one time, and order by the highest timestamp.
The result should be:
m_id | v_id | timestamp
------------------------
6 | 1 | 1333635349
34 | 1 | 1333635336
And i wrote this query:
SELECT * FROM table GROUP BY m_id ORDER BY timestamp DESC
But, the results are:
m_id | v_id | timestamp
------------------------
34 | 1 | 1333635323
6 | 1 | 1333635317
I think it causes because it first does GROUP_BY and then ORDER the results.
Any ideas? Thank you.
One way to do this that correctly uses group by:
select l.*
from table l
inner join (
select
m_id, max(timestamp) as latest
from table
group by m_id
) r
on l.timestamp = r.latest and l.m_id = r.m_id
order by timestamp desc
How this works:
selects the latest timestamp for each distinct m_id in the subquery
only selects rows from table that match a row from the subquery (this operation -- where a join is performed, but no columns are selected from the second table, it's just used as a filter -- is known as a "semijoin" in case you were curious)
orders the rows
If you really don't care about which timestamp you'll get and your v_id is always the same for a given m_i you can do the following:
select m_id, v_id, max(timestamp) from table
group by m_id, v_id
order by max(timestamp) desc
Now, if the v_id changes for a given m_id then you should do the following
select t1.* from table t1
left join table t2 on t1.m_id = t2.m_id and t1.timestamp < t2.timestamp
where t2.timestamp is null
order by t1.timestamp desc
Here is the simplest solution
select m_id,v_id,max(timestamp) from table group by m_id;
Group by m_id but get max of timestamp for each m_id.
You can try this
SELECT tbl.* FROM (SELECT * FROM table ORDER BY timestamp DESC) as tbl
GROUP BY tbl.m_id
SQL>
SELECT interview.qtrcode QTR, interview.companyname "Company Name", interview.division Division
FROM interview
JOIN jobsdev.employer
ON (interview.companyname = employer.companyname AND employer.zipcode like '100%')
GROUP BY interview.qtrcode, interview.companyname, interview.division
ORDER BY interview.qtrcode;
I felt confused when I tried to understand the question and answers at first. I spent some time reading and I would like to make a summary.
The OP's example is a little bit misleading.
At first I didn't understand why the accepted answer is the accepted answer.. I thought that the OP's request could be simply fulfilled with
select m_id, v_id, max(timestamp) as max_time from table
group by m_id, v_id
order by max_time desc
Then I took a second look at the accepted answer. And I found that actually the OP wants to express that, for a sample table like:
m_id | v_id | timestamp
------------------------
6 | 1 | 11
34 | 2 | 12
34 | 3 | 13
6 | 4 | 14
6 | 5 | 15
he wants to select all columns based only on (group by)m_id and (order by)timestamp.
Then the above sql won't work. If you still don't get it, imagine you have more columns than m_id | v_id | timestamp, e.g m_id | v_id | timestamp| columnA | columnB |column C| .... With group by, you can only select those "group by" columns and aggreate functions in the result.
By far, you should have understood the accepted answer.
What's more, check row_number function introduced in MySQL 8.0:
https://www.mysqltutorial.org/mysql-window-functions/mysql-row_number-function/
Finding top N rows of every group
It does the simlar thing as the accepted answer.
Some answers are wrong. My MySQL gives me error.
select m_id,v_id,max(timestamp) from table group by m_id;
#abinash sahoo
SELECT m_id,v_id,MAX(TIMESTAMP) AS TIME
FROM table_name
GROUP BY m_id
#Vikas Garhwal
Error message:
[42000][1055] Expression #2 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'testdb.test_table.v_id' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
Why make it so complicated? This worked.
SELECT m_id,v_id,MAX(TIMESTAMP) AS TIME
FROM table_name
GROUP BY m_id
Just you need to desc with asc. Write the query like below. It will return the values in ascending order.
SELECT * FROM table GROUP BY m_id ORDER BY m_id asc;