I am trying to get the second last records use mysql.
I did some research, some sample has fix gap between numbers or date. But my situation is that the contract_id is not always +1 from the previous one. Anyone ideas? Thank you so much.
merchant_id contract_id start_date
10 501 2016-05-01
10 506 2016-06-01
13 456 2015-12-01
13 462 2016-01-01
14 620 2016-06-01
14 642 2016-07-01
14 656 2016-07-05
merchant_id Second_last_contract_id
10 501
13 456
14 642
contract_id != previous contract_id + X. (The X is not fixed)
'start_date' tell us the contracts creating order.
Here's one option using user-defined variables to establish a row number per group of merchants and then filtering on the 2nd in each group ordered by contracts:
select *
from (
select *,
#rn:=if(#prevMerchantId=merchantid,
#rn+1,
if(#prevMerchantId:=merchantid, 1, 1)
) as rn
from yourtable cross join (select #rn:=0, #prevMerchantId:=null) t
order by merchantId, contractid desc
) t
where rn = 2
SQL Fiddle Demo
Here's another option, filtering the results of GROUP_CONCAT() using SUBSTRING_INDEX():
SELECT merchant_id,
SUBSTRING_INDEX(SUBSTRING_INDEX(
GROUP_CONCAT(contract_id ORDER BY start_date DESC),
',', 2), ',', -1) AS Second_last_contract_id
FROM the_table
GROUP BY merchant_id
See it on sqlfiddle.
Related
Using MariaDB and trying to see if I can get pull original rankings for each row of a table based on the create date.
For example, imagine a scores table that has different scores for different users and categories (lower score is better in this case)
id
leaderboardId
userId
score
submittedAt ↓
rankAtSubmit
9
15
555
50.5
2022-01-20 01:00:00
2
8
15
999
58.0
2022-01-19 01:00:00
3
7
15
999
59.1
2022-01-15 01:00:00
3
6
15
123
49.0
2022-01-12 01:00:00
1
5
15
222
51.0
2022-01-10 01:00:00
1
4
14
222
87.0
2022-01-09 01:00:00
1
5
15
555
51.0
2022-01-04 01:00:00
1
The "rankAtSubmit" column is what I'm trying to generate here if possible.
I want to take the best/smallest score of each user+leaderboard and determine what the rank of that score was when it was submitted.
My attempt at this failed because in MySQL you cannot reference outer level columns more than 1 level deep in a subquery resulting in an error trying to reference t.submittedAt in the following query:
SELECT *, (
SELECT ranking FROM (
SELECT id, RANK() OVER (PARTITION BY leaderboardId ORDER BY score ASC) ranking
FROM scores x
WHERE x.submittedAt <= t.submittedAt
GROUP BY userId, leaderboardId
) ranks
WHERE ranks.id = t.id
) rankAtSubmit
FROM scores t
Instead of using RANK(), I was able to accomplish this by with a single subquery that counts the number of users that have a score that is lower than and submitted before the given score.
SELECT id, userId, score, leaderboardId, submittedAt,
(
SELECT COUNT(DISTINCT userId) + 1
FROM scores t2
WHERE t2.userId = t.userId AND
t2.leaderboardId = t.leaderboardId AND
t2.score < t.score AND
t2.submittedAt <= t.submittedAt
) AS rankAtSubmit
FROM scores t
What I understand from your question is you want to know the minimum and maximum rank of each user.
Here is the code
SELECT userId, leaderboardId, score, min(rankAtSubmit),max(rankAtSubmit)
FROM scores
group BY userId,
leaderboardId,
scorescode here
The table provided shows all new users signing up on a specific date in the format YYYY-MM-DD.
Your query should output the change from one month to the next. Because the first month has no preceding month, your output should skip that row. Your output should look like the following table.
My table data
Table data:
ID DateJoined
1 2017-01-06
2 2017-01-12
3 2017-01-16
4 2017-01-25
5 2017-02-05
6 2017-02-07
7 2017-02-21
8 2017-03-05
9 2017-03-07
10 2017-03-14
11 2017-03-16
12 2017-03-25
13 2017-03-25
14 2017-03-25
15 2017-03-25
16 2017-03-26
17 2017-04-05
18 2017-04-14
19 2017-04-21
20 2017-05-07
23 2017-05-14
24 2017-05-16
25 2017-05-25
26 2017-05-25
27 2017-05-25
28 2017-05-25
Enter image description here
I want this output:
count all records from every month and subtract it from the next month record.
This is my query:
SELECT
MONTH(L.joindate),
COUNT(L.joindate) - COUNT(R.joindate),
MONTH(R.joindate),
COUNT(R.joindate)
FROM
userlog AS L
LEFT JOIN
userlog AS R
ON MONTH(R.joindate)= (SELECT MIN(MONTH(joindate)) FROM userlog WHERE MONTH(joindate) < MONTH(L.joindate))
GROUP BY (MONTH(L.joindate)),(MONTH(R.joindate));
Use lag(), available in MySQL 8.0:
select date_format(joindate, '%Y-%m-01') joinmonth,
count(*) - lag(count(*), 1, 0) over(order by date_format(joindate, '%Y-%m-01')) m2m
from userlog
group by joinmonth
Note that I changed the logic to truncate dates to the first of month to use date_format().
In earlier versions, you can use a correlated subquery:
select date_format(joindate, '%Y-%m-01') joinmonth,
count(*) - (
select count(*)
from userlog l1
where l1.joindate >= date_format(l.joindate, '%Y-%m-01') - interval 1 month
and l1.joindate < date_format(l.joindate, '%Y-%m-01')
) m2m
from userlog l
group by joinmonth
LIMIT 12 OFFSET 1
You need to use Lag. Also, since it says you need to skip the first row so I have used the not null condition. I believe this query should work.
select
Month,
MonthToMonthChange
from
(
select
m_name as Month,
(total_id - diff) as MonthToMonthChange
from
(
select
total_id,
m_name,
Lag(total_id, 1) OVER(
ORDER BY
m_num ASC
) AS diff
from
(
select
MonthNAME(DateJoined) m_name,
Month(DateJoined) m_num,
count(*) total_id
from
maintable
Group by
m_name,
m_num
) as first_subquery
) as second_subquery
) as final_query
where
MonthToMonthChange IS NOT NULL;
select
MONTHNAME(UL1.DateJoined) as MONTH,
count(UL1.DateJoined) -
(
select count(UL2.DateJoined)
from tablename UL2
where MONTH(UL2.DateJoined )=MONTH(UL1.DateJoined) -1
) as MonthToMonthChange
from tablename UL1
where Month(UL1.DateJoined)!=1
Group by MONTHNAME(UL1.DateJoined)
order by UL1.DateJoined ASC;
https://i.stack.imgur.com/BXXDb.png
I tried this and it worked
select date_format(DateJoined, CONCAT('%M')) as Month,
count(*) - lag(count(*), 1, 0) over(order by date_format(DateJoined, CONCAT('%m'))) MonthToMonthChange
from maintable_OKLOT
group by Month
limit 12 offset 1
I have a table with records that are updated every minute with a decimal value (10,2). To ignore measure errors I want to have the number that has been inserted the most.
Therefor I tried:
SELECT date_time,max(sensor1),count(ID)
FROM `weigh_data
group by day(date_time),sensor1
This way I get the number of records
Datetime sensor1 count(ID)
2020-03-19 11:49:12 33.22 3
2020-03-19 11:37:47 33.36 10
2020-03-20 07:32:02 32.54 489
2020-03-20 00:00:43 32.56 891
2020-03-20 14:20:51 32.67 5
2020-03-21 07:54:16 32.50 1
2020-03-21 00:00:58 32.54 1373
2020-03-21 01:15:16 32.56 9
2020-03-22 08:35:12 32.52 2
2020-03-22 00:00:40 32.54 575
2020-03-22 06:50:54 32.58 1
What I actually want is for each day one row which has the highest count(ID)
Anyone can help me out on this?
With newer MySQL (8.0 and later) you can use the RANK window function to rank the rows according to the count.
Note that this will return all "ties" which means if there are 100 readings of X and 100 readings of Y (and 100 is the max), both X and Y will be returned.
WITH cte AS (
SELECT
DATE(date_time), sensor1,
RANK() OVER (PARTITION BY DATE(date_time) ORDER BY COUNT(*) DESC) rnk
FROM `weigh_data` GROUP BY DATE(date_time), sensor1
)
SELECT * FROM cte WHERE rnk=1
If you just want to pick one (non deterministic) of the ties, you can instead use ROW_NUMBER in place of RANK
A DBfiddle to test with.
Here is a solution based on a correlated subquery, that works in all versions of MySQL:
select w.*
from weigh_data w
where w.datetime = (
select w1.datetime
from weigh_data w1
where w1.datetime >= date(w.datetime) and w1.datetime < date(w.datetime) + interval 1 day
order by sensor1 desc
limit 1
)
Just like the window function solution using rank(), this allows top ties.
For performance, you want an index on (datetime, sensor1).
I have the following query:
SELECT
DATE(`timeStamp`),COUNT(*)
FROM
`wf`.sh`
WHERE
(DATE(`timeStamp`) >= curdate()- INTERVAL 31 DAY)
GROUP BY
DATE(`timeStamp`)
HAVING
COUNT(DATE(`timeStamp`)) > 0
ORDER BY
DATE(`timeStamp`) ASC;
The purpose of this query is to retrieve the amount of users online in my system per day, in the space of a month.
Example dataset:
uID timeStamp
1 2016-11-28 00:27:01
1 2016-11-28 01:10:15
1234 2016-11-28 02:50:00
2 2016-11-28 06:11:09
47 2016-11-28 08:32:48
1246 2016-11-28 09:51:47
In its current format, this query returns the count of rows with duplicate dates, for example:
timeStamp COUNT(*)
2017-01-29 256
2017-01-30 224
2017-01-31 240
2017-02-01 95
2017-02-02 136
I have another field uID; I need to modify my query so that GROUP also ignores rows with a duplicate uID field for each day. I tried creating another GROUP BY but was given an error that 'incorrect GROUP BY clause' (or something of that nature).
Can this be done via pure MySQL?
You can use a subselect
SELECT
visitDate,COUNT(*)
FROM
(SELECT DISTINCT DATE(`timeStamp`) as visitDate, uID FROM `wf`.sh`) alias_t
WHERE
(visitDate >= curdate()- INTERVAL 31 DAY)
GROUP BY
visitDate
HAVING
COUNT(visitDate) > 0
ORDER BY
visitDate ASC;
I have an MySQL table, similar to this example:
c_id date value
66 2015-07-01 1
66 2015-07-02 777
66 2015-08-01 33
66 2015-08-20 200
66 2015-08-21 11
66 2015-09-14 202
66 2015-09-15 204
66 2015-09-16 23
66 2015-09-17 0
66 2015-09-18 231
What I need to get is count of periods where dates are in row. I don't have fixed start or end date, there can be any.
For example: 2015-07-01 - 2015-07-02 is one priod, 2015-08-01 is second period, 2015-08-20 - 2015-08-21 is third period and 2015-09-14 - 2015-09-18 as fourth period. So in this example there is four periods.
SELECT
SUM(value) as value_sum,
... as period_count
FROM my_table
WHERE cid = 66
Cant figure this out all day long.. Thx.
I don't have enough reputation to comment to the above answer.
If all you need is the NUMBER of splits, then you can simply reword your question: "How many entries have a date D, such that the date D - 1 DAY does not have an entry?"
In which case, this is all you need:
SELECT
COUNT(*) as PeriodCount
FROM
`periods`
WHERE
DATE_ADD(`date`, INTERVAL - 1 DAY) NOT IN (SELECT `date` from `periods`);
In your PHP, just select the "PeriodCount" column from the first row.
You had me working on some crazy stored procedure approach until that clarification :P
I should get deservedly flamed for this, but anyway, consider the following...
DROP TABLE IF EXISTS my_table;
CREATE TABLE my_table
(date DATE NOT NULL PRIMARY KEY
,value INT NOT NULL
);
INSERT INTO my_table VALUES
('2015-07-01',1),
('2015-07-02',777),
('2015-08-01',33),
('2015-08-20',200),
('2015-08-21',11),
('2015-09-14',202),
('2015-09-15',204),
('2015-09-16',23),
('2015-09-17',0),
('2015-09-18',231);
SELECT x.*
, SUM(y.value) total
FROM
( SELECT a.date start
, MIN(c.date) end
FROM my_table a
LEFT
JOIN my_table b
ON b.date = a.date - INTERVAL 1 DAY
LEFT
JOIN my_table c
ON c.date >= a.date
LEFT
JOIN my_table d
ON d.date = c.date + INTERVAL 1 DAY
WHERE b.date IS NULL
AND c.date IS NOT NULL
AND d.date IS NULL
GROUP
BY a.date
) x
JOIN my_table y
ON y.date BETWEEN x.start AND x.end
GROUP
BY x.start;
+------------+------------+-------+
| start | end | total |
+------------+------------+-------+
| 2015-07-01 | 2015-07-02 | 778 |
| 2015-08-01 | 2015-08-01 | 33 |
| 2015-08-20 | 2015-08-21 | 211 |
| 2015-09-14 | 2015-09-18 | 660 |
+------------+------------+-------+
4 rows in set (0.00 sec) -- <-- This is the number of periods
there is a simpler way of doing this, see here SQLfiddle:
SELECT min(date) start,max(date) end,sum(value) total FROM
(SELECT #i:=#i+1 i,
ROUND(Unix_timestamp(date)/(24*60*60))-#i diff,
date,value
FROM tbl, (SELECT #i:=0)n WHERE c_id=66 ORDER BY date) t
GROUP BY diff
This select groups over the same difference between sequential number and date value.
Edit
As Strawberry remarked quite rightly, there was a flaw in my apporach, when a period spans a month change or indeed a change into the next year. The unix_timestamp() function can cure this though: It returns the seconds since 1970-1-1, so by dividing this number by 24*60*60 you get the days since that particular date. The rest is simple ...
If you only need the count, as your last comment stated, you can do it even simpler:
SELECT count(distinct diff) period_count FROM
(SELECT #i:=#i+1 i,
ROUND(Unix_timestamp(date)/(24*60*60))-#i diff,
date,value
FROM tbl,(SELECT #i:=0)n WHERE c_id=66 ORDER BY date) t
Tnx. #cars10 solution worked in MySQL, but could not manage to get period count to echo in PHP. It returned 0. Got it working tnx to #jarkinstall. So my final select looks something like this:
SELECT
sum(coalesce(count_tmp,coalesce(count_reserved,0))) as sum
,(SELECT COUNT(*) FROM my_table WHERE cid='.$cid.' AND DATE_ADD(date, INTERVAL - 1 DAY) NOT IN (SELECT date from my_table WHERE cid='.$cid.' AND coalesce(count_tmp,coalesce(count_reserved,0))>0)) as periods
,count(*) as count
,(min(date)) as min_date
,(max(date)) as max_date
FROM my_table WHERE cid=66
AND coalesce(count_tmp,coalesce(count_reserved,0))>0
ORDER BY date;