How To Display Value by Max Date / Last Date of Year - mysql

I need a query to display a value (saldo) following max date / the last date of year, but I have tried but it and it's always showing a value from the first date.
Here is my query:
SELECT saldo, MAX(tgl_lap) FROM laporan GROUP BY DATE_FORMAT(tgl_lap,'%Y','%m') ORDER BY DATE_FORMAT(tgl_lap,'%Y','%m') DESC
my result:
And my expectated result should be only showing values like in the red box below:

WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY YEAR(tgl_lap) ORDER BY tgl_lap DESC) rn
FROM laporan
)
SELECT *
FROM cte
WHERE rn = 1

select *
From laporan t
where t.tgl_lap =
(
select max(a.tgl_lap)
from laporan a
where year(t.tgl_lap) = year(a.tgl_lap)
)
According to the sample data, considering there would be only one record per date, the sub query will return max date of the year and main query will return complete record of the last or max date of the year.

Related

Group by and find amount difference sql same table

I have table like this below :
And i want the result like : sum the amount group by date and find the difference between the result
If you are using MySQL 8.0 then you can use lag() to achieve expected output.
select
date,
amount,
coalesce((amount - last_val), amount) as diff
from
(
select
date,
amount,
lag(amount) over (order by date) as last_val
from
(
select
date,
sum(amount) as amount
from myTable
group by
date
) subq
) subo
order by
date

Average timediff of 2nd and 3rd datetimes for a group

I need to find the average time in days between a customer's second order and third order
I know that I need to use the timestampdiff but am quite at a loss for how to select the second and third dates and need some sort of nest.
SELECT CustomerID,
OrderDate,
diff,
avg(timestampdiff(day, start_date, end_date)) AS average_days
FROM () o3
WHERE date3, date2
ORDER BY CustomerID, OrderDate;
Table
To achieve your desired result, you first need to calculate ROW_NUMBER from your data PARTITION BY CustmerId. Then keep rows only with RowNumber IN (2,3) and then get the DateDiff between two days. The following query will help getting your desired results-
SELECT CustomerID,datediff(MAX(OrderDate),MIN(OrderDate))
FROM
(
SELECT *,
#row_num :=IF(#prev_value = concat_ws('',CsutomerID),#row_num+1,1)AS RowNumber
, #prev_value := concat_ws('',CsutomerID)
FROM your_table A
ORDER BY CustomerID,OrderDate
)B
WHERE B.RowNumber IN (2,3)
GROUP BY CustomerID;

How to find which year do values tend to increase in ? in SQL

Basically I have a table like this:
Table Time:
ID.......Date
1......08/26/2016
1......08/26/2016
2......05/29/2016
3......06/22/2016
4......08/26/2015
5......05/23/2015
5......05/23/2015
6......08/26/2014
7......04/26/2014
8......08/26/2013
9......03/26/2013
The query should return like this
Year........CountNum
2016........4
2015........3
To find out which year does its value tend to increase in. I notice that I want to display the years that have more values (number of row in this case) than the previous year.
What I've done so far
SELECT Year, count(*) as CountNum
FROM Time
GROUP BY Year
ORDER BY CountNum DESC;
I don't know how to get the year from date format. I tried year(Date) function, but I got Null data.
Please help!
It should works fine.
select year(date), count(*) as countNum
from time
group by year(date)
order by countNum
Join the grouped data to itself with 1 year offset:
select
a.*
from
(
select year(`Date`) as _year, count(*) as _n
from time group by 1
) a
left join
(
select year(`Date`) as _year, count(*) as _n
from time group by 1
) b
on a._year = b._year-1
where a._n > b._n
order by 1

How to Group a table and get results for a row based on the previous rows' data

I have a lookup table that relates dates and people associated with those dates:
id, user_id,date
1,1,2014-11-01
2,2,2014-11-01
3,1,2014-11-02
4,3,2014-11-02
5,1,2014-11-03
I can group these by date(day):
SELECT DATE_FORMAT(
MIN(date),
'%Y/%m/%d 00:00:00 GMT-0'
) AS date,
COUNT(*) as count
FROM user_x_date
GROUP BY ROUND(UNIX_TIMESTAMP(created_at) / 43200)
But, how can get the number of unique users, that have now shown up previously? For instance this would be a valid result:
unique, non-unique, date
2,0,2014-11-01
1,1,2014-11-02
0,1,2014-11-03
Is this possibly without having to rely on a scripting language to keep track of this data?
I think this query will do what you want, at least it seems to work for your limited sample data.
The idea is to use a correlated sub-query to check if the user_id has occurred on a date before the date of the current row and then do some basic arithmetic to determine number of unique/non-unique users for each date.
Please give it a try.
select
sum(u) - sum(n) as "unique",
sum(n) as "non-unique",
date
from (
select
date,
count(user_id) u,
case when exists (
select 1
from Table1 i
where i.user_id = o.user_id
and i.date < o.date
) then 1 else 0
end n
from Table1 o
group by date, user_id
) q
group by date
order by date;
Sample SQL Fiddle
I didn't include the id column in the sample fiddle as it's not needed (or used) to produce the result and won't change anything.
This is the relevant question: "But, how can get the number of unique users, that have now shown up previously?"
Calculate the first time a person shows up, and then use that for the aggregation:
SELECT date, count(*) as FirstVisit
FROM (SELECT user_id, MIN(date) as date
FROM user_x_date
GROUP BY user_id
) x
GROUP BY date;
I would then use this as a subquery for another aggregation:
SELECT v.date, v.NumVisits, COALESCE(fv.FirstVisit, 0) as NumFirstVisit
FROM (SELECT date, count(*) as NumVisits
FROM user_x_date
GROUP BY date
) v LEFT JOIN
(SELECT date, count(*) as FirstVisit
FROM (SELECT user_id, MIN(date) as date
FROM user_x_date
GROUP BY user_id
) x
GROUP BY date
) fv
ON v.date = fv.date;

MySQL - How to select rows with the min(timestamp) per hour of a given date

I have a table of production readings and need to get a result set containing a row for the min(timestamp) for EACH hour.
The column layout is quite simple:
ID,TIMESTAMP,SOURCE_ID,SOURCE_VALUE
The data sample would look like:
123,'2013-03-01 06:05:24',PMPROD,12345678.99
124,'2013-03-01 06:15:17',PMPROD,88888888.99
125,'2013-03-01 06:25:24',PMPROD,33333333.33
126,'2013-03-01 06:38:14',PMPROD,44444444.44
127,'2013-03-01 07:12:04',PMPROD,55555555.55
128,'2013-03-01 10:38:14',PMPROD,44444444.44
129,'2013-03-01 10:56:14',PMPROD,22222222.22
130,'2013-03-01 15:28:02',PMPROD,66666666.66
Records are added to this table throughout the day and the source_value is already calculated, so no sum is needed.
I can't figure out how to get a row for the min(timestamp) for each hour of the current_date.
select *
from source_readings
use index(ID_And_Time)
where source_id = 'PMPROD'
and date(timestamp)=CURRENT_DATE
and timestamp =
( select min(timestamp)
from source_readings use index(ID_And_Time)
where source_id = 'PMPROD'
)
The above code, of course, gives me one record. I need one record for the min(hour(timestamp)) of the current_date.
My result set should contain the rows for IDs: 123,127,128,130. I've played with it for hours. Who can be my hero? :)
Try below:
SELECT * FROM source_readings
JOIN
(
SELECT ID, DATE_FORMAT(timestamp, '%Y-%m-%d %H') as current_hour,MIN(timestamp)
FROM source_readings
WHERE source_id = 'PMPROD'
GROUP BY current_hour
) As reading_min
ON source_readings.ID = reading_min.ID
SELECT a.*
FROM Table1 a
INNER JOIN
(
SELECT DATE(TIMESTAMP) date,
HOUR(TIMESTAMP) hour,
MIN(TIMESTAMP) min_date
FROM Table1
GROUP BY DATE(TIMESTAMP), HOUR(TIMESTAMP)
) b ON DATE(a.TIMESTAMP) = b.date AND
HOUR(a.TIMESTAMP) = b.hour AND
a.timestamp = b.min_date
SQLFiddle Demo
With window function:
WITH ranked (
SELECT *, ROW_NUMBER() OVER(PARTITION BY HOUR(timestamp) ORDER BY timestamp) rn
FROM source_readings -- original table
WHERE date(timestamp)=CURRENT_DATE AND source_id = 'PMPROD' -- your custom filter
)
SELECT * -- this will contain `rn` column. you can select only necessary columns
FROM ranked
WHERE rn=1
I haven't tested it, but the basic idea is:
1) ROW_NUMBER() OVER(PARTITION BY HOUR(timestamp) ORDER BY timestamp)
This will give each row a number, starting from 1 for each hour, increasing by timestamp. The result might look like:
|rest of columns |rn
123,'2013-03-01 06:05:24',PMPROD,12345678.99,1
124,'2013-03-01 06:15:17',PMPROD,88888888.99,2
125,'2013-03-01 06:25:24',PMPROD,33333333.33,3
126,'2013-03-01 06:38:14',PMPROD,44444444.44,4
127,'2013-03-01 07:12:04',PMPROD,55555555.55,1
128,'2013-03-01 10:38:14',PMPROD,44444444.44,1
129,'2013-03-01 10:56:14',PMPROD,22222222.22,2
130,'2013-03-01 15:28:02',PMPROD,66666666.66,1
2) Then on the main query we select only rows with rn=1, in other words, rows that has lowest timestamp in each hourly partition (1st row after sorted by timestamp in each hour).