I'm looking for a query that will calculate rank based on a condition, as following:
The "rank index" is the calculation I'm looking for - if the difference between previous time and current time is less than 6 hours, than the counter should remain the same. If more than 6 hours have past than promote the index by 1.
Any ideas?
Based on your explanation, the last value should be 4 not 5.
Use lag() and a cumulative sum. Assuming the datetime column is stored as a date:
select t.*,
sum(case when prev_datetime > datetime - interval '6' hour then 0 else 1 end) over
(order by datetime) as rank_index
from (select t.*,
lag(datetime) over (order by datetime) as prev_datetime
from t
) t;
Note: If you want this for each key1/key2 combination, then you want to include partition by key1, key2 in the window specifications.
I have a table of accident date I want to calculate the maximum between the difference of date i and date i + 1 which are in the same column. when we declare an accident date, I want to find the record of days without accidents.
You can use lag(). Assuming a table structure like mytable(dt), where dt is of a date-like datatype, you would do:
select max(diff)
from (select dt - lag(dt) over(order by dt) diff from mytable) t
I have a sample data, i want to get the Difference in month over month data 'Lag' column for only row B
If there always is just one row per month and id, then just use lag(). You can wrap this in a case expression so it only applies to id 'B'.
select
id,
date,
data,
case when id = 'B'
then data - lag(data) over(partition by id order by date)
end lag_diff
from mytable
seems it is too long ago that I needed create own SQL Statements. I have a table (GAS_COUNTER) with timestamps (TS) and values (VALUE).
There are hundreds of entries per day, but I only need the latest of the day. I tried different ways but never get what I need.
Edit
Thanks for the fast replies, but some do not meet my needs (I need the latest value of each day in the table) and some don't work. My best own statement was:
select distinct (COUNT),
from
(select
extract (DAY_OF_YEAR from TS) as COUNT,
extract (YEAR from TS) as YEAR,
extract (MONTH from TS) as MONTH,
extract (DAY from TS) as DAY,
VALUE as VALUE
from GAS_COUNTER
order by COUNT)
but the value is missing. If I put it in the first select all rows return. (logical correct as every line is distinct)
Here an example of the Table content:
TS VALUE
2015-07-25 08:47:12.663 0.0
2015-07-25 22:50:52.155 2.269999999552965
2015-08-10 11:18:07.667 52.81999999284744
2015-08-10 20:29:20.875 53.27999997138977
2015-08-11 10:27:21.49 54.439999997615814
2nd Edit and solution
select TS, VALUE from GAS_COUNTER
where TS in (
select max(TS) from GAS_COUNTER group by extract(DAY_OF_YEAR from TS)
)
This one would give you the very last record:
select top 1 * from GAS_COUNTER order by TS desc
Here is one that would give you last records for every day:
select VALUE from GAS_COUNTER
where TS in (
select max(TS) from GAS_COUNTER group by to_date(TS,'yyyy-mm-dd')
)
Depending on the database you are using you might need to replace/adjust to_date(TS,'yyyy-mm-dd') function. Basically it should extract date-only part from the timestamp.
Select the max value for the timestamp.
select MAX(TS), value -- or whatever other columns you want from the record
from GAS_COUNTER
group by value
Something like this would window the data and give you the last value on the day - but what happens if you get two TS the same? Which one do you want?
select *
from ( select distinct cast( TS as date ) as dt
from GAS_COUNTER ) as gc1 -- distinct days
cross apply (
select top 1 VALUE -- last value on the date.
from GAS_COUNTER as gc2
where gc2.TS < dateadd( day, 1, gc1.dt )
and gc2.TS >= gc1.dt
order by gc2.TS desc
) as x
I would appreciate a little expert help please.
in an SQL SELECT statement I am trying to get the last day with data per month for the last year.
Example, I am easily able to get the last day of each month and join that to my data table, but the problem is, if the last day of the month does not have data, then there is no returned data. What I need is for the SELECT to return the last day with data for the month.
This is probably easy to do, but to be honest, my brain fart is starting to hurt.
I've attached the select below that works for returning the data for only the last day of the month for the last 12 months.
Thanks in advance for your help!
SELECT fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,fd.column_name
FROM super_table fd,
(SELECT TRUNC(daterange,'MM')-1 first_of_month
FROM (
select TRUNC(sysdate-365,'MM') + level as DateRange
from dual
connect by level<=365)
GROUP BY TRUNC(daterange,'MM')) fom
WHERE fd.cust_id = :CUST_ID
AND fd.coll_date > SYSDATE-400
AND TRUNC(fd.coll_date) = fom.first_of_month
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,
TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)
You probably need to group your data so that each month's data is in the group, and then within the group select the maximum date present. The sub-query might be:
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY YEAR(coll_date) * 100 + MONTH(coll_date);
This presumes that the functions YEAR() and MONTH() exist to extract the year and month from a date as an integer value. Clearly, this doesn't constrain the range of dates - you can do that, too. If you don't have the functions in Oracle, then you do some sort of manipulation to get the equivalent result.
Using information from Rhose (thanks):
SELECT MAX(coll_date) AS last_day_of_month
FROM Super_Table AS fd
GROUP BY TO_CHAR(coll_date, 'YYYYMM');
This achieves the same net result, putting all dates from the same calendar month into a group and then determining the maximum value present within that group.
Here's another approach, if ANSI row_number() is supported:
with RevDayRanked(itemDate,rn) as (
select
cast(coll_date as date),
row_number() over (
partition by datediff(month,coll_date,'2000-01-01') -- rewrite datediff as needed for your platform
order by coll_date desc
)
from super_table
)
select itemDate
from RevDayRanked
where rn = 1;
Rows numbered 1 will be nondeterministically chosen among rows on the last active date of the month, so you don't need distinct. If you want information out of the table for all rows on these dates, use rank() over days instead of row_number() over coll_date values, so a value of 1 appears for any row on the last active date of the month, and select the additional columns you need:
with RevDayRanked(cust_id, server_name, coll_date, rk) as (
select
cust_id, server_name, coll_date,
rank() over (
partition by datediff(month,coll_date,'2000-01-01')
order by cast(coll_date as date) desc
)
from super_table
)
select cust_id, server_name, coll_date
from RevDayRanked
where rk = 1;
If row_number() and rank() aren't supported, another approach is this (for the second query above). Select all rows from your table for which there's no row in the table from a later day in the same month.
select
cust_id, server_name, coll_date
from super_table as ST1
where not exists (
select *
from super_table as ST2
where datediff(month,ST1.coll_date,ST2.coll_date) = 0
and cast(ST2.coll_date as date) > cast(ST1.coll_date as date)
)
If you have to do this kind of thing a lot, see if you can create an index over computed columns that hold cast(coll_date as date) and a month indicator like datediff(month,'2001-01-01',coll_date). That'll make more of the predicates SARGs.
Putting the above pieces together, would something like this work for you?
SELECT fd.cust_id,
fd.server_name,
fd.instance_name,
TRUNC(fd.coll_date) AS coll_date,
fd.column_name
FROM super_table fd,
WHERE fd.cust_id = :CUST_ID
AND TRUNC(fd.coll_date) IN (
SELECT MAX(TRUNC(coll_date))
FROM super_table
WHERE coll_date > SYSDATE - 400
AND cust_id = :CUST_ID
GROUP BY TO_CHAR(coll_date,'YYYYMM')
)
GROUP BY fd.cust_id,fd.server_name,fd.instance_name,TRUNC(fd.coll_date),fd.column_name
ORDER BY fd.server_name,fd.instance_name,TRUNC(fd.coll_date)