Getting date closest to variable - sql

I am working on a report application for rotating equipment. We calculate the status of the equipment every 3 hours and store the calculation in an Oracle database.
Before the report presented only the "current state" (where we only retrieved the latest calculation). But now the user is going to have the option to scroll back in time and look at earlier calculations.
SELECT *
FROM G_RunningHoursEvent
WHERE GRTE_TagName='#!VARIABLE_TAGNAME#' AND
GRTE_ValuesUpdate IN (SELECT MAX(GRTE_ValuesUpdate)
FROM G_RunningHoursEvent
WHERE GRTE_TagName='#!VARIABLE_TAGNAME#')
SQL is not my strongest suit and most of colleagues are on holiday. I need some tips on how I can extract the latest calculation closest to the new variable #!VARIABLE_TIME#
Is there someone that have any suggestions ?

If I understand correctly, the user will select a datetime value and you need to select the Event record which is closest to that time.
The following query does some magic in the sub-query to establish the record which is closest to the #!VARIABLE_TIME# value.
select
from g_runninghoursevent
where grte_tagname = '#!VARIABLE_TAGNAME#'
and grte_valuesupdate in
( select grte_valuesupdate
from ( with data as
( select grte_valuesupdate
, abs(grte_valuesupdate - #!VARIABLE_TIME#) tdiff
from g_runninghoursevent
where grte_tagname = '#!VARIABLE_TAGNAME#` )
select grte_valuesupdate
, rank() over (order by tdiff asc) rnk
from data )
where rnk = 1 )
Notes
The ABS() call means the sub-query will return the record with the smallest difference, either before or after the input parameter.
The sub-query uses IN rather than equality because you might get a tie for the smallest difference
I have assumed #!VARIABLE_TIME# is a datetime; if it is a string you will need to cast it to a date with teh appropriate mask.

This query should solve your problem:
SELECT *
FROM G_RunningHoursEvent
WHERE GRTE_TagName='#!VARIABLE_TAGNAME#' AND
TRUNC(GRTE_ValuesUpdate) = (SELECT TRUNC(GRTE_ValuesUpdate)
FROM (SELECT GRTE_ValuesUpdate
FROM G_RunningHoursEvent
WHERE GRTE_TagName='#!VARIABLE_TAGNAME#'
ORDER BY GRTE_ValuesUpdate DESC)
WHERE ROWNUM = 1);
Doing this you are ordering the G_RunningHoursEvent records by date having the closest record (in time) in the 1st row.
With the other SELECT, you just extract this 1st record.

Related

How Can I Retrieve The Earliest Date and Status Per Each Distinct ID

I have been trying to write a query to perfect this instance but cant seem to do the trick because I am still receiving duplicated. Hoping I can get help how to fix this issue.
SELECT DISTINCT
1.Client
1.ID
1.Thing
1.Status
MIN(1.StatusDate) as 'statdate'
FROM
SAMPLE 1
WHERE
[]
GROUP BY
1.Client
1.ID
1.Thing
1.status
My output is as follows
Client Id Thing Status Statdate
CompanyA 123 Thing1 Approved 12/9/2019
CompanyA 123 Thing1 Denied 12/6/2019
So although the query is doing what I asked and showing the mininmum status date per status, I want only the first status date. I have about 30k rows to filter through so whatever does not run overload the query and have it not run. Any help would be appreciated
Use window functions:
SELECT s.*
FROM (SELECT s.*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY statdate) as seqnum
FROM SAMPLE s
WHERE []
) s
WHERE seqnum = 1;
This returns the first row for each id.
Use whichever of these you feel more comfortable with/understand:
SELECT
*
FROM
(
SELECT *, ROW_NUMBER() OVER (PARTITION BY id ORDER BY statusdate) as rn
FROM sample
WHERE ...
) x
WHERE rn = 1
The way that one works is to number all rows sequentially in order of StatusDate, restarting the numbering from 1 every time ID changes. If you thus collect all the number 1's togetyher you have your set of "first records"
Or can coordinate a MIN:
SELECT
*
FROM
sample s
INNER JOIN
(SELECT ID, MIN(statusDate) as minDate FROM sample WHERE ... GROUP BY ID) mins
ON s.ID = mins.ID and s.StatusDate = mins.MinDate
WHERE
...
This one prepares a list of all the ID and the min date, then joins it back to the main table. You thus get all the data back that was lost during the grouping operation; you cannot simultaneously "keep data" and "throw away data" during a group; if you group by more than just ID, you get more groups (as you have found). If you only group by ID you lose the other columns. There isn't any way to say "GROUP BY id, AND take the MIN date, AND also take all the other data from the same row as the min date" without doing a "group by id, take min date, then join this data set back to the main dataset to get the other data for that min date". If you try and do it all in a single grouping you'll fail because you either have to group by more columns, or use aggregating functions for the other data in the SELECT, which mixes your data up; when groups are done, the concept of "other data from the same row" is gone
Be aware that this can return duplicate rows if two records have identical min dates. The ROW_NUMBER form doesn't return duplicated records but if two records have the same minimum StatusDate then which one you'll get is random. To force a specific one, ORDER BY more stuff so you can be sure which will end up with 1

Get Max(date) or latest date with 2 conditions or group by or subquery

I only have basic SQL skills. I'm working in SQL in Navicat. I've looked through the threads of people who were also trying to get latest date, but not yet been able to apply it to my situation.
I am trying to get the latest date for each name, for each chemical. I think of it this way: "Within each chemical, look at data for each name, choose the most recent one."
I have tried using max(date(date)) but it needs to be nested or subqueried within chemical.
I also tried ranking by date(date) DESC, then using LIMIT 1. But I was not able to nest this within chemical either.
When I try to write it as a subquery, I keep getting an error on the ( . I've switched it up so that I am beginning the subquery a number of different ways, but the error returns near that area always.
Here is what the data looks like:
1
Here is one of my failed queries:
SELECT
WELL_NAME,
CHEMICAL,
RESULT,
APPROX_LAT,
APPROX_LONG,
DATE
FROM
data_all
ORDER BY
CHEMICAL ASC,
date( date ) DESC (
SELECT
WELL_NAME,
CHEMICAL,
APPROX_LAT,
APPROX_LONG,
DATE
FROM
data_all
WHERE
WELL_NAME = WELL_NAME
AND CHEMICAL = CHEMICAL
AND APPROX_LAT = APPROX_LAT
AND APPROX_LONG = APPROX_LONG,
LIMIT 2
)
If someone does have a response, it would be great if it is in as lay language as possible. I've only had one coding class. Thanks very much.
Maybe something like this?
SELECT WELL_NAME, CHEMICAL, MAX(DATE)
FROM data_all
GROUP BY WELL_NAME, CHEMICAL
If you want all information, then use the ANSI-standard ROW_NUMBER():
SELECT da.*
FROM (SELECT da.*
ROW_NUMBER() OVER (PARTITION BY chemical, name ORDER BY date DESC) as senum
FROM data_all da
) da
WHERE seqnum = 1;

SQL find nearest date without going over, or return the oldest record

I have a view in SQL Server with prices of items over time. My users will be passing a date variable and I want to return the closest record without going over, or if no such record exists return the oldest record present. For example, with the data below, if the user passes April for item A it will return the March record and for item B it will return the June record.
I've tried a lot of variations with Union All and Order by but keep getting a variety of errors. Is there a way to write this using a Case Statement?
example:
case when min(Month)>Input Date then min(Month)
else max(Month) where Month <= Input Date?
Sincere apologies for attaching sample dataset as an image, I couldn't get it to format right otherwise.
Sample Dataset
You can use SELECT TOP (1) with order by DATE DESC + Item type + date comparison to get the latest. ORDER BY will order records by date, then you get the latest either this month (if exists) or earlier months.
Here's a rough outline of a query (without more of your table it's hard to be exact):
WITH CTE AS
(
SELECT
ITEM,
PRICE,
MIN(ACTUAL_DATE) OVER (PARTITION BY ITEM ORDER BY ITEM) AS MIN_DATE,
MAX(INPUT_DATE<=ACTUAL_DATE) OVER (PARTITION BY ITEM ORDER BY ITEM,ACTUAL_DATE) AS MATCHED_DATE
FROM TABLE
)
SELECT
CTE.ITEM,
CTE.PRICE,
CASE
WHEN
CTE.MATCHED_DATE IS NOT NULL
THEN
CTE.MATCHED_DATE
ELSE
CTE.MIN_DATE
END AS MOSTLY_MATCHED_DATE
FROM CTE
GROUP BY
CTE.ITEM,
CTE.PRICE
The idea is that in a Common Table Expression, you use the PARTITION BY function to identify the key date for each item, record by record, and then you do a test in aggregate to pull either your matched record or your default record.

Latest date and time in SQL without ORDER BY

I'm trying to find a way to display the last event held (last date and time) in an events table whilst displaying all the columns for that event without using ORDER BY.
For example:
SELECT * from Events
where dateheld in (select max(dateheld) from events)
AND starttime in (select max(starttime) from events)
When I put MAX starttime, it displays nothing. When I put MIN starttime it works but displays the earliest time of that date and not the latest.
I guess you could print out your records, throw them down the stairs, and the ones that go farthest have the "lightest" dates. You cannot sort without order by. It's like wanting water that isn't wet. Unless your data naturally comes out in the order you want, you MUST sort.
Of course, if you want only the record that has the absolute most recent date, and don't need more than just that one record, then
SELECT yourdatetimefield, ...
FROM yourtable
HAVING yourdatetimefield = MAX(yourdatetimefield)
If you are only looking for the latest item:
EDIT gets a little more complicated when you have seperate date and time fields, but this should work. This is a ridiculous kludge for a situation where date and time should be stored in one field.
SELECT *
FROM (
SELECT *
FROM Events
WHERE dateTime = (SELECT MAX(dateheld) FROM Events)
) temp
WHERE starttime = (SELECT MAX(starttime) FROM (
SELECT *
FROM Events
WHERE dateTime = (SELECT MAX(dateheld) FROM Events)
) temp 2 )

SQL Average Inter-arrival Time, Time Between Dates

I have a table with sequential timestamps:
2011-03-17 10:31:19
2011-03-17 10:45:49
2011-03-17 10:47:49
...
I need to find the average time difference between each of these(there could be dozens) in seconds or whatever is easiest, I can work with it from there. So for example the above inter-arrival time for only the first two times would be 870 (14m 30s). For all three times it would be: (870 + 120)/2 = 445 (7m 25s).
A note, I am using postgreSQL 8.1.22 .
EDIT: The table I mention above is from a different query that is literally just a one-column list of timestamps
Not sure I understood your question completely, but this might be what you are looking for:
SELECT avg(difference)
FROM (
SELECT timestamp_col - lag(timestamp_col) over (order by timestamp_col) as difference
FROM your_table
) t
The inner query calculates the distance between each row and the preceding row. The result is an interval for each row in the table.
The outer query simply does an average over all differences.
i think u want to find avg(timestamptz).
my solution is avg(current - min value). but since result is interval, so add it to min value again.
SELECT avg(target_col - (select min(target_col) from your_table))
+ (select min(target_col) from your_table)
FROM your_table
If you cannot upgrade to a version of PG that supports window functions, you
may compute your table's sequential steps "the slow way."
Assuming your table is "tbl" and your timestamp column is "ts":
SELECT AVG(t1 - t0)
FROM (
-- All this silliness would be moot if we could use
-- `` lead(ts) over (order by ts) ''
SELECT tbl.ts AS t0,
next.ts AS t1
FROM tbl
CROSS JOIN
tbl next
WHERE next.ts = (
SELECT MIN(ts)
FROM tbl subquery
WHERE subquery.ts > tbl.ts
)
) derived;
But don't do that. Its performance will be terrible. Please do what
a_horse_with_no_name suggests, and use window functions.