Oracle - Counting timestamps where difference between timestamps greater than 1 hour - oracle

I have a worklog table that contains the following fields:
worklog_id,
agent_name,
ticket_number,
timestamp,
worklog_notes.
I would like to be able to count the number of worklog entries made where if the agent_name, ticket_number and timestamp (date) are the same the worklog entry is only counted if the time between the two entries is greater than 1 hour.
Example: John Smith make three worklog entries on ticket 12345. The first timestamp is "10/11/2012 9:11:44 AM", the second timestamp is "10/11/2012 9:36:16 AM" and the third timestamp is "10/11/2012 11:18:20 AM". In this example I would only want to give the agent credit for two worklog entries as the first two were less than an hour apart.
I've tried getting the logic to work using a "where" sub-query, but cannot get it working. Would anyone have any example they could provide? Thanks! :)

Does this get what you want? The first entry by a given agent on a ticket should always be counted, and entries after that should only be counted if at least an hour has elapsed since the prior entry.
select agent_name, ticket_number, count(*) from (
select agent_name, ticket_number, timestamp,
lag(timestamp) over
(partition by agent_name, ticket_number order by timestamp) prev_timestamp
)
from worklog
where (prev_timestamp is null
or (timestamp - prev_timestamp) >= interval '1' hour
)
group by agent_name, ticket_number
I'm not sure this is exactly what you want -- if an agent keeps adding entries within an hour of the prior entry, none of them will be counted except the first. So someone who adds a lot of updates gets penalized.
Maybe what you really want is to count the number of distinct hours in which an update was made:
select agent_name, ticket_number, count(distinct to_char(timestamp,'DD-MON-YYYY HH24')
from worklog
group by agent_name, ticket_number

Related

How to SELECT the MAX Time Difference Between Any 2 Consecutive Rows Per Value?

Just had a user answer this correctly for TSQL, but wondering how best to achieve this now in SQL Developer/PLSQL seeing as there is no DATEDIFF function.
Table I want to query on has some 'CODE' values, which can naturally have multiple primary key records ('OccsID') in a table 'Occs'. There is also a datetime column called 'CreateDT' for each OccsID.
Just want to find the maximum possible time variance between any 2 consecutive rows in 'Occs', per 'CODE'.
If you subtract the "next" date and "this" date (using the LEAD analytic function), you'll get the date difference. Then fetch the maximum difference per code. Something like this:
with diff as
(select occsid,
code,
nvl(lead(createdt) over (partition by code order by createdt), createdt) - createdt date_diff
from test
)
select code,
max(date_diff)
from diff
group by code;
Assuming that this T-SQL version works for you (from the prior question)
SELECT x.code, MAX(x.diff_sec) FROM
(
SELECT
code,
DATEDIFF(
SECOND,
CreateDT,
LEAD(CreateDT) OVER(PARTITION BY CODE ORDER BY CreateDT) --next row's createdt
) as diff_sec
FROM Occs
)x
GROUP BY x.code
The simplest option is just to subtract the two dates to get a difference in days. You can then multiply to get the difference in hours, minutes, or seconds
SELECT x.code, MAX(x.diff_day), MAX(x.diff_sec)
FROM
(
SELECT
code,
CreateDT -
LEAD(CreateDT) OVER(PARTITION BY CODE ORDER BY CreateDT) as diff_day,
24*60*60* (CreateDT -
LEAD(CreateDT) OVER(PARTITION BY CODE ORDER BY CreateDT)) as diff_sec
FROM Occs
)x
GROUP BY x.code

Why does my total session (aggregated using EXTRACT MONTH) is less than total session if I broke down by the date?

I'm trying to generate my total session by month. I've tried using two different ways.
I'm using date field for the first column
I'm using month field that is extracted from date field using EXTRACT(MONTH FROM date) AS month
I have tried using below code for the 1st one:
with
session1 as(
select date,
session_id
from table
where date >= '2019-05-20' AND date <= '2019-05-21')
SELECT date_key, COUNT(DISTINCT session_id) AS sessions from session1
GROUP BY 1
For the 2nd one I tried using this code:
with
session1 as(
select date,
session_id
from table
where date >= '2019-05-20' AND date <= '2019-05-21')
SELECT EXTRACT (MONTH FROM date_key) AS month, COUNT(DISTINCT session_id) AS sessions from session1
GROUP BY 1
For the result, I got the output as per below:
20 May: 1,548 Sessions; 21 May: 1,471 Sessions; Total: 3,019
May: 2,905
So, there's 114 session discrepancy and I'd like to know why.
Thank you in advance.
For simplicity sake - let's say there is only one session during two consecutive days. So if you will count by day and then sum result - you will get 2 sessions, while if you will count distinct sessions for whole two days - you will get just 1 session
Hope this shows you the reason why - you are counting some sessions twice on different days - maybe when they go over end of one and start of next day
The following query should show you which sessions_ids occur on both dates.
select session_id, count(distinct date) as num_dates
from table
where date >= '2019-05-20' AND date <= '2019-05-21'
group by 1
having num_dates > 1
This is either a data processing issue, or your session definition is allowed to span multiple days. Google Analytics, for example, traditionally ends a session and begins a new session at midnight. Other sessionization schemes might not impose this restriction.

Oracle, count of distinct items by date

I need to get a count of items by date from field1 grouped by last update time. What I am looking for is how many times an item from field1 appears on a specific date for the last 30 days where field 2 = 0. This will be run every day so the date will roll. Field1 will be a number >0, field2 will be any number (negative and positive), last_upd_time will be a system time when the last update occurred. I don't need the time, only the date. My current query that returns all of the data is:
select field1, field2, trunc(last_upd_time)
from table
where field2 = '0' and last_upd_time >= SYSDATE - 30
I have attempted to use count, group by, and group by/having. Not saying I was using them correctly, but I did try.
Try this:
SELECT TO_CHAR(last_upd_time,'DD-MM-YYYY') last_upd_time, COUNT(DISTINCT field1)
FROM table WHERE field2='0' AND
last_upd_time>=SYSDATE - INTERVAL '30' DAY
GROUP BY TO_CHAR(last_upd_time,'DD-MM-YYYY');

Oracle Daily count/average over a year

I'm pulling two pieces of information over a specific time period, but I would like to fetch the daily average of one tag and the daily count of another tag. I'm not sure how to do daily averages over a specific time period, can anyone provide some advice? Below were my first ideas on how to handle this however to change every date would be annoying. Any help is appreciated thanks
SELECT COUNT(distinct chargeno), to_char(chargetime, 'mmddyyyy') AS chargeend
FROM batch_index WHERE plant=1 AND chargetime>to_date('2012-06-18:00:00:00','yyyy-mm-dd:hh24:mi:ss')
AND chargetime<to_date('2012-07-19:00:00:00','yyyy-mm-dd:hh24:mi:ss')
group by chargetime;
The working version of the daily sum
SELECT to_char(bi.chargetime, 'mmddyyyy') as chargtime, SUM(cv.val)*0.0005
FROM Charge_Value cv, batch_index bi WHERE cv.ValueID =97
AND bi.chargetime<=to_date('2012-07-19','yyyy-mm-dd')
AND bi.chargeno = cv.chargeno AND bi.typ=1
group by to_char(bi.chargetime, 'mmddyyyy')
seems like in the first one you want to change the group to the day - not the time... (plus i dont think you need to specify all those 0's for seconds..)
SELECT COUNT(distinct chargeno), to_char(chargetime, 'mmddyyyy') AS chargeend
FROM batch_index WHERE plant=1 AND chargetime>to_date('2012-06-18','yyyy-mm-dd')
AND chargetime<to_date('2012-07-19','yyyy-mm-dd')
group by to_char(chargetime, 'mmddyyyy') ;
not 100% I'm following your question, but if you just want to do aggregates (sums, avg), then do just that. I threw in the rollup just in case that is what you were looking for
with fakeData as(
select trunc(level *.66667) nr
, trunc(2*level * .33478) lvl --these truncs just make the doubles ints
,trunc(sysdate+trunc(level*.263784123)) dte --note the trunc, this gets rid of the to_char to drop the time
from dual
connect by level < 600
) --the cte is just to create fake data
--below is just some aggregates that may help you
select sum(nr) daily_sum_of_nr
, avg(nr) daily_avg_of_nr
, count(distinct lvl) distinct_lvls_per_day
, count(lvl) count_of_nonNull_lvls_per_day
, dte days
from fakeData
group by rollup(dte)
--if you want the query to supply a total for the range, you may use rollup ( http://psoug.org/reference/rollup.html )

Adding one month to saved date(oracle)

I have a table A which contains a Date type attribute. I want to write a query to select the date in another table B with value one month after the value in A.Any one know how to do it in oracle?
uhm... This was the first hit on google:
http://psoug.org/reference/date_func.html
It seems you're looking for the "add_months" function.
You need to use the ADD_MONTHS function in Oracle.
http://www.techonthenet.com/oracle/functions/add_months.php
Additional info: If you want to use this function with today's date you can use ADD_MONTHS(SYSDATE, 1) to get one month from now.
The question is to select a date_field from table b where date_field of table b is one month ahead of a date_field in table a.
An additional requirement must be taken into consideration which is currently unspecified in the question. Are we interested in whole months (days of month not taken into consideration) or do we want to include the days which might disqualify dates that are one month ahead but only by a couple of days (example: a=2011-04-30 and b=2011-05-01, b is 1 month ahead but only by 1 day).
In the first case, we must truncate both dates to their year and month values:
SELECT TRUNC( TO_DATE('2011-04-22','yyyy-mm-dd'), 'mm') as trunc_date
FROM dual;
gives:
trunc_date
----------
2011-04-01
In the second case we don't have to modify the dates.
At least two approaches can be used to solve the initial problem:
First one revolves around adding one month to the date_field in table a and finding a row in table b with a matching date.
SELECT b.date_field
FROM tab_a as a
,tab_b as b
WHERE ADD_MONTHS( TRUNC( a.date_field, 'mm' ), 1) = TRUNC( b.date_field, 'mm' )
;
Note the truncated dates. Leaving this out will require a perfect day to day match between dates.
The second approaches is based on calculating the difference in months between two dates and picking a calculation that gives a 1 month difference.
SELECT b.date_field
FROM tab_a as a
,tab_b as b
WHERE months_between( TRUNC( b.date_field, 'mm') , TRUNC(a.date_field, 'mm') ) = 1
The order of the fields in months_between is important here. In the provided example:
for b.date_field one month ahead of a.date_field the value is 1
for b.date_field one month before a.date_field the value is -1 (negative one)
Reversing the order will also reverse the results.
Hope this answers your question.

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