https://i.stack.imgur.com/3vREY.jpg
I have included an example in above link. Can you assist with helping me find the days between each ID IF the TYPE is 'C' and the TYPE_ID is 123?
For example -- The number of days between ID 313031 is 1,146 days (between 2/20/2018 and 1/1/2015) since we would be ignoring the other ID because it has a TYPE A.
First you have to determine the records you want compared. You could for instance use MIN(date) and MAX(date), but this really depends on your data.
Next, to determine the difference you can simply subtract the dates
So for instance, assuming you want to compare the first and last date for a given ID:
Select ID,
Min(Date),
MAX(Date),
Min(Date) - MAX(Date) as difference_in_days
From your_table
WHERE TYPE = 'C' and TYPE_ID = 123
GROUP BY ID
this will work :
select id,max1-min1 from (select max(date) as max1,min(date) as min1,id from tablename
where type='C' group by id )
Related
I have an oracle DB.
My table has ID and DATE columns (and more).
I would like to select for every ID the next available record after a certain date. For only one ID the query would be:
SELECT * FROM my_table
WHERE id = 1 AND date >= '01.01.2018'
(just ignoring the to_date() function)
How would that look like for multiple IDs? And I do want to SELECT *.
Thanks!
We can use ROW_NUMBER here:
SELECT ID, date -- and maybe other columns
FROM
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY date) rn
FROM my_table
WHERE date >= date '2018-01-01'
) t
WHERE rn = 1
The idea here is to assign a row number to each ID partition, starting with the earliest date which occurs after the cutoff you specify. The first record from each partition would then be the immediate next date, assuming it exists.
I am trying to find whether a person (id = A3) is continuously active in a program at least five months or more in a given year (2013). Any suggestion would be appreciated. My data look like as follows:
You simply use group by and a conditional expression:
select id,
(case when count(ActiveMonthYear) >= 5 then 'YES!' else 'NAW' end)
from table t
where ListOfTheMonths between '201301' and '201312'
group by id;
EDIT:
I suppose "continuously" doesn't just mean any five months. For that, there are various ways. I like the difference of row numbers approach
select distinct id
from (select t.*,
(row_number() over (partition by id order by ListOfTheMonths) -
count(ActiveMonthYear) over (partition by id order by ListOfTheMonths)
) as grp
from table t
where ListOfTheMonths between '201301' and '201312'
) t
where ActiveMonthYear is not null
group by id, grp
having count(*) >= 5;
The difference in the subquery is constant for groups of consecutive active months. This is then used a grouping. The result is a list of all ids that meet this criteria. You can add a where for a particular id (do it in the subquery).
By the way, this is written using select distinct and group by. This is one of the rare cases where these two are appropriately used together. A single id could have two periods of five months in the same year. There is no reason to include that person twice in the result set.
Each of my rows have a date. I want the database to keep the good date. But I am in a situation where I want only the first date. But I still want all the other rows. So I would like to fill the date column with all the same date in my result.
For an example (Because I don't think I expressed myself well)
I have this:
name value date
a 10 5/13
b 14 2/13
c 20 1/13
a 11 7/13
a 5 8/13
b 8 9/13
I want it to become like this in the result:
name value date
a 26 5/13
b 22 5/13
c 20 5/13
I searched for this information but I only find the way to select the first row.
for now I'm doing
SELECT name, SUM(value), date FROM table
ORDER BY name
And I'm kind of clueless for what to do next.
Thanks :)
Databases don't have a concept of "first". Here is an attempt, but no guarantees unless you have a way of ordering to determine first:
select name, sum(value), const.date
from table cross join
(select top 1 date from table) const
group by name, const.date
If you only want to do this for a query, to provide this aggregated data for some specific client requirement, then #freshPrince's answer is appropriate. But if want to actually modify the data in the table itself, and prevent the issue from arising again, then you need to change the schema.
Create Table newTable(
name varChar(30) not null,
date datetime not null,
value decimal(10,2) not null default(0),
primary key (name, date) )
Insert newTable (name, date, value)
Select name, SUM(value), Min(date)
FROM currentTable
Group By Name
and delete the old table... then rename the new table to whatever...
You will also have to modify the process used to insert new rows so that instread of always inserting a new row, it updates the existing row for a specified name and date if it already exists...
Your question is slightly confusing since your desired result is showing a date that does not exists with either b or c but if that is the result that you want want you could use something similar to the following:
select name, sum(value) value, d.date
from yt
cross join
(
select min(date) date
from yt
where name = (select min(name)
from yt)
) d
group by name, d.date;
See SQL Fiddle with Demo
But it seems like you actually would want the min(date) for each name:
select name, sum(value) value, min(date)
from yt
group by name;
See SQL Fiddle with Demo.
If the order of the date should be the determined by the name then you could use:
select t.name, sum(value) value, d.date
from yt t
cross join
(
select top 1 name, date
from yt
order by name, date
) d
group by t.name, d.date;
See Demo
I have a really big query in which makes some troubles for me because one join can return several rows. I only want the latest row (identified by a date field) in this result set, but I cant seem to put together the correct query to make it work.
The query I need MAX date from is:
SELECT custid,reason,date FROM OPT opt WHERE opt.custid = 167043;
Teh custid is really found through a join, but for simplicity I've added it to the where clause here. This query produces the following result:
custid grunn date
167043 "Test 1" 19.10.2005 12:33:18
167043 "Test 2" 28.11.2005 16:23:35
167043 "Test 3" 14.06.2010 15:43:16
How can I retrieve only one record from this resultset? And that record is the one with the highest date? Ultimately Im putting this into a big query which does alot of joins, so hopefully I can use this example into my bigger query.
You can do this:
SELECT * FROM
( SELECT custid,reason,date FROM OPT opt WHERE opt.custid = 167043
ORDER BY date DESC
)
WHERE ROWNUM = 1;
You can solve it by using analytic functions. Try something like this:
select custid
,reason
,date
from (select custid
,reason
,date
,row_number() over(partition by cust_id order by date desc) as rn
from opt)
where rn = 1;
This is how it works: The resultset is divided into groups of cust_id (partition by). In each group, the rows will be sorted by the date column in descending order (order by). Each row within the group will be assigned a sequence number (row_number) from 1 to N.
This way the row with the highest value for date will be assigned 1, the second latest 2, third latest 3 etc..
Finally, I just pick the rows with nr = 1, which basically filters out the other rows.
Or another way using the LAST function in its aggregate form.
with my_source_data as (
select 167043 as custid, 'Test 1' as reason, date '2010-10-01' as the_date from dual union all
select 167043 as custid, 'Test 2' as reason, date '2010-10-02' as the_date from dual union all
select 167043 as custid, 'Test 3' as reason, date '2010-10-03' as the_date from dual union all
select 167044 as custid, 'Test 1' as reason, date '2010-10-01' as the_date from dual
)
select
custid,
max(reason) keep (dense_rank last order by the_date) as reason,
max(the_date)
from my_source_data
group by custid
I find this quite useful as it rolls the process of finding the last row and the value all into one. The use of MAX (or another aggregate function such as MIN) in case that the combination of the grouping and the order by is not deterministic.
This function will basically take the contents of the column based on the grouping, order it by the ordering given then take the last value.
rather than using row_number() I think it's better to select what you actually want to select (e.g. the last date)
SELECT custid
, reason
, date
from
(
SELECT custid
, reason
, date
, max(opt.date) over (partition by opt.custid order by opt.date) last_date
FROM OPT opt
WHERE opt.custid = 167043;
)
where date = last_date
both solutions with ROW_NUMBER and KEEP are good. I would tend to prefer ROW_NUMBER when retrieving a large number of columns, and keep KEEP for one or two columns, otherwise you will have to deal with duplicates and the statement will get pretty unreadable.
For a small number of columns however, KEEP should perform better
I have a table which looks like that:
As You see, there are some date duplicates, so how to select only one row for each date in that table?
the column 'id_from_other_table' is from INNER JOIN with the table above
There are multiple rows with the same date, but the time is different. Therefore, DISTINCT start_date will not work. What you need is: cast the start_date to a DATE (so the TIME part is gone), and then do a DISTINCT:
SELECT DISTINCT CAST(start_date AS DATE) FROM table;
Depending on what database you use, the type name for DATE is different.
Do you need any other information except the date? If not:
SELECT DISTINCT start_date FROM table;
You mention that there are date duplicates, but it appears they're quite unique down to the precision of seconds.
Can you clarify what precision of date you start considering dates duplicate - day, hour, minute?
In any case, you'll probably want to floor your datetime field. You didn't indicate which field is preferred when removing duplicates, so this query will prefer the last name in alphabetical order.
SELECT MAX(owner_name),
--floored to the second
dateadd(second,datediff(second,'2000-01-01',start_date),'2000-01-01') AS StartDate
From MyTable
GROUP BY dateadd(second,datediff(second,'2000-01-01',start_date),'2000-01-01')
Select Distinct CAST(FLOOR( CAST(start_date AS FLOAT ) )AS DATETIME) from Table
If you want to select any random single row for particular day, then
SELECT * FROM table_name GROUP BY DAY(start_date)
If you want to select single entry for each user per day, then
SELECT * FROM table_name GROUP BY DAY(start_date),owner_name
here is the solution for your query returning only one row for each date in that table
here in the solution 'tony' will occur twice as two different start dates are there for it
SELECT * FROM
(
SELECT T1.*, ROW_NUMBER() OVER(PARTITION BY TRUNC(START_DATE),OWNER_NAME ORDER BY 1,2 DESC ) RNM
FROM TABLE T1
)
WHERE RNM=1
You have to convert the "DateTime" to a "Date". Then you can easier select just one for the given date no matter the time for that date.