Get smallest date for each element in access query - sql

So I have a table containing different elements and dates.
It basically looks like this:
actieElement beginDatum
1 1/01/2010
1 1/01/2010
1 10/01/2010
2 1/02/2010
2 3/02/2010
What I now need is the smallest date for every actieElement.
I've found a solution using a simple GROUP BY statement, but that way the query loses its scope and you can't change anything anymore.
Without the GROUP BY statement I get multiple dates for every actieElement because certain dates are the same.
I thought of something like this, but it also does not work as it would give the subquery more then 1 record:
SELECT s1.actieElement, s1.begindatum
FROM tblActieElementLink AS s1
WHERE (((s1.actieElement)=(SELECT TOP 1 (s2.actieElement)
FROM tblActieElementLink s2
WHERE s1.actieElement = s2.actieElement
ORDER BY s2.begindatum ASC)));

Try this
SELECT s1.actieElement, s1.begindatum
FROM tblActieElementLink AS s1
WHERE s1.begindatum =(SELECT MIN(s2.begindatum)
FROM tblActieElementLink s2
WHERE s1.actieElement = s2.actieElement
);

SELECT DISTINCT T1.actieElement, T1.beginDatum
FROM tblActieElementLink AS T1
INNER JOIN (
SELECT T2.actieElement,
MIN(T2.beginDatum) AS smallest_beginDatum
FROM tblActieElementLink AS T2
GROUP
BY T2.actieElement
) AS DT1
ON T1.actieElement = DT1.actieElement
AND T1.beginDatum = DT1.smallest_beginDatum;

Add a DISTINCT clause to your SELECT.

Related

SQL: How to join two rows on the same table, based on the same timestamp?

This is a data output table in postgreSQL.
I am looking for a way to join rows 7 and 8 on the timestamp condition to remove the [null] values.
This is the expected output I am looking for.
So that I will only have one row when the timestamp matches.
--------------------------------------------------------
05:32:33 | Pump2Stop | 49
--------------------------------------------------------
Any idea how can this be done?
assuming you only ever have one record per timestamp with a non-null value in a specific column, try this:
SELECT to_timestamp, MAX(str_v), MAX(long_v)
FROM table
GROUP BY to_timestamp;
You can use a self-join
select stp.to_timestamp,
stp.str_v,
strt.long_v
from the_table stp
join the_table strt
on stp.to_timestamp = strt.to_timestamp
and strt.str_v is null
where stp.str_v = 'Pump2Stop'
If I understand correctly, you want all rows -- with the specified rows combined. Based on your comment, I think this does what you want:
select t1.to_timestamp, t1.str_v
coalesce(t2.long_v, t1.long_v)
from t t1 left join
t t2
on t.to_timestamp = t2.to_timestamp and
t.str_v in ('Pump2Stop', 'Pump2Stop')
where t1.str_v is not null;
You can also use a window function:
select t.to_timestamp, t.str_v,
coalesce(long_v, imputed_long_v) as long_v
from (select t.*,
max(long_v) over (partition by to_timestamp) as imputed_long_v
from t
) t
where str_v is not null;

Return a NULL value if Date not in CTE

I have a query that counts the number of records imported for every day according to the current date. The only problem is that the count only returns when records have been imported and NULLS are ignored
I have created a CTE with one column in MSSQL that lists dates in a certain range e.g. 2019-01-01 - today.
The query that i've currently got is like this:
SELECT TableName, DateRecordImported, COUNT(*) AS ImportedRecords
FROM Table
WHERE DateRecordImported IN (SELECT * FROM DateRange_CTE)
GROUP BY DateRecordImported
I get the results fine for the dates that exist in the table for example:
TableName DateRecordImported ImportedRecords
______________________________________________
Example 2019-01-01 165
Example 2019-01-02 981
Example 2019-01-04 34
Example 2019-01-07 385
....
but I need a '0' count returned if the date from the CTE is not in the Table. Is there a better alternative to use in order to return a 0 count or does my method need altering slightly
You can do LEFT JOIN :
SELECT C.Date, COUNT(t.DateRecordImported) AS ImportedRecords
FROM DateRange_CTE C LEFT JOIN
table t
ON t.DateRecordImported = C.Date -- This may differ use actual column name instead
GROUP BY C.Date; -- This may differ use actual column name instead
Move the position of the CTE from a subquery to the FROM:
SELECT T.TableName,
DT.PCTEDateColumn} AS DateRecordImported,
COUNT(T.{TableIDColumn}) AS ImportedRecords
FROM DateRange_CTE DT
LEFT JOIN [Table] T ON DT.{TEDateColumn} = T.DateRecordImported
GROUP BY DT.{CTEDateColumn};
You'll need to replace the values in braces ({})
You can try this
SELECT TableName, DateRecordImported,
case when DateRecordImported is null
then '0'
else count(*) end AS ImportedRecords
FROM Table full join DateRange_CTE
on Table.DateRecordImported DateRange_CTE.ImportedDate
group by DateRecordImported,ImportedDate
(ImportedDate is name of column of CTE)

Grouping records on consecutive dates

If I have following table in Postgres:
order_dtls
Order_id Order_date Customer_name
-------------------------------------
1 11/09/17 Xyz
2 15/09/17 Lmn
3 12/09/17 Xyz
4 18/09/17 Abc
5 15/09/17 Xyz
6 25/09/17 Lmn
7 19/09/17 Abc
I want to retrieve such customer who has placed orders on 2 consecutive days.
In above case Xyz and Abc customers should be returned by query as result.
There are many ways to do this. Use an EXISTS semi-join followed by DISTINCT or GROUP BY, should be among the fastest.
Postgres syntax:
SELECT DISTINCT customer_name
FROM order_dtls o
WHERE EXISTS (
SELEST 1 FROM order_dtls
WHERE customer_name = o.customer_name
AND order_date = o.order_date + 1 -- simple syntax for data type "date" in Postgres!
);
If the table is big, be sure to have an index on (customer_name, order_date) to make it fast - index items in this order.
To clarify, since Oto happened to post almost the same solution a bit faster:
DISTINCT is an SQL construct, a syntax element, not a function. Do not use parentheses like DISTINCT (customer_name). Would be short for DISTINCT ROW(customer_name) - a row constructor unrelated to DISTINCT - and just noise for the simple case with a single expression, because Postgres removes the pointless row wrapper for a single element automatically. But if you wrap more than one expression like that, you get an actual row type - an anonymous record actually, since no row type is given. Most certainly not what you want.
What is a row constructor used for?
Also, don't confuse DISTINCT with DISTINCT ON (expr, ...). See:
Select first row in each GROUP BY group?
Try something like...
SELECT `order_dtls`.*
FROM `order_dtls`
INNER JOIN `order_dtls` AS mirror
ON `order_dtls`.`Order_id` <> `mirror`.`Order_id`
AND `order_dtls`.`Customer_name` = `mirror`.`Customer_name`
AND DATEDIFF(`order_dtls`.`Order_date`, `mirror`.`Order_date`) = 1
The way I would think of it doing it would be to join the table the date part with itselft on the next date and joining it with the Customer_name too.
This way you can ensure that the same customer_name done an order on 2 consecutive days.
For MySQL:
SELECT distinct *
FROM order_dtls t1
INNER JOIN order_dtls t2 on
t1.Order_date = DATE_ADD(t2.Order_date, INTERVAL 1 DAY) and
t1.Customer_name = t2.Customer_name
The result you should also select it with the Distinct keyword to ensure the same customer is not displayed more than 1 time.
For postgresql:
select distinct(Customer_name) from your_table
where exists
(select 1 from your_table t1
where
Customer_name = your_table.Customer_name and Order_date = your_table.Order_date+1 )
Same for MySQL, just instead of your_table.Order_date+1 use: DATE_ADD(your_table.Order_date , INTERVAL 1 DAY)
This should work:
SELECT A.customer_name
FROM order_dtls A
INNER JOIN (SELECT customer_name, order_date FROM order_dtls) as B
ON(A.customer_name = B.customer_name and Datediff(B.Order_date, A.Order_date) =1)
group by A.customer_name

SQL: Query new rows of a new date

i have table as shown here in this picture --> http://www.directupload.net/file/d/3710/lj7etq5j_png.htm
I need the correct Query to get only data_id 10.
The query should be like this: Compare the latest date rows (2014-08-08) with the earliest date rows (2014-08-06). If there is a row on 2014-08-08 which is NOT at 2014-08-06, this row should returned.
I already tried it with self-joins and Sub-Selects, but i did't get it work.
Thx for any help!
Maybe something like this is what you're looking for?
select * from Table1
where
data not in (
select data from Table1
where dataOfDate = (select min(dataofdate) from Table1)
)
and dataOfDate = (select max(dataofdate) from Table1)
The first where clause compares the data field of the returned rows to the data field in the set of oldest rows and the second where clause limits the set of rows the the newest.
Note that I'm only comparing rows based on thedatafield, so you might have to change the query if you want to includenextTableIdin the comparison.
Here is a sample SQL Fiddle.
How about something like this:
SELECT d1.* FROM Dates d1 LEFT JOIN Dates d2 ON d1.nextTableId = d2.nextTableId WHERE d1.dataofDate = '2014-08-08' AND d2.dataofDate = '2014-08-06' AND d2.data_id IS NULL;
SELECT `data_id`
FROM `my_table`
WHERE `dataOfDate` = (SELECT MAX(`dataOfDate`) FROM `my_table`)
AND `nextTableId` NOT IN (
SELECT `nextTableId` FROM `my_table` WHERE `dataOfDate` = (SELECT MIN(`dataOfDate`) FROM `my_table`)
)
select all rows with max date
that don't have values amongst rows with min date
edit ops, too late

Getting SUM from 2 different tables into one result

I have been trying to get this to work for 12 hrs now and I cannot :-( Can someone please show me how I can get the ssnumber to group and get the total for each ssnumber.
Here is what I have now. In Table number 1 I have this code
SELECT
UNIT_NO, SUM(RATEB) AS TOTALRTE
FROM TABLE1
WHERE
TRUCK_PAID = 1
AND PICK_UP_DATE >= '(fromdate)'
AND PICK_UP_DATE <= '(todate)'
GROUP BY
UNIT_NO
ORDER BY
UNIT_NO
But table number 2 is where the ssnumber column is, so what I'm trying to do is the rateB sum from all of the loads for each unit_no and then group them and then go into table number 2 and group the ssnumber with the unit number from table number 1 and sum the rateB from table number 1.
Something like this (see below) but its not working :-(
SELECT
UNIT_NO, SUM(RATEB)
FROM
TABLE1
WHERE
TRUCK_PAID = 1
AND PICK_UP_DATE >= '(fromdate)'
AND PICK_UP_DATE <= '(todate)'
GROUP BY
UNIT_NO
JOIN
TABLE TABLE1.UNIT_NO = TABLE2.UNIT_NO GROUP BY TABLE2.SS_NUM
or
SELECT
UNIT_NO, SUM(RATEB) AS TOTALRATE
FROM
TABLE1
GROUP BY
UNIT_NO
JOIN
TRUCKS ON (TABLE1.UNIT_NO = TABLE2.UNIT_NO)
GROUP BY
TABLE2.SSNUMBER
Thank you guys so much for any help...
As requested, it is hard to really understand what you are trying to accomplish without more info about table2 and maybe an example of what you are expecting. However, what I got from your description is that you are trying to accomplish something like this?
SELECT UNIT_NO, TOTALRTE, TOTALLDSRTE
FROM
(
SELECT UNIT_NO,SUM(RATEB) AS TOTALRTE
FROM LOADS
GROUP BY UNIT_NO
) AS tbl1
JOIN
(
SELECT SS_NUM, SUM(RATEB) AS TOTALLDSRTE
FROM LOADS
GROUP BY SS_NUM
) AS tbl2
ON tbl1.UNIT_NO = tbl2.SS_NUM
I would suggest instead of getting data from two select queries in one select query, try to fetch them as separate queries. This saves a lot of time. That, or you can create a table for the result and update the result of each query into the table.