I have an SQLite DB with date, month, year fields in integers (I believe they should have used a date field but the choice wasn't mine). I would like to select the row whose date value is the latest. What is the best query to do that?
select * from your_table
order by year, month, `date` desc
limit 1
Related
I have a google bigquery table with orders with a DATE column and other columns related to the orders. The starting date of the dataset is from 2021-01-01 (yyyy-mm-dd).
My aim is to filter on the DATE column from last year and this year to the previous iso week. For this, I used the ISOWEEK to create a new column:
WITH
last_week_last_year AS (
SELECT
DATE,
EXTRACT(ISOWEEK FROM DATE) AS isoweek,
FROM
`orders`
WHERE
EXTRACT(ISOWEEK FROM DATE) = EXTRACT(ISOWEEK FROM CURRENT_DATE())-1
GROUP BY 1, 2
ORDER BY DATE
)
SELECT * FROM last_week_last_year
This query results as the following table:
The issue is that when I filter on the original orders table by the DATE from the last_week_last_year table I get all the orders back instead of just the filtered version.
My method to filter is WHERE DATE IN (SELECT DATE FROM last_week_last_year) as seen below.
SELECT
*
FROM
`orders`
WHERE
DATE IN (SELECT DATE FROM last_week_last_year)
ORDER BY DATE DESC;
A snapshot of resulting table. It contains all of the records from 2021-01-01 until the latest day.
How can I make sure that on the latter query the table is filtered based on the first query's dates in DATE column?
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 table with the following columns:
Date
Skills,
Customer ID
I want to find out Date(x), Customers, Count of Customers in between Date(x) and Date(x)+6
Can somebody guide me how to make this query, or can I create this function in SQL Server?
If I understand you correctly, you want something like this:
(take care, can be bad syntax, because i "work" only with oracle. But I think that it should work)
select date, customer_id, COUNT(*)
from your_table --add your table
where date between getdate() and DATEADD(day, 6, getdate())
-- between current database system date and +6 day
group by date, customer id
order by COUNT (*) desc -- if you want, you can order your result - ASC||DESC
If you have data on each date, then perhaps this is what you want:
select date, count(*),
sum(count(*)) over (order by date rows between 6 preceding and current row) as week_count
from t
group by date;
customer Date location
1 25Jan2018 texas
2 15Jan2018 texas
3 12Feb2018 Boston
4 19Mar2017 Boston.
I am trying to find out count of customers group by yearmon of Date column.Date column is of text data type
eg: In jan2018 ,the count is 2
I would do something like the following:
SELECT
date_part('year', formattedDate) as Year
,date_part('month', formattedDate) as Month
,count(*) as CustomerCountByYearMonth
FROM
(SELECT to_date(Date,'DDMonYYYY') as formattedDate from <table>) as tbl1
GROUP BY
date_part('year', formattedDate)
,date_part('month', formattedDate)
Any additional formatting for dates could be done on the inner query that will allow for adjustments in case some single digit days need to be padded or a month has four letters instead of three etc.
By converting to date type, you can properly order by date type and not alphabetical etc.
Optionally:
SELECT
Year
,Month
,count(*) as CustomerCountByYearMonth
FROM
(SELECT
date_part('year', to_date(Date,'DDMonYYYY')) as Year
,date_part('month', to_date(Date,'DDMonYYYY')) as Month
FROM <table>) as tbl1
GROUP BY
Year
,Month
You shouldn't store dates in a text column...
select substring(Date, length(Date)-6), count(*)
from tablename
group by substring(Date, length(Date)-6)
I thought #Jarlh asked a good question -- what about dates like January 1? Is it 01Jan2019 or 1Jan2019? If it can be either, perhaps a regex would work.
select
substring (date from '\d+(\D{3}\d{4})') as month,
count (distinct customer)
from t
group by month
The 'distinct customer' also presupposes you may have the same customer listed in the same month, but you only want to count it once. If that's not the case, just remove 'distinct.'
And, if you wanted the output in date format:
select
to_date (substring (date from '\d+(\D{3}\d{4})'), 'monyyyy') as month,
count (distinct customer)
from t
group by month
If it is a date column, you can truncate the date:
select date_trunc('month', date) as yyyymm, count(*)
from t
group by yyyymm
order by yyyymm;
I really read that the type was date. For a string, just use string functions:
select substr(date, 3, 7) as mmmyyyy, count(*)
from t
group by mmmyyyy;
Unfortunately, ordering doesn't work in this case. You should really be storing dates using the proper type.
I have a table with 3 columns: id, date and amount, but I would like to get accumulated SUM for each date (Last column).
Do you have an easy solution how to add this column?
I am trying with this:
SELECT date, sum(amount) as accumulated
FROM table group by date
WHERE max(date);
Should I user OVER() for this?
Use a window function to the total for each day:
SELECT date,
amount,
sum(amount) over (partition by date) as accumulated
FROM the_table;
However this will only work, if your dates all have the same time part (in Oracle a DATE column also contains a time). To make sure you ignore the time part, use trunc() to make sure all time parts are normalized to 00:00:00
SELECT date,
amount,
sum(amount) over (partition by trunc(date)) as accumulated
FROM the_table;
Use This:
SELECT T.ID, T.DATE, T.AMOUNT, (SELECT SUM(S.AMOUNT) FROM TABLE S WHERE S.DATE=T.DATE) ACCUMULATED
from
table T
This will give you the records from the table with a sum for all records for the date.