I have a table with entries which has a DATE field. Each entry has a distinct date.
I'd like to select all the entries from the last month registred in the database. How?
I tried:
SELECT *
FROM registries
WHERE reg_date = DATE_FORMAT(MAX(reg_date), "%m")`
...without success
If you wanted the last 30 days, this will work
SELECT * FROM `registries`
WHERE `reg_date` > DATE_SUB( NOW(), INTERVAL 30 DAY )
Based on OMG Ponies' query with corrections:
SELECT
r.*
FROM
registries AS r
JOIN (
SELECT
MAX(t.reg_date) AS max_date
FROM
registries AS t) AS t
ON DATE_FORMAT(t.max_date, '%Y-%m') = DATE_FORMAT(r.reg_date, '%Y-%m')
Though the performance of the query wouldn't be excellent, since it would operate the JOIN on two calculated values.
I believe it can still perform decently unless you start hitting millions of records.
On the other hand, you could probably run it faster by querying first for the MAX(reg_date)
SELECT
CONCAT(DATE_FORMAT(MAX(r.reg_date), "%Y-%m"), '-01') AS first_day
FROM
registries AS r
And then injecting the result in a query:
SELECT
r.*
FROM
registries AS r
WHERE
r.reg_date BETWEEN '<first_day>' AND LAST_DAY('<first_day>')
With first_day as a place holder for the previous' query result.
Provided you indexed reg_date, this should run pretty fast.
Note: LAST_DAY is a MySQL only function.
This will give you all records for last month (May):
SELECT [col,] DATEDIFF(TIMESTAMP,
2010-05-01 00:00) dif1,
DATEDIFF(TIMESTAMP, 2010-05-31 00:00)
dif2 FROM tablename HAVING dif1 >= 0
AND dif2 <= 0
Related
I have a PL-SQL table with a structure as shown in the example below:
I have customers (customer_number) with insurance cover start and stop dates (cover_start_date and cover_stop_date). I also have dates of accidents for those customers (accident_date). These customers may have more than one row in the table if they have had more than one accident. They may also have no accidents. And they may also have a blank entry for the cover stop date if their cover is ongoing. Sorry I did not design the data format, but I am stuck with it.
I am looking to calculate the number of accidents (num_accidents) and number of customers (num_customers) in a given time period (period_start), and from that the number of accidents-per-customer (which will be easy once I've got those two pieces of information).
Any ideas on how to design a PL-SQL function to do this in a simple way? Ideally with the time periods not being fixed to monthly (for example, weekly or fortnightly too)? Ideally I will end up with a table like this shown below:
Many thanks for any pointers...
You seem to need a list of dates. You can generate one in the query and then use correlated subqueries to calculate the columns you want:
select d.*,
(select count(distinct customer_id)
from t
where t.cover_start_date <= d.dte and
(t.cover_end_date > d.date + interval '1' month or t.cover_end_date is null)
) as num_customers,
(select count(*)
from t
where t.accident_date >= d.dte and
t.accident_date < d.date + interval '1' month
) as accidents,
(select count(distinct customer_id)
from t
where t.accident_date >= d.dte and
t.accident_date < d.date + interval '1' month
) as num_customers_with_accident
from (select date '2020-01-01' as dte from dual union all
select date '2020-02-01' as dte from dual union all
. . .
) d;
If you want to do arithmetic on the columns, you can use this as a subquery or CTE.
I have a SQL query in PostgreSQL 9.4 that, while more complex due to the tables I am pulling data from, boils down to the following:
SELECT entry_date, user_id, <other_stuff>
FROM <tables, joins, etc>
GROUP BY entry_date, user_id
WHERE <whatever limits I want, such as limiting the date range or users>
With the result that I have one row per user, per day for which I have data. In general, this query would be run for an entry_date period of one month, with the desired result of having one row per day of the month for each user.
The problem is that there may not be data for every user every day of the month, and this query only returns rows for days that have data.
Is there some way to modify this query so it returns one row per day for each user, even if there is no data (other than the date and the user) in some of the rows?
I tried doing a join with a generate_series(), but that didn't work - it can make there be no missing days, but not per user. What I really need would be something like "for each user in list, generate series of (user,date) records"
EDIT: To clarify, the final result that I am looking for would be that for each user in the database - defined as a record in a user table - I want one row per date. So if I specify a date range of 5/1/15-5/31/15 in my where clause, I want 31 rows per user, even if that user had no data in that range, or only had data for a couple of days.
generate_series() was the right idea. You probably did not get the details right. Could work like this:
WITH cte AS (
SELECT entry_date, user_id, <other_stuff>
FROM <tables, joins, etc>
GROUP BY entry_date, user_id
WHERE <whatever limits I want>
)
SELECT *
FROM (SELECT DISTINCT user_id FROM cte) u
CROSS JOIN (
SELECT entry_date::date
FROM generate_series(current_date - interval '1 month'
, current_date - interval '1 day'
, interval '1 day') entry_date
) d
LEFT JOIN cte USING (user_id, entry_date);
I picked a running time window of one month ending "yesterday". You did not define your "month" exactly.
Assuming entry_date to be data type date.
Simpler for your updated requirements
To get results for every user in a users table (and not for a current selection) and for your given time range, it gets simpler. You don't need the CTE:
SELECT *
FROM (SELECT user_id FROM users) u
CROSS JOIN (
SELECT entry_date::date
FROM generate_series(timestamp '2015-05-01'
, timestamp '2015-05-31'
, interval '1 day') entry_date
) d
LEFT JOIN (
SELECT entry_date, user_id, <other_stuff>
FROM <tables, joins, etc>
GROUP BY entry_date, user_id
WHERE <whatever>
) t USING (user_id, entry_date);
Why this particular way to call generate_series()?
Generating time series between two dates in PostgreSQL
And best use ISO 8601 date format (YYYY-MM-DD) which works regardless of locale settings.
many queries are by week, month or quarter when the base table date is either date or timestamp.
in general, in group by queries, does it matter whether using
- functions on the date
- a day table that has extraction pre-calculated
note: similar question as DATE lookup table (1990/01/01:2041/12/31)
for example, in postgresql
create table sale(
tran_id serial primary key,
tran_dt date not null default current_date,
sale_amt decimal(8,2) not null,
...
);
create table days(
day date primary key,
week date not null,
month date not null,
quarter date non null
);
-- week query 1: group using funcs
select
date_trunc('week',tran_dt)::date - 1 as week,
count(1) as sale_ct,
sum(sale_amt) as sale_amt
from sale
where date_trunc('week',tran_dt)::date - 1 between '2012-1-1' and '2011-12-31'
group by date_trunc('week',tran_dt)::date - 1
order by 1;
-- query 2: group using days
select
days.week,
count(1) as sale_ct,
sum(sale_amt) as sale_amt
from sale
join days on( days.day = sale.tran_dt )
where week between '2011-1-1'::date and '2011-12-31'::date
group by week
order by week;
to me, whereas the date_trunc() function seems more organic, the the days table is easier to use.
is there anything here more than a matter of taste?
-- query 3: group using instant "immediate" calendar table
WITH calender AS (
SELECT ser::date AS dd
, date_trunc('week', ser)::date AS wk
-- , date_trunc('month', ser)::date AS mon
-- , date_trunc('quarter', ser)::date AS qq
FROM generate_series( '2012-1-1' , '2012-12-31', '1 day'::interval) ser
)
SELECT
cal.wk
, count(1) as sale_ct
, sum(sa.sale_amt) as sale_amt
FROM sale sa
JOIN calender cal ON cal.dd = sa.tran_dt
-- WHERE week between '2012-1-1' and '2011-12-31'
GROUP BY cal.wk
ORDER BY cal.wk
;
Note: I fixed an apparent typo in the BETWEEN range.
UPDATE: I used Erwin's recursive CTE to squeeze out the duplicated date_trunc(). Nested CTE galore:
WITH calendar AS (
WITH RECURSIVE montag AS (
SELECT '2011-01-01'::date AS dd
UNION ALL
SELECT dd + 1 AS dd
FROM montag
WHERE dd < '2012-1-1'::date
)
SELECT mo.dd, date_trunc('week', mo.dd + 1)::date AS wk
FROM montag mo
)
SELECT
cal.wk
, count(1) as sale_ct
, sum(sa.sale_amt) as sale_amt
FROM sale sa
JOIN calendar cal ON cal.dd = sa.tran_dt
-- WHERE week between '2012-1-1' and '2011-12-31'
GROUP BY cal.wk
ORDER BY cal.wk
;
Yes, it is more than a matter of taste. The performance of the query depends on the method.
As a first approximation, the functions should be faster. They don't require joins, doing the read in a single table scan.
However, a good optimizer could make effective use of a lookup table. It would know the distribution of the target values. And, an in memory join could be quite fast.
As a database design, I think having a calendar table is very useful. Some information such as holidays just isn't going to work as a function. However, for most ad hoc queries the date functions are fine.
1. Your expression:
... between '2012-1-1' and '2011-12-31'
doesn't work. Basic BETWEEN requires the left argument to be less than or equal to the right argument. Would have to be:
... BETWEEN SYMMETRIC '2012-1-1' and '2011-12-31'
Or it's just a typo and you mean something like:
... BETWEEN '2011-1-1' and '2011-12-31'
It's unclear to me, what your queries are supposed to retrieve. I'll assume you want all weeks (Monday to Sunday) that start in 2011 for the rest of this answer. This expression generates exactly that in less than a microsecond on modern hardware (works for any year):
SELECT generate_series(
date_trunc('week','2010-12-31'::date) + interval '7d'
,date_trunc('week','2011-12-31'::date) + interval '6d'
, '1d')::date
*Note that the ISO 8601 definition of the "first week of a year is slightly different.
2. Your second query does not work at all. No GROUP BY?
3. The question you link to did not deal with PostgreSQL, which has outstanding date / timestamp support. And it has generate_series() which can obviate the need for a separate "days" table in most cases - as demonstrated above. Your query would look like this:
In the meantime #wildplasser provided an example query that was supposed to go here.
By popular* demand, a recursive CTE version - which is actually not that far from being a serious alternative!
* and by "popular" I mean #wildplasser's very serious request.
WITH RECURSIVE days AS (
SELECT '2011-01-01'::date AS dd
,date_trunc('week', '2011-01-01'::date )::date AS wk
UNION ALL
SELECT dd + 1
,date_trunc('week', dd + 1)::date AS wk
FROM days
WHERE dd < '2011-12-31'::date
)
SELECT d.wk
,count(*) AS sale_ct
,sum(s.sale_amt) AS sale_amt
FROM days d
JOIN sale s ON s.tran_dt = d.dd
-- WHERE d.wk between '2011-01-01' and '2011-12-31'
GROUP BY 1
ORDER BY 1;
Could also be written as (compare to #wildplasser's version):
WITH RECURSIVE d AS (
SELECT '2011-01-01'::date AS dd
UNION ALL
SELECT dd + 1 FROM d WHERE dd < '2011-12-31'::date
), days AS (
SELECT dd, date_trunc('week', dd + 1)::date AS wk
FROM d
)
SELECT ...
4. If performance is of the essence, just make sure, that you do not apply functions or calculations to the values of your table. This prohibits the use of indexes and is generally very slow, because every row has to be processed. That's why your first query is going to suck with big table. When ever possible, apply calculations to the values you filter with, instead.
Indexes on expressions are one way around this. If you had an index like
CREATE INDEX sale_tran_dt_week_idx ON sale (date_trunc('week', tran_dt)::date);
.. your first query could be very fast again - at some cost for write operations for index maintenance.
I want to query statistics using SQL from 3 different days (in a row). The display would be something like:
15 users created today, 10 yesterday, 12 two days ago
The SQL would be something like (for today):
SELECT Count(*) FROM Users WHERE created_date >= '2012-05-11'
And then I would do 2 more queries for yesterday and the day before.
So in total I'm doing 3 queries against the entire database. The format for created_date is 2012-05-11 05:24:11 (date & time).
Is there a more efficient SQL way to do this, say in one query?
For specifics, I'm using PHP and SQLite (so the PDO extension).
The result should be 3 different numbers (one for each day).
Any chance someone could show some performance numbers in comparison?
You can use GROUP BY:
SELECT Count(*), created_date FROM Users GROUP BY created_date
That will give you a list of dates with the number of records found on that date. You can add criteria for created_date using a normal WHERE clause.
Edit: based on your edit:
SELECT Count(*), created_date FROM Users WHERE created_date>='2012-05-09' GROUP BY date(created_date)
The best solution is to use GROUP BY DAY(created_date). Here is your query:
SELECT DATE(created_date), count(*)
FROM users
WHERE created_date > CURRENT_DATE - INTERVAL 3 DAY
GROUP BY DAY(created_date)
This would work I believe though I have no way to test it:
SELECT
(SELECT Count(*) FROM Users WHERE created_date >= '2012-05-11') as today,
(SELECT Count(*) FROM Users WHERE created_date >= '2012-05-10') as yesterday,
(SELECT Count(*) FROM Users WHERE created_date >= '2012-05-11') as day_before
;
Use GROUP BY like jeroen suggested, but if you're planning for other periods you can also set ranges like this:
SELECT SUM(IF(created_date BETWEEN '2012-05-01' AND NOW(), 1, 0)) AS `this_month`,
SUM(IF(created_date = '2012-05-09', 1, 0)) AS `2_days_ago`
FROM ...
As noted below, SQLite doesn't have IF function but there is CASE instead. So this way it should work:
SELECT SUM(CASE WHEN created_date BETWEEN '2012-05-01' AND NOW() THEN 1 ELSE 0 END) AS `this_month`,
SUM(CASE created_date WHEN '2012-05-09' THEN 1 ELSE 0 END) AS `2_days_ago`
FROM ...
I need to schedule some items in a postgres query based on a requested delivery date for an order. So for example, the order has a requested delivery on a Monday (20120319 for example), and the order needs to be prepared on the prior working day (20120316).
Thoughts on the most direct method? I'm open to adding a dates table. I'm thinking there's got to be a better way than a long set of case statements using:
SELECT EXTRACT(DOW FROM TIMESTAMP '2001-02-16 20:38:40');
This gets you previous business day.
SELECT
CASE (EXTRACT(ISODOW FROM current_date)::integer) % 7
WHEN 1 THEN current_date-3
WHEN 0 THEN current_date-2
ELSE current_date-1
END AS previous_business_day
To have the previous work day:
select max(s.a) as work_day
from (
select s.a::date
from generate_series('2012-01-02'::date, '2050-12-31', '1 day') s(a)
where extract(dow from s.a) between 1 and 5
except
select holiday_date
from holiday_table
) s
where s.a < '2012-03-19'
;
If you want the next work day just invert the query.
SELECT y.d AS prep_day
FROM (
SELECT generate_series(dday - 8, dday - 1, interval '1d')::date AS d
FROM (SELECT '2012-03-19'::date AS dday) x
) y
LEFT JOIN holiday h USING (d)
WHERE h.d IS NULL
AND extract(isodow from y.d) < 6
ORDER BY y.d DESC
LIMIT 1;
It should be faster to generate only as many days as necessary. I generate one week prior to the delivery. That should cover all possibilities.
isodow as extract parameter is more convenient than dow to test for workdays.
min() / max(), ORDER BY / LIMIT 1, that's a matter of taste with the few rows in my query.
To get several candidate days in descending order, not just the top pick, change the LIMIT 1.
I put the dday (delivery day) in a subquery so you only have to input it once. You can enter any date or timestamp literal. It is cast to date either way.
CREATE TABLE Holidays (Holiday, PrecedingBusinessDay) AS VALUES
('2012-12-25'::DATE, '2012-12-24'::DATE),
('2012-12-26'::DATE, '2012-12-24'::DATE);
SELECT Day, COALESCE(PrecedingBusinessDay, PrecedingMondayToFriday)
FROM
(SELECT Day, Day - CASE DATE_PART('DOW', Day)
WHEN 0 THEN 2
WHEN 1 THEN 3
ELSE 1
END AS PrecedingMondayToFriday
FROM TestDays) AS PrecedingMondaysToFridays
LEFT JOIN Holidays ON PrecedingMondayToFriday = Holiday;
You might want to rename some of the identifiers :-).