I have an SQL table with columns (month, year, and yearmonth)
Month year
5 2020
6 2020
11 2020
I want to insert yearmonth column values in this format
yearmonth
202005
202006
202011
I've tried couple of Datetime functions, but it didn't work.
I think that it is simpler to use string functions:
select t.*, concat(year, lpad(month, 2, '0')) as yearmonth
from mytable t
lpad() adds the "missing" 0 to the month part if needed, which you can then concatenate with the year.
On the other hand, if you want a numeric result, arithmetics is the way to go:
select t.*, year * 100 + month as yearmonth
from mytable t
If you want a string, I might use:
select cast(year * 100 + month as char)
Related
I have a table called months - this contains all 12 months of the calendar, the IDs correspond to the month number.
I will be running a query to retrieve 2 or 3 sequential months from this table, e.g
April & May
June, July, August
December & January
However I want to ensure that whenever December are January and retrieved, that it retrieves them in that order, and not January - December. Here is what I have tried:
SELECT * FROM `months`
WHERE start_date BETWEEN <date1> AND <date2>
ORDER BY
FIELD(id, 12, 1)
This works for December & January, but now when I try to retrieve January & February it does those in the wrong order, i.e "February - January" - I'm guessing because we specified 1 in the ORDER BY as the last value.
Anybody know the correct way to achieve this? As I mentioned this should also work for 3 months, so for example "November, December, January" and "December, January, February" should all be retrieved in that order.
If you want December first, but the other months in order, then:
order by (id = 12) desc, id
MySQL treats booleans as numbers, with "1" for true and "0" for false. The desc puts the 12s first.
EDIT:
To handle the more general case, you can use window functions. Assuming the numbers are consecutive, then the issue is trickier. This will work for 2 and 3-month spans:
order by (case min(id) over () > 1 then id end),
(case when id > 6 1 else 2 end),
id
I'm reluctant to think about a more general solution based only on months. After all, you can just use:
order by start_date
Or, if you have an aggregation query:
order by min(start_date)
to solve the real problem.
This is not "mysql solution" properly :
with cte (id, month) AS (
select id, month from months
union all
select id, month from months
)
, cte1 (id, month, r) as (select id, month, row_number() over() as r from cte )
select * from cte1
where id in (12, 1)
and r >= 12 order by r limit 2 ;
DECLARE
#monthfrom int = 12,
#monthto int = 1;
with months as (select 1 m
union all
select m+1 from months where m<12)
select m
from months
where m in (#monthfrom,#monthto)
order by
case when #monthfrom>#monthto
then
m%12
else
m
end
result:
12
1
Basically in MySQL this can be done the same way:
set #from =12;
set #to =1;
with recursive months(m) as (
select 1 m
union all
select m+1 from months where m<12)
select *
from months
where m in (#from,#to)
order by case when #from>#to then m%12 else m end;
I want the count of records in every month and total count of records from start upto that month.
For ex.,
I have a table that looks like this:
#id,created#
1,'2016-01-01'
2,'2011-02-02'
3,'2011-02-09'
4,'2011-02-05'
5,'2011-03-07'
6,'2011-03-08'
How do I select and group these so the output is:
#Month, new, total#
Jan 2016, 1, 1
Feb 2016, 3, 4
Mar 2016, 2, 6
Thanks very much.
Here you go:
SELECT DATE_FORMAT(`created`,'%M %Y') AS month, COUNT(*) AS count,
(SELECT count(*) FROM test WHERE MONTH(created) <= MONTH(t.created)) AS total
FROM test t
GROUP BY MONTH(created);
Here's the SQL Fiddle.
Using single table read:
SELECT
CONCAT(LEFT(MONTHNAME(dt), 3), ' ', YEAR(dt)) month,
new,
#total:=#total + new total
FROM
(SELECT
created - INTERVAL DAY(created) - 1 DAY dt, COUNT(*) new
FROM
t
GROUP BY created - INTERVAL DAY(created) - 1 DAY
ORDER BY dt) t
CROSS JOIN
(SELECT #total:=0) t2
Demo
I want to subtract 2 periods of dates to compare data of current year vs previous year and my query is not working.
If I remove a period it works, but not sure how to pull 2 periods.
Thanks
SELECT year(date) as year, WEEK(ADDDATE(date, 5-DAYOFWEEK(date)), 3) AS 'Week', DATE,
FROM TABLE
WHERE year(date) in (2015, 2016)
AND (DATE BETWEEN "2016-10-23" AND "2016-11-12")
AND (DATE BETWEEN "2015-11-06" AND "2015-11-12")
ORDER BY DATE, year(date)
Please try this one.
SELECT year(date) as year, WEEK(ADDDATE(date, 5-DAYOFWEEK(date)), 3) AS 'Week', DATE,
FROM TABLE
WHERE
((DATE BETWEEN "2016-10-23" AND "2016-11-12")) OR
((DATE BETWEEN "2015-11-06" AND "2015-11-12"))
ORDER BY DATE, year(date)
I think you want or:
SELECT year(date) as year, WEEK(ADDDATE(date, 5-DAYOFWEEK(date)), 3) AS Week, DATE
FROM TABLE
WHERE DATE BETWEEN '2016-10-23' AND '2016-11-12' OR
DATE BETWEEN '2015-11-06' AND '2015-11-12';
Notes:
You don't need the date() comparison in the where clause.
You should use single quotes for date and string constants.
You should not use single quotes for column aliases.
SELECT year(DATE) AS year
,WEEK(ADDDATE(DATE, 5 - DAYOFWEEK(DATE)), 3) AS 'Week'
,DATE
FROM TABLE
WHERE (
(
DATE BETWEEN '2016-10-23'
AND '2016-11-12'
)
OR (
DATE BETWEEN '2015-11-06'
AND '2015-11-12'
)
)
Here's the table structure and some sample data:
pID.....month.....year
27 .....3 .....2008
27 .....12 .....2012
31 .....6 .....2008
99 .....1 .....2006
42 .....1 .....2009
pID is the practiceID and month and year represent the date period they've entered data for. I need to grab the number of practices that have entered data for the first time in Oct 2012, Nov 2012, Dec 2012 and so on.
I tried the following query for Oct 2012:
SELECT *
FROM
IPIPKDIS
where
practiceID NOT IN (
SELECT practiceID
from
IPIPKDIS
where
year < 2012 and month < 10
)
and year = 2012
and month = 10
and measureCatRecID = 2
ORDER BY year, month;
but it's grabbing months and year less than 10/2012.
If I run the queries isolated (not as subquery) they both work fine.
Any ideas?
This summary query will yield the first (smallest) date in the table for each value of practiceID.
SELECT practiceID,
MIN(STR_TO_DATE( CONCAT(year, ' ', month), '%Y %m')) first_date
FROM IPIPKDIS
GROUP BY practiceID
If you want to retrieve then the whole row for the first reported month, you'd do a nested query like this:
SELECT *
FROM IPIPKDIS I
JOIN (
SELECT practiceID,
MIN(STR_TO_DATE( CONCAT(year, ' ', month), '%Y %m')) first_date
FROM IPIPKDIS
GROUP BY practiceID
) first ON ( first.practiceID = I.practiceID
AND STR_TO_DATE( CONCAT(I.year, ' ', I.month), '%Y %m') = first.first_date)
The trick to the second query is to use the JOIN to extract just the first-month rows from your table. We use date arithmetic to do the date comparisons.
For example I have a table with fields:
id date
1 2001-01-01
2 2001-01-05
.................
N 2011-12-31
How get i get all months last days from this table?
example:
if i have dates like 2001-05-31 and 2001-06-01
i need only 2001-05-31 not both
You can do SELECT LAST_DAY for example the below returns Oct. 31st. 2010
SELECT LAST_DAY('2010-10-10');
select max(date_format(date, '%d')) as last_day_of_the_month
from table
group by date_format(date, '%Y%m')
maybe this would work better?
select DISTINCT(LAST_DAY(date)) from table GROUP BY date_format(date, '%Y%m')
SELECT id, date
FROM `table`
WHERE DATE( date ) = LAST_DAY( date )
The DATE function over field is for filter the date without time, use only if you have a datetime column.
The query get all rows with date = last day of month.
select subdate(adddate(subdate(`date`, day(`date`) - 1), interval 1 month), 1)
Note: This is the "hard way". See #harper89's answer - it's better :)
I found a solution. but this query is very slow on large tables. so I am still looking for a better solution
select DISTINCT(LAST_DAY(date)) from table;
select max(date)
from table
group by year(date), month(date)