This is a follow-up question on Get age from the birthday field with type date using SQL. I have a date field in a MySQL database for the birthday of a user and get the age using this query:
SELECT
ROUND(DATEDIFF(
Cast((SELECT NOW()) as Date),
Cast(birthday as Date)
) / 365, 0) as age
FROM member
Now, I need to select the number of people in different age groups. For example, I need to know how many people are in the age group 13-17, 18-21, 22-25, 26-35, 36-50, 51-MAX.
Is that possible using MySQL?
I have thought of UNIONs, like this:
SELECT
ROUND(DATEDIFF(
Cast((SELECT NOW()) as Date),
Cast(birthday as Date)
) / 365, 0) as age,
1 as agegroup
FROM member WHERE age >=13 AND age <=17
UNION
SELECT
ROUND(DATEDIFF(
Cast((SELECT NOW()) as Date),
Cast(birthday as Date)
) / 365, 0) as age
2 as agegroup
FROM member WHERE age >=18 AND age <=21
But that would be long and ugly. There must be a better way!
select AgeGroup
, count(*)
from (
select case
when age between 13 and 17 then 1
when age between 18 and 21 then 2
...
end as AgeGroup
from (
SELECT ROUND(DATEDIFF(Cast(NOW() as Date),
Cast(birthday as Date)) / 365, 0) as age
FROM YourTable
) as SubQueryAlias
) as SubQueryAlias2
group by
AgeGroup
Another possible solution:-
SELECT AgeRange.MinAge, AgeRange.MaxAge, COUNT(*)
FROM
(
SELECT 13 AS MinAge, 17 AS MaxAge
UNION SELECT 18, 21
UNION SELECT 22, 25
UNION SELECT 26, 35
UNION SELECT 36, 50
UNION SELECT 51, 9999
) AgeRange
INNER JOIN YourTable
ON ROUND(DATEDIFF(CAST(NOW() as DATE), CAST(birthday as DATE)) / 365, 0) BETWEEN AgeRange.MinAge AND AgeRange.MaxAge
GROUP BY AgeRange.MinAge, AgeRange.MaxAge
Possibly easier to expand if needs be, or to move to using date ranges from a table (so the resulting report could be updated by users easily if required).
If you had the age as a column in a table you would do it like this:
SELECT
SUM(CASE WHEN age < 10 THEN 1 ELSE 0 END) AS under10,
SUM(CASE WHEN 10<age AND age <19 THEN 1 ELSE 0 END) AS age10to19,
.
.
.
FROM table
There are likely to be minor changes because age isn't in its own column or if you want extra or different ranges. I'm sure you can work them out yourself!
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 am very new to SQL and I need to write a query that selects data for a specific part. However, It should select only the part that is the most recent(given by date) and anything that is only 30 days prior to it. Please consider the table below:
PartID | Part_NAME | DATE
-----------------------------
1 AAA 6/16/2015
2 BBB 6/15/2015
3 AAA 6/11/2015
4 AAA 1/1/2008
I need a query that gives me:
PartID | Part_NAME | DATE
-----------------------------
1 AAA 6/16/2015
3 AAA 6/11/2015
I have tried:
select * from ( select * from sales_table where Part_NAME = 'AAA') where DATE BETWEEN (max(DATE) and (max(DATE)-30))
I have read some articles saying that I cannot use WHERE and functions like max() together and advised me to use group by or having but it didn't work for me as well. Thank you.
IF you want data from the last 30 days of the current day, you can do :
SELECT *
FROM sales_table
WHERE
[DATE] >= DATEADD(DAY, -30,GETDATE())
AND [DATE] <= GETDATE()
AND Part_NAME = 'AAA'
IF you want data from the last 30 days from the last date of sale of each Part_NAME (this will take the max recorded date of sale for each Part_NAME and get the last 30 days records of each one of them.)
SELECT *
FROM (
SELECT *,
MAX([DATE]) OVER(PARTITION BY Part_NAME ORDER BY PartID) AS RecentDate
FROM sales_table
) D
WHERE
[DATE] >= DATEADD(DAY, -30, RecentDate)
AND [DATE] <= RecentDate
AND Part_NAME = 'AAA'
You can accomplish by using datediff and getdate() and a subquery.
SELECT * FROM (
SELECT *,DATEDIFF(DD,[DATE],GETDATE()) AS DAYSBETWEEN FROM sales_table
) AS X
WHERE DAYSBETWEEN <= 30
If you want data from the last 30 days, it would be:
select st.*
from sales_table st join
(select top (1) st2.*
from sales_table st2
order by st2.date desc
) st2
on st2.part_name = st.part_name and
st.date >= dateadd(day, -30, cast(getdate() as date));
I have a one database and saved one year data. I want to access all data in monthwise. like (01-01-2017-01-30-2017). My column type Datetime
date_table
01-01-2016
01-02-2016
02-02-2017
.....
.....
I Want to show
date_table count
01-01-2016 to 30-01-2016 5
01-02-2016 t0 28-02-2016 10
Have a derived table that returns the year and month. GROUP BY its result and count:
select y, m, count(*)
from
(
select year(date_column) y , month(date_column) m
from tablename
)
group by y, m
I don't know MySQL very well, so you'll have to format the output by yourself.
You can use mysql's built in function LAST_DAY, like this:
Just replace here "t", with your table name
SELECT CONCAT( CONCAT(mnth, '-01'), ' to ', LAST_DAY(CONCAT(mnth, '-01')) ) AS date_table, cnt FROM (
select substr(cast(date_table as CHAR), 1, 7) AS mnth, count(*) AS cnt from t
group by substr(cast(date_table as CHAR), 1, 7)
) outertable
If your Date is of type DateTime in SQL-Server
SELECT DATENAME(MONTH,TimeStamp) AS [MONTH]
, DATEPART(YY,TimeStamp) [YEAR]
, SUM(*) Count
FROM Table
WHERE TimeStamp IS NOT NULL
GROUP BY DATENAME(MONTH,TimeStamp), DATEPART(YY,TimeStamp)
Result would be like this
MONTH YEAR Count
December2015 237
October 2015 51
August 2016 0
January 2016 9
March 2016 160
good morning,
I have a nagging issue I cannot really solve.. I have a database table like this, showing resources spent (value) in a date range by person:
id,name,startdate,enddate,value
--------------------------------
10,John,2012-01-14,2012-10-30,200000
11,Jack,2012-02-01,2012-08-01,70000
12,John,2012-05-01,2012-06-01,2000
I need a query that creates the result like this, summarizing the 'value' by month, taking partial months into account
month, name, value
------------------
2012-01, John, 9000
2012-02, John, 18000
2012-03, John, 18000
2012-04, John, 18000
2012-05, John, 20000
2012-06, John, 18000
2012-07, John, 18000
2012-08, John, 18000
2012-01, John, 18000
2012-02, Jack, 10000
2012-03, Jack, 10000
2012-04, Jack, 10000
2012-05, Jack, 10000
2012-06, Jack, 10000
2012-07, Jack, 10000
2012-08, Jack, 0
Now I know how I'd do this procedurally (like with PHP) with a loop: get the daily amount, then check month by month how many days spent there according to the range and multiply it by the daily amount.
thanks
peter
If you don't have a calendar table and can't create one, you can simulate a virtual calendar table in your query. Here's a query that should answer your question, that makes use of such a virtual table:
select m.startmonth,
e.name,
coalesce(sum(r.value *
datediff(case when adddate(m.startmonth, interval 1 month) <
r.enddate
then adddate(m.startmonth, interval 1 month)
else r.enddate end,
case when m.startmonth > r.startdate
then m.startmonth else r.startdate end) /
datediff(r.enddate,r.startdate)),0) valueshare
from
(select cast('2012-01-01' as date) startmonth union all
select cast('2012-02-01' as date) startmonth union all
select cast('2012-03-01' as date) startmonth union all
select cast('2012-04-01' as date) startmonth union all
select cast('2012-05-01' as date) startmonth union all
select cast('2012-06-01' as date) startmonth union all
select cast('2012-07-01' as date) startmonth union all
select cast('2012-08-01' as date) startmonth union all
select cast('2012-09-01' as date) startmonth union all
select cast('2012-10-01' as date) startmonth) m
cross join employees e
left join resources_spent r
on r.enddate > m.startmonth and
r.startdate < adddate(m.startmonth, interval 1 month) and
r.name = e.name
group by m.startmonth, e.name
order by 2,1
SQLFiddle here.
I think you need a calendar table with one row for each date. Other fields would be whatever is useful to you, such as fiscal periods, holidays, whatever.
Then, for your report, you could create a temp table and populate it like this:
insert into YourTempTable
(id, date, amount)
select id, c.datefield, amount
from YourTable join Calendar c on datefield >= startdate
and datefield <= enddate
where whatever
From there, you select from YourTempTable and YourTable, joining on the id.
I have a MySQL table ATT which has EMP_ID,ATT_DATE,ATT_STATUS with ATT_STATUS with different values 1-Present,2-Absent,3-Weekly-off. I want to find out those EMP_ID's which have status 2 consecutively for 10 days in a given date range.
Please help
Please have a try with this:
SELECT EMP_ID FROM (
SELECT
IF((#prevDate!=(q.ATT_DATE - INTERVAL 1 DAY)) OR (#prevEmp!=q.EMP_ID) OR (q.ATT_STATUS != 2), #rownum:=#rownum+1, #rownum:=#rownum) AS rownumber, #prevDate:=q.ATT_DATE, #prevEmp:=q.EMP_ID, q.*
FROM (
SELECT
EMP_ID
, ATT_DATE
, ATT_STATUS
FROM
org_tb_dailyattendance, (SELECT #rownum:=0, #prevDate:='', #prevEmp:=0) vars
WHERE ATT_DATE BETWEEN '2013-01-01' AND '2013-02-15'
ORDER BY EMP_ID, ATT_DATE, ATT_STATUS
) q
) sq
GROUP BY EMP_ID, rownumber
HAVING COUNT(*) >= 10
The logic is, to first sort the table by employee id and the dates. Then introduce a rownumber which increases only if
the days are not consecutive or
the employee id is not the previous one or
the status is not 2
Then I just grouped by this rownumber and counted if there are 10 rows in each group. That should be the ones who were absent for 10 days or more.
Have you tried something like this
SELECT EMP_ID count(*) as consecutive_count min(ATT_DATE)
FROM (SELECT * FROM ATT ORDER BY EMP_ID)
GROUP BY EMP_ID, ATT_DATE
WHERE ATT_STATUS = 2
HAVING consecutive_count > 10