counting age in sql and view the data in sql table - mysql

Question

|---------------------|
|DUN|NAME |BIRTHDAY |
--------------------
|A |MELAYU |21/2/2014|
|A |CINA |21/2/2002|
|B |CINA |21/2/2011|
|-------------------- |
My table name is maklumat. I want view the data based on age using sql. below are the example table to view.
|---------------------|
|DUN|AGE<3 |AGE 3-6 |
--------------------
|A |1 |1 |
|B |0 |1 |
|-------------------- |
THIS IS WHAT I'VE TRIED
SELECT jantina,dun, count(tarikh_lahir > 1954-04-07) as 60KEBAWAH, count(1950-04-07>=tarikh_lahir <= 1954-04-07) as 60HINGGA64, count(1945-04-07>=tarikh_lahir <= 1949-04-07) as 65HINGGA69, count(1940-04-07>=tarikh_lahir <= 1944-04-07) as 70HINGGA74,count(1935-04-07>=tarikh_lahir <= 1939-04-07) as 75HINGGA79,count(1930-04-07>=tarikh_lahir <= 1934-04-07) as 80HINGGA85, count(1925-04-07>=tarikh_lahir <= 1929-04-07) as 86HINGGA89, count(1920-04-07>=tarikh_lahir <= 1924-04-07) as 90HINGGA94, count(1915-04-07>=tarikh_lahir <= 1919-04-07) as 95HINGGA99, count(1910-04-07>=tarikh_lahir <= 1914-04-07) as 100HINGGA104,count(tarikh_lahir <1910-04-07) as 60KEBAWAH
FROM maklumat_ahli
WHERE jantina = 'lelaki' AND (
kematian_tarikh IS NULL
AND bayaran_pertama IS NULL
AND bayaran_kedua IS NULL
)
GROUP BY dun

Answer 1

SELECT M.DUN, COUNT(M.AGE) AS 'AGE<3', COUNT(M1.AGE) AS 'AGE 3-6' FROM MAKLUMAT M
LEFT JOIN MAKLUMAT M1 ON M.DUN = M1.DUN AND M1.AGE BETWEEN 3 AND 6
WHERE M.AGE < 3
GROUP BY M.DUN

Answer 2

First calculate the age in years. This is this year minus the birth year. Exception: The birthday date is yet to come this year. In this case we must subtract one. With the ages thus found we can start to count, which we can do with the sum function by adding one for every match.
select
dun,
sum(age < 3) as age0to2,
sum(age between 3 and 6) as age3to6
from
(
select
dun,
year(now()) - year(birthday) -
case when month(now()) * 100 + day(now()) < month(birthday) * 100 + day(birthday) then
1
else
0
end
as age
from maklumat
)
group by dun;

Answer 3

Here is why your select statement fails: MySQL knows a boolean datatype, which is zero for FALSE and non-zero for TRUE. When evaluating an expression the result is either 0 or 1 (or NULL).
The expression (tarikh_lahir > 1954-04-07) is 1 for birth dates after the date stated, 0 for birth dates before and only NULL when the birth date is NULL. Count counts all non-null values, zeros and ones alike, i.e. all dates given. This is not what you want.
Instead of COUNT you can use SUM, to sum the positive results (i.e. the 1s). Besides: The dates should be strings as far as I know, otherwise they would be treated as numeric expressions, i.e. 1954-04-07 = 1954 minus 04 minus 07.
SELECT
jantina,
dun,
sum(tarikh_lahir > '1954-04-07') as 60KEBAWAH,
sum('1950-04-07' >= tarikh_lahir <= '1954-04-07') as 60HINGGA64,
sum('1945-04-07' >= tarikh_lahir <= '1949-04-07') as 65HINGGA69,
sum('1940-04-07' >= tarikh_lahir <= '1944-04-07') as 70HINGGA74,
sum('1935-04-07' >= tarikh_lahir <= '1939-04-07') as 75HINGGA79,
sum('1930-04-07' >= tarikh_lahir <= '1934-04-07') as 80HINGGA85,
sum('1925-04-07' >= tarikh_lahir <= '1929-04-07') as 86HINGGA89,
sum('1920-04-07' >= tarikh_lahir <= '1924-04-07') as 90HINGGA94,
sum('1915-04-07' >= tarikh_lahir <= '1919-04-07') as 95HINGGA99,
sum('1910-04-07' >= tarikh_lahir <= '1914-04-07') as 100HINGGA104,
sum(tarikh_lahir <'1910-04-07') as 60KEBAWAH
FROM maklumat_ahli
WHERE jantina = 'lelaki'
AND kematian_tarikh IS NULL
AND bayaran_pertama IS NULL
AND bayaran_kedua IS NULL
GROUP BY jantina, dun;
BTW: Are names starting with digits allowed in MySQL? Otherwise you will have to find other names for 60HINGGA64 etc. or use quotes.