Question

 |---------------------|
 |DUN|NAME   |BIRTHDAY |
  --------------------
 |A  |MELAYU |21/2/2014|
 |A  |CINA   |21/2/2002|
 |B  |CINA   |21/2/2011|
 |-------------------- |

我的表名是maklumat。我想用sql查看基于年龄的数据。下面是要查看的示例表。

 |---------------------|
 |DUN|AGE<3  |AGE 3-6  |
  --------------------
 |A  |1      |1        |
 |B  |0      |1        |
 |-------------------- |

这就是我做过的事情

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

首先计算年龄。这是今年减去出生年份。例外：今年的生日日期尚未到来。在这种情况下，我们必须减去一个。通过这样发现的年龄，我们可以开始计算，我们可以通过为每个匹配添加一个sum函数来做。

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

这就是你的select语句失败的原因：MySQL知道一个布尔数据类型，对于FALSE为零，对于TRUE为非零。在计算表达式时，结果为0或1（或NULL）。

表达式（tarikh_lahir＆gt; 1954-04-07）在所述日期之后的出生日期为1，出生日期之前为0，出生日期为NULL时仅为NULL。计数所有非空值，零和1，即所有给定的日期。这不是你想要的。

您可以使用SUM代替COUNT来对正结果（即1）求和。此外：据我所知，日期应该是字符串，否则它们将被视为数字表达式，即1954-04-07 = 1954减去04减去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：MySQL中是否允许以数字开头的名字？否则，您必须找到60HINGGA64等的其他名称或使用引号。

在sql中计算年龄并查看sql表中的数据

3 个答案: