Question

我想弄清楚如何将学生年龄转换为年龄+月。例如，如果学生的出生日期是2000年1月1日，则学生年龄将显示为：16和6个月

我能够编写查询脚本，以便学生年龄显示为整数，但我不知道如何包含月份。我需要改变什么？

ALTER procedure [dbo].[RP_Student_Attendance] (@perf_code varchar(10))
as

-- set @perf_code = '3496A'

SELECT
    i.short_name, 
    i.text1 as perf_desc, 
    i.text2 perf_dates, 
    i.text3 perf_times,
    p.perf_no, 
    h.[Role], 
    h.customer_no,
    c.sex, b.birthday, 
    g.grade_level, 
    --(DATEDIFF(yy,b.birthday,GETDATE())) as student_age, 
    student_age = CASE

WHEN dateadd(year, datediff (year, b.birthday, GETDATE()),b.birthday) > GETDATE()
THEN datediff (year, b.birthday, GETDATE()) - 1
ELSE datediff (year, b.birthday, GETDATE())end,
    c.fname + ' ' + c.lname as student_name

FROM T_INVENTORY i
JOIN T_PERF p on i.inv_no = p.perf_no
JOIN T_TICKET_HISTORY h on p.perf_no = h.perf_no and [Role] = 4
JOIN T_CUSTOMER c on h.customer_no = c.customer_no 
LEFT JOIN (select customer_no, key_value as birthday from TX_CUST_KEYWORD where keyword_no = 1) b on c.customer_no = b.customer_no
LEFT JOIN (select customer_no, key_value as grade_level from TX_CUST_KEYWORD where keyword_no = 428) g on c.customer_no = g.customer_no
WHERE p.perf_code = @perf_code

Answer 1

如果你不关心精确度，这会让你进入球场：

mysql> select floor(datediff(now(), '2000-01-01') / 365) as years, ceil(mod(datediff(now(), '2000-01-01'), 365) / 30) as months;
+-------+--------+
| years | months |
+-------+--------+
|    16 |      6 |
+-------+--------+

我再说一遍，如果你不关心精度，只使用。这可能很容易报告“7”个月，而实际上是“6”。

年龄格式 - 年和月

1 个答案: