如何获得年龄在18至23岁之间的记录表格。请帮助.. [英] how to get record form table having age between 18 to 23 years.. Please Help..

问题描述

我有sql表成员，如： -

id name dob

1 Ashwin 18/02/1989

2 Bhushan 30/12/1986

3 sachin 15/03/1991

4 ash 02/11/1980

5 kevin 03 / 03/1993





如何获得年龄介于18至23岁之间的记录表..请帮助..

I have sql table member like:-
id name dob
1 Ashwin 18/02/1989
2 Bhushan 30/12/1986
3 sachin 15/03/1991
4 ash 02/11/1980
5 kevin 03/03/1993


how to get record form table having age between 18 to 23 years.. Please Help..

推荐答案

您应该使用 DATEDIFF [< a href =http://technet.microsoft.com/en-us/library/ms189794.aspxtarget =_ blanktitle =新窗口> ^ ]功能。

You should use DATEDIFF[^] function.

SELECT id, name, dob
FROM member
WHERE DATEDIFF(yy, dob, GETDATE()) BETWEEN 18 AND 23

或 DATEADD [ ^ ]功能：

or DATEADD[^] function:

select * from member
where DOB between DATEADD(YEAR, -23, getdate()) and DATEADD(YEAR, -18, getdate())

但更好日期：）

but datediff better :)

尝试以下脚本..

try the below script..

Create table member
(
mno int identity(1,1) primary key,
name varchar(50),
dob varchar(50)
)
insert into member(name,dob) values('Ashwin','18/02/1989')
insert into member(name,dob) values('Bhushan','30/12/1986')
insert into member(name,dob) values('Sachin','15/03/1991')
insert into member(name,dob) values('Ash','02/11/1980')
insert into member(name,dob) values('Kevin','03/03/1993')

SELECT mno id,name,dob, DATEDIFF(YEAR,CONVERT(DATETIME, dob,105),GETDATE()) Age 
from member
where DATEDIFF(YEAR,CONVERT(DATETIME, dob,105),GETDATE()) between 18 and 23
drop table member

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