根据出生日期计算年龄 [英] calculate age based on date of birth
本文介绍了根据出生日期计算年龄的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这里我想显示用户的日期:
if(isset($ _ GET ['id']))
{
$ id = intval($ _ GET ['id ']);
$ dnn = mysql_fetch_array($ dn); $ dn = mysql_query('select username,email,skype,avatar,date,signup_date,gender from users where id ='。$ id。''');
$ dnn = mysql_fetch_array($ dn);
echo{$ dnn ['date']};
。
。
。
解决方案
PHP = 5.3.0 < sup>
#面向对象
$ from = new DateTime('1970-02 -01');
$ to = new DateTime('today');
echo $ from-> diff($ to) - > y;
#procedureural
echo date_diff(date_create('1970-02-01'),date_create('today')) - > y;
functions: date_create()
, date_diff()
< h1> MySQL = 5.0.0
SELECT TIMESTAMPDIFF(YEAR ,'1970-02-01',CURDATE())AS age
functions: TIMESTAMPDIFF()
, CURDATE()
I have a table of users in sql and they each have birth dates. I want to convert their date of birth to their age (years only), e.g. date: 15.03.1999 age: 14 and 15.03.2014 will change to age: 15
Here I want to show the date of the user:
if(isset($_GET['id']))
{
$id = intval($_GET['id']);
$dnn = mysql_fetch_array($dn);$dn = mysql_query('select username, email, skype, avatar, date, signup_date, gender from users where id="'.$id.'"');
$dnn = mysql_fetch_array($dn);
echo "{$dnn['date']}";
.
.
.
解决方案
PHP >= 5.3.0
# object oriented
$from = new DateTime('1970-02-01');
$to = new DateTime('today');
echo $from->diff($to)->y;
# procedural
echo date_diff(date_create('1970-02-01'), date_create('today'))->y;
functions: date_create()
, date_diff()
MySQL >= 5.0.0
SELECT TIMESTAMPDIFF(YEAR, '1970-02-01', CURDATE()) AS age
functions: TIMESTAMPDIFF()
, CURDATE()
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