我有三个整数,包含日期、月份和年份。我想把它转换成日期格式。
$date = 20;
$month = 8;
$year = 1989;
我已经应用了下面的公式,但当我回显$mydate时,它显示08-08-1990
$mydate = date("d-m-Y",mktime(0,0,0,$date,$month,$year));
我找不到我在哪里犯错误!
更改month
和day
位置反向
$mydate = date("d-m-Y",mktime(0,0,0,$month,$date,$year));
输出
20-08-1989
mktime语法mktime(hour,minute,second,month,day,year)
它将像
$mydate = date("d-m-Y",mktime(0,0,0,$month,$date,$year));
mktime()
会像一样对待他们
mktime(hours,minutes,seconds,month,day,year);
像一样更改
$mydate = date("d-m-Y",mktime(0,0,0,$month,$date,$year));
mktime语法为:mktime(小时、分钟、秒、月、日、年)
mktime将为您完成!
$date = date("d-m-Y",mktime(0,0,0,$month,$date,$year));
您可以使用strtotime函数。
例如
echo $a=strtotime("2009-03-18");
echo "<br />";
echo date("Y-m-d",$a);
这样做,它可能会帮助你
$originalDate = $year."-".$month."-".$date;
$newDate = date("d-m-Y", strtotime($originalDate));
echo $newDate;