$date ='20101015';
如何将日期转换为$year = 2010
年,$month = 10
月,$day =15
日
谢谢
substr
,例如:$year = substr($date,0,4); # extract 4 char starting at position 0.
$month = substr($date,4,2); # extract 2 char starting at position 4.
$day = substr($date,6); # extract all char starting at position 6 till end.
如果你的原始字符串有前导或尾随空格,那么这将失败,因此最好提供 substr
去掉空格的输入。所以在调用 substr
之前,你可以这样做:
$date = trim($date);
你可以使用 sscanf
一次性完成所有操作。
sscanf
— 根据指定格式从字符串中解析输入示例:
list($y, $m, $d) = sscanf('20101015', '%4d%2d%2d');
或者
sscanf('20101015', '%4d%2d%2d', $y, $m, $d);
http://www.w3schools.com/php/func_string_substr.asp
$year=substr($date,0,4);
$month=substr($date,4,2);
$day=substr($date,6,2);