php var 转换为年份和月份

Question

$date ='20101015';

如何转换为 $year = 2010, $month = 10,$day =15

谢谢

Answer 1

您可以将 PHP 子字符串函数用作substr：

$year  = substr($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.

如果您的原始字符串作为前导或尾随空格，这将失败，因此最好将substr修剪后的输入作为。因此，在致电之前，substr您可以执行以下操作：

$date = trim($date);