将结果存储在var而不是printf中

我有这个PHP脚本,从日期开始获取年龄:

$bday = new DateTime('1987-04-21');
$today = new DateTime(date('Y-m-d'));
$diff = $today->diff($bday);
printf('%d Years, %d Months', $diff->y, $diff->m);

现在我不想直接printf,我想把结果放在var中以便在别处使用它.怎么样？

只需替换printf为sprintf这样

$var = sprintf('%d Years, %d Months', $diff->y, $diff->m);

sprintf的文档