【发布时间】:2015-01-09 22:00:55
【问题描述】:
我有一个字符串变量名$mydate,意思是10 january 2014
$mydate="10-01-2014";
我想把它转换成字符串变量也是'2014-01-10'
嗨,半疯狂, 我把你的解决方案是这样的:
foreach($report_data['summary'] as $key=>$row) {
$substrdate=substr($row['payment_type'],-16); //i have see the result is 10-01-2014
$originalDate = '10-01-2014';
try {
$date = DateTime::createFromFormat('d-m-Y', $originalDate);
//echo $date->format('Y-m-d');
} catch(Exception $e) {
die("Error converting date. Exception caught: " . $e->getMessage());
}
$summary_data_row[] = array('data'=>'<span style="color:'.$color.'">'.$date->format('Y-m-d').'</span>', 'align'=>'right');
$summary_data_row[] = array('data'=>'<span style="color:'.$color.'">'.$row['comment'].'</span>', 'align'=>'right');
}//end of foreach
它运行良好,直到我用具有相同值的 $substrdate 替换变量 $originalDate -> '10-01-2014' 为什么它不再工作了?
【问题讨论】:
-
echo date('Y-m-d')aj mbak。 Bs dipisah pake-,|,/dll. -
Skrg kendala ny apa stlh pk
$substrdate? ga keluar atau formatnya salah ?
标签: php date datetime-format