[英]How to convert 2014-06-24T22:33:49.180Z format into Y-m-d H:i:s in PHP
I received Fatal error: Call to a member function format() on a non-object
error when I tried to use format on a json converted date string of 2014-06-24T22:37:13.151Z
(specifically from new Date()
) 我收到Fatal error: Call to a member function format() on a non-object
当我尝试在2014-06-24T22:37:13.151Z
的json转换日期字符串上使用格式时(特别是来自new Date()
) Fatal error: Call to a member function format() on a non-object
错误Fatal error: Call to a member function format() on a non-object
)
$objData->myCodeCreateDate="2014-06-24T22:39:34.652Z"
$date = DateTime::createFromFormat('j-M-Y', $objData->myCodeCreateDate);
echo $date->format('Y-m-d H:i:s');
That looks like the way Amazon sends back its timestamp, which follows this format in UTC time: 看起来像亚马逊发回其时间戳的方式,该时间戳遵循UTC时间的这种格式:
Ymd\\TH:i:su\\Z
.Ymd\\TH:i:su\\Z
You can edit your createFromFormat
format to the above, and it should work. 您可以将createFromFormat
格式编辑为上述格式,并且该格式应该可以使用。
Here's an example with your data: 这是您的数据的示例:
$date_string = "2014-06-24T22:37:13.151Z";
$date_object = DateTime::createFromFormat('Y-m-d\TH:i:s.u\Z', $date_string);
echo $date_object->format('Y-m-d H:i:s');
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