问题描述
我正在从Excel工作表中读取数据并将其插入MySQL表中.在此过程中,我将使用PHP方法strtotime()
将时间戳(字符串)转换为日期时间.
I am reading the data from an excel sheet and inserting into MySQL table. In this process, I am converting the timestamp (string) to datetime using the PHP method strtotime()
.
$timestamp = date('Y-m-d H:i:sa',strtotime(trim($value->Time)));
在某些情况下,此操作将失败.
This is failing in some cases.
示例:
echo date('Y-m-d H:i:s',strtotime('04-13-2018 0:00:53'));
echo date('Y-m-d H:i:s',strtotime('04-12-2018 0:00:53'));
输出:
1970-01-01 00:00:00
2018-12-04 00:00:53
有人可以帮我解决这个问题吗?
Can anyone help me how can I solve this issue?
失败的示例字符串:
04-12-2018 0:00:53
04-12-2018 0:01:53
04-12-2018 0:02:53
04-12-2018 0:03:53
04-12-2018 0:04:53
04-12-2018 0:05:53
04-12-2018 0:06:53
04-12-2018 0:07:53
04-12-2018 0:08:54
04-12-2018 0:09:54
04-12-2018 0:10:53
04-12-2018 0:11:53
04-12-2018 0:12:53
04-12-2018 0:13:53
04-12-2018 0:14:53
04-12-2018 0:15:53
04-12-2018 0:16:53
04-12-2018 0:17:53
04-12-2018 0:18:53
04-12-2018 0:19:54
04-12-2018 0:20:54
04-12-2018 0:21:54
04-12-2018 0:22:53
04-12-2018 0:23:54
04-12-2018 0:24:53
04-12-2018 0:25:54
推荐答案
您的格式不是解析器可以理解的格式.
在您的情况下,13
不是月".因此解析器到目前为止还不了解.
In your case 13
is not a "month". So the parser doesn't understand to date.
您应使用 DateTime::createFromFormat()
:
$date = DateTime::createFromFormat('m-d-Y H:i:s','04-13-2018 0:00:53');
echo $date->format('Y-m-d H:i:s');
输出:
2018-04-13 00:00:53
请注意,格式也可以是:'m-d-Y G:i:s'
加上G
表示一个小时的24小时格式,不带前导零" .
Note that the format could also be: 'm-d-Y G:i:s'
with G
for "24-hour format of an hour without leading zeros".
这篇关于PHP strtotime():转换后显示"1970-01-01"的日期的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!