来自 strtotime 的 PHP 日期和当前时间
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PHP Date from strtotime with current time
提问by user979331
I have been looking online for this answer and have come up empty...I am extremely tired so I thought I would give this a go....
我一直在网上寻找这个答案，结果是空的......我非常累所以我想我会试一试......
I have a variable that has a date from a textbox
我有一个变量，它有来自文本框的日期
$effectiveDate=$_REQUEST['effectiveDate'];
What I am trying to do is take this date and add the current time
我想要做的是取这个日期并添加当前时间
date('Y-m-d H:i:s', strtotime($effectiveDate))
When I echo this out I get 1969-12-31 19:00:00
当我回应这一点时，我得到 1969-12-31 19:00:00
Is this possible? Can someone point me in the right direction?
这可能吗？有人可以指出我正确的方向吗？
回答by user979331
I found a solution to my problem....
我找到了解决我的问题的方法......
$currentDate = date("Y-m-d");
$currentTime = date("H:i:s");
$currentDate = date("Y-m-d H:i:s", strtotime($currentDate . $currentTime));
echo $currentDate;
This takes a date from variable in one format and takes the date from another variable in another format and puts them together :)
这从一种格式的变量中获取日期，并从另一种格式中的另一个变量中获取日期并将它们放在一起:)
Thanks everyone for their time.....
感谢大家的时间......
DateTime::createFromFormat
would also work but only if you have PHP 5.3 or higher...(I think)
也可以，但前提是你有 PHP 5.3 或更高版本......（我认为）
回答by David Barnes
The effectiveDate string is not in a format that strtotime recognizes, so strtotime returns false which is interpreted as 0 which causes the date to be displayed as January 1, 1970 at 00:00:00, minus your time zone offset.
EffectiveDate 字符串不是 strtotime 识别的格式，因此 strtotime 返回 false ，它被解释为 0，这导致日期显示为 1970 年 1 月 1 日 00:00:00，减去您的时区偏移。
回答by Niet the Dark Absol
The result you see is caused by the entered date not being in a format recognised by strtotime. The most likely case I can think of without knowing the format you used is that you used the US order of putting the month and day the wrong way around - this confuses strtotime, because if it accepts both then it can't distinguish February 3rd and March 2nd, so it has to reject US-formatted dates.
您看到的结果是由于输入的日期不是strtotime. 在不知道您使用的格式的情况下，我能想到的最可能的情况是您使用了美国的顺序，将月份和日期以错误的方式放置 - 这strtotime令人困惑，因为如果它同时接受，则无法区分 2 月 3 日和 3 月第二，所以它必须拒绝美国格式的日期。
The most reliable format for strtotimeis YYYY-MM-DD HH:ii:ss, as it is unambigous.
最可靠的格式strtotime是 YYYY-MM-DD HH:ii:ss，因为它是明确的。
回答by Marcos Fedato
The date is just a timestamp, it is not object-oriented and i don't like it.
日期只是一个时间戳，它不是面向对象的，我不喜欢它。
You can use the DateTime object.
您可以使用 DateTime 对象。
The object-oriented best way is:
面向对象的最佳方式是：
$effectiveDate=$_REQUEST['effectiveDate'];
// here you must pass the original format to pass your original string to a DateTimeObject
$dateTimeObject = DateTime::createFromFormat('Y-m-d H:i:s', $effectiveDate);
// here you must pass the desired format
echo $dateTimeObject->format('Y-m-d H:i:s');
