php 从日期/时间字符串中获取时间
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Get time from a date/time string
提问by Ross
I have a date value stored in a variable. I need to extract the time part of the value in to a separate variable and then add/subtract time from it.
我有一个存储在变量中的日期值。我需要将值的时间部分提取到一个单独的变量中，然后从中添加/减去时间。
The date variable is set with date('YmdHis'), giving (for example) 20110805124000 for August 5th 2011, 12:40:00
日期变量设置为 date('YmdHis')，给出（例如）2011 年 8 月 5 日 12:40:00 的 20110805124000
From the value 20110805124000 (which is stored in the variable $fulltime), I need to store the time only in the format 12:40 (so ignoring the year, month, day and seconds and adding the colon between the hour and minute) in a variable called $shorttime. I then need to add a number of hours to that time (so for example +3 hours would change the value in the $shorttime variable to 15:40). The number of hours I need to add is stored in a variable called $addtime, and this value could be a negative number.
从值 20110805124000（存储在变量 $fulltime 中），我只需要以 12:40 格式存储时间（因此忽略年、月、日和秒，并在小时和分钟之间添加冒号）一个名为 $shorttime 的变量。然后，我需要为该时间添加几个小时（例如，+3 小时会将 $shorttime 变量中的值更改为 15:40）。我需要添加的小时数存储在一个名为 $addtime 的变量中，该值可能是一个负数。
Is this easily doable? Could anyone help?
这容易实现吗？有人可以帮忙吗？
Thanks :)
谢谢 ：）
回答by Bindiya Patoliya
$time = '2013-01-22 10:45:45';
echo $time = date("H:i:s",strtotime($time));
It will give the time 10:45:45from datetime.
它将给出10:45:45日期时间的时间。
回答by Saad Imran.
<?PHP
$addhours = 3;
$date = DateTime::createFromFormat('YmdHis', '20110805124000');
$shorttime = $date->format("H:i");
$newdate = $date->add(DateInterval::createFromDateString($addhours . "hours"));
$newtime = $newdate->format("H:i");
echo $shorttime . "<br />";
echo $newtime . "<br />";
?>
for your reference:
供你参考：
http://www.php.net/manual/en/datetime.createfromformat.php
http://www.php.net/manual/en/datetime.createfromformat.php
http://www.php.net/manual/en/dateinterval.createfromdatestring.php
http://www.php.net/manual/en/dateinterval.createfromdatestring.php
