php 添加 30 分钟到日期
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adding 30 minutes to date
提问by RussellHarrower
So what I need to do is add 30 minutes to the following
所以我需要做的是在下面添加 30 分钟
date("Ymdhis");
I tried this
我试过这个
+strtotime("+30 minutes");
however it does not seem to like it. I wondering what the correct why to do this is.
然而它似乎并不喜欢它。我想知道为什么这样做是正确的。
回答by Josh
Your method of using strtotimeshould work.
您的使用方法strtotime应该有效。
<?php
echo date("Y/m/d H:i:s", strtotime("now")) . "\n";
echo date("Y/m/d H:i:s", strtotime("+30 minutes"));
?>
Output
输出
2012/03/22 10:55:45
2012/03/22 11:25:45 // 30 minutes later
However your method of adding time probably isn't correct. The above will work to add 30 minutes to the current time. Suppose you want to add 30 minutes from a given time, $t, then use strtotime's second parameter, which is used as a base for the calculation of relative dates.
但是，您添加时间的方法可能不正确。以上将工作增加 30 分钟到当前时间。假设您要从给定时间添加 30 分钟$t，然后使用strtotime的第二个参数，该参数用作计算相对日期的基础。
date("Y/m/d H:i:s", strtotime("+30 minutes", $t));
回答by SuperNoob
I tested this code but it doesn't work for me:
我测试了这段代码，但它对我不起作用：
$t = date();
date("Y/m/d h:i:s", strtotime("+30 minutes", $t));
Here's my solution
这是我的解决方案
//This is where you put the date, but I use the current date for this example
$date = date("Y-m-d H:i:s");
//Convert the variable date using strtotime and 30 minutes then format it again on the desired date format
$add_min = date("Y-m-d H:i:s", strtotime($date . "+30 minutes"));
echo $date . "<br />"; //current date or whatever date you want to put in here
echo $add_min; //add 30 minutes
回答by staticsan
strtotime()accepts a second parameter which is its starting point.
strtotime()接受作为起点的第二个参数。
If you have date("Ymdhis", $somedate)and wanted to add 30 minutes to it, you can do
date("Ymdhis", strtotime("+30 minutes", $someddate))
如果你有date("Ymdhis", $somedate)并想增加 30 分钟，你可以这样做
date("Ymdhis", strtotime("+30 minutes", $someddate))
回答by Adam
Try something like.
尝试类似的东西。
$Start = "12:00:00";
$Minutes = 30;
$To = date("H:i:s", strtotime($Start)+($Minutes*60));
回答by sikander
Use this function:
使用这个功能：
date("Ymdhis", strtotime("+30 minutes"))
回答by Jacob S
<?php
print date("Y-m-d h:i:s", (time() + (60*30)) );
?>
回答by matt
Not sure what your entire code looks like, but:
不确定你的整个代码是什么样的，但是：
date("Ymdhis");
is returning a string. So it doesn't make sense to add the result of
正在返回一个字符串。所以添加结果没有意义
strtotime("+30 minutes");
(which is an integer) to that string.
（这是一个整数）到那个字符串。
You either want
你要么想要
strtotime("+30 minutes");
by itself, or
本身，或
date("Ymdhis", strtotime("+30 minutes"));
to get the formatted string.
获取格式化的字符串。
回答by mvdnes
Do you mean date("Ymdhis", strtotime("+30 minutes"));? This will represent the date that is 30 minutes in the future.
你的意思是date("Ymdhis", strtotime("+30 minutes"));？这将代表未来 30 分钟的日期。
