在 PHP 中获取当天的日期
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Getting the date for current day in PHP
提问by anurag-jain
I want to get the date for current day in php. what i tried is here...
我想在 php 中获取当天的日期。我试过的就在这里...
echo $x."<br>";
echo date("D",$x)."<br>";
But the output was
但输出是
21-02-10
Thu
It is giving correct date but not the correct day value.Why..?
它给出了正确的日期，但不是正确的日期值。为什么......？
What I want day is the date for monday for the current week which can be generated on any day of the week. so what I did was, I'm taking the today's day and comparing with (Mon,Tue.... Sun) and respectively creating a timestamp using
我想要的日期是本周的星期一日期，可以在一周中的任何一天生成。所以我所做的是，我将今天的一天与 (Mon,Tue .... Sun) 进行比较，并分别使用
case "Mon":
$startdate1=date("d-m-y");
$parts = explode('-',$startdate1);
$startdate2 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+1),$parts[2]));
$startdate3 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+2),$parts[2]));
$startdate4 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+3),$parts[2]));
$startdate5 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+4),$parts[2]));
$startdate6 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+5),$parts[2]));
$startdate7 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+6),$parts[2]));
$dates=array(1 => $startdate1,$startdate2,$startdate3,$startdate4,$startdate5,$startdate6,$startdate7);
$i=1;
while( $i <= 7 )
{
echo $dates[$i];
$i++;
}
break;
$date is the final array respective to today that has to be returned. Is there any other better method to do this operation.
$date 是必须返回的与今天相关的最终数组。有没有其他更好的方法来做这个操作。
采纳答案by anurag-jain
What i did to resolve it is used the date format ('d-m-Y') instead of ('d-m-y') in date function, which was causing the problem. Hence strtotime accepted the format and gave the correct result for
我为解决它所做的是在日期函数中使用日期格式 ('dm-Y') 而不是 ('dm-y')，这导致了问题。因此 strtotime 接受了格式并给出了正确的结果
$t=date('d-m-Y');
echo date("D",strtotime($t));
回答by chanchal
I tried this to get current day.
我试过这个来获得当前日期。
echo date('l'); // output: current day.
回答by David Snabel-Caunt
How about this:
这个怎么样：
//today is monday
if (1 == date('N')){
$monday = time();
}else{
$monday = strtotime('last Monday');
}
for ($i = 0; $i < 7; $i++){
echo date('d-m-Y', $monday) . '<br>';
$monday = strtotime('tomorrow', $monday);
}
First find Monday, if it is not today, then print 7 dates
先找星期一，如果不是今天，则打印7个日期
回答by codeholic
What I want day is the date for monday for the current week which can be generated on any day of the week.
我想要的日期是本周的星期一日期，可以在一周中的任何一天生成。
That's what you want. $mdayis the month day of this week's Monday. Nevermind if it's not positive, mktimewill handle that right. $mondayhas the timestamp of the Monday's midnight.
那就是你想要的。$mday是本周星期一的月日。没关系，如果它不是积极的，mktime会处理好。$monday具有星期一午夜的时间戳。
$now = getdate();
$mday = $now['mday'] - ($now['wday'] + 6) % 7;
$monday = mktime(0, 0, 0, $now['mon'], $mday, $now['year']);
echo(date('d-m-y', $monday));
回答by Mustafa ELnagar
I use the function date and path to it the "D" that refere to the current day , and it works with me
我使用函数日期和指向它的路径“D”，它指的是当天，它与我一起工作
$today = date("D");
and to get the full info about the current date
并获取有关当前日期的完整信息
$today = date("D M j G:i:s T Y"); // Sat Mar 10 17:16:18 MST 2001
回答by poke
what i tried is here...
echo date("D",$x)."<br>";
我试过的就在这里...
echo date("D",$x)."<br>";
dateexpects a timestamp (int) value as the second parameter. Your $xis a string containing an ambiguous date format. Convert that date into a timestamp first, using strptimeor strtotimeand use the datefunction correctly to get the correct day value.
date期望时间戳 (int) 值作为第二个参数。您$x是一个包含不明确日期格式的字符串。首先将该日期转换为时间戳，使用strptimeorstrtotime并date正确使用该函数以获得正确的日期值。
Regarding your second part, you don't need to (and shouldn't) check the day name to calculate the correct Monday, Tuesday etc. A more efficient approach is for exampleusing strtotimeto get last Mondayetc.
关于您的第二部分，您不需要（也不应该）检查日期名称来计算正确的星期一、星期二等。更有效的方法是例如使用strtotimegetlast Monday等。
回答by Gordon
You are likely passing a string as timestamp
您可能将字符串作为时间戳传递
echo $x."<br>";
echo date("D",$x)."<br>";
Remove $xand it will output the correct day or change it to
删除$x，它将输出正确的日期或将其更改为
$x = '21-02-2010';
echo date('D', strtotime($x));
