php 将日期转换为日期名称，例如 Mon、Tue、Wed
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Convert date to day name e.g. Mon, Tue, Wed
提问by user1444027
I have a variable that outputs the date in the following format:
我有一个以以下格式输出日期的变量：
201308131830
Which is 2013 - Aug - 13 - 18:30
这是 2013 年 - 八月 - 13 - 18:30
I am using this to retrieve the day name but it is getting the wrong days:
我正在使用它来检索日期名称，但它的日期错误：
$dayname = date('D', strtotime($longdate));
Any idea what I am doing wrong?
知道我做错了什么吗？
回答by John Conde
This is what happens when you store your dates and times in a non-standard format. Working with them become problematic.
当您以非标准格式存储日期和时间时，就会发生这种情况。与他们合作成为问题。
$datetime = DateTime::createFromFormat('YmdHi', '201308131830');
echo $datetime->format('D');
回答by Albert Naa
Your code works for me
你的代码对我有用
$date = '15-12-2016';
$nameOfDay = date('D', strtotime($date));
echo $nameOfDay;
Use l instead of D, if you prefer the full textual representation of the name
如果您更喜欢名称的全文表示，请使用 l 而不是 D
回答by Rayees Pk
echo date('D', strtotime($date));
echo date('l', strtotime($date));
Result
结果
Tue
Tuesday
回答by Sammitch
Your code works for me.
你的代码对我有用。
$input = 201308131830;
echo date("Y-M-d H:i:s",strtotime($input)) . "\n";
echo date("D", strtotime($input)) . "\n";
Output:
输出：
2013-Aug-13 18:30:00
Tue
Howeverif you pass 201308131830as a numberit is 50 to 100x larger than can be represented by a 32-bit integer. [dependent on your system's specific implementation] If your server/PHP version does not support 64-bit integers then the number will overflow and probably end up being output as a negative number and date()will default to Jan 1, 1970 00:00:00 GMT.
但是，如果您201308131830作为数字传递，它会比 32 位整数表示的大 50 到 100 倍。[取决于您系统的具体实现] 如果您的服务器/PHP 版本不支持 64 位整数，那么该数字将溢出并可能最终被输出为负数并date()默认为Jan 1, 1970 00:00:00 GMT.
Make sure whatever source you are retrieving this data from returns that date as a string, and keep it as a string.
确保您从中检索此数据的任何来源将该日期作为字符串返回，并将其保留为字符串。
回答by plaidcorp
Seems like you could use date() and the lowercase "L" format character in the following way:
似乎您可以通过以下方式使用 date() 和小写“L”格式字符：
$weekday_name = date("l", $timestamp);
Works well for me, here is the doc: http://php.net/manual/en/function.date.php
对我来说效果很好，这里是文档：http: //php.net/manual/en/function.date.php
回答by Alvaro
You can not use strtotimeas your time format is not within the supported date and time formats of PHP.
您不能使用，strtotime因为您的时间格式不在PHP 支持的日期和时间格式范围内。
Therefor, you have to create a valid date format first making use of createFromFormat function.
因此，您必须首先使用createFromFormat 函数创建有效的日期格式。
//creating a valid date format
$newDate = DateTime::createFromFormat('YmdHi', $longdate);
//formating the date as we want
$finalDate = $newDate->format('D');
回答by ARN
For other language use day of week to recognize day name
对于其他语言，请使用星期几来识别日期名称
For example for Persian use below code
例如波斯语使用下面的代码
$dayName = getDayName(date('w', strtotime('2019-11-14')));
function getDayName($dayOfWeek) {
switch ($dayOfWeek){
case 6:
return '????';
case 0:
return '?? ????';
case 1:
return '?? ????';
case 2:
return '?? ????';
case 3:
return '???? ????';
case 4:
return '??? ????';
case 5:
return '????';
default:
return '';
}
}
More info : https://www.php.net/manual/en/function.date.php
