在 PHP 中格式化日期字符串
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/3654101/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Format a date string in PHP
提问by cannyboy
If I have a string which represents a date, like "2011/07/01" (which is 1st July 2011) , how would I output that in more readable forms, like:
如果我有一个表示日期的字符串，例如 "2011/07/01" (即 2011 年 7 月 1 日)，我将如何以更易读的形式输出它，例如：
1 July 2011
1 Jul 2011 (month as three letters)
And also, how could I make it intelligently show date ranges like "2011/07/01" to "2011/07/11" as
而且，我怎样才能让它智能地将日期范围显示为“2011/07/01”到“2011/07/11”
1 - 11 July 2001
(without repeating the 'July' and '2011' in this case)
（在这种情况下不重复“七月”和“2011年”）
回答by NullUserException
You can convert your date to a timestamp using strtotime()and then use date()on that timestamp. On your example:
您可以使用将日期转换为时间戳strtotime()，然后date()在该时间戳上使用。在你的例子中：
$date = date("j F Y", strtotime("2011/07/01")); // 1 July 2011
$date = date("j M Y", strtotime("2011/07/01")); // 1 Jul 2011
回答by Richard Fearn
As NullUserException mentioned, you can use strtotimeto convert the date strings to timestamps. You can output 'intelligent' ranges by using a different date format for the first date, determined by comparing the years, months and days:
正如 NullUserException 提到的，您可以使用strtotime将日期字符串转换为时间戳。您可以通过对第一个日期使用不同的日期格式来输出“智能”范围，通过比较年、月和日来确定：
$date1 = "2011/07/01";
$date2 = "2011/07/11";
$t1 = strtotime($date1);
$t2 = strtotime($date2);
// get date and time information from timestamps
$d1 = getdate($t1);
$d2 = getdate($t2);
// three possible formats for the first date
$long = "j F Y";
$medium = "j F";
$short = "j";
// decide which format to use
if ($d1["year"] != $d2["year"]) {
$first_format = $long;
} elseif ($d1["mon"] != $d2["mon"]) {
$first_format = $medium;
} else {
$first_format = $short;
}
printf("%s - %s\n", date($first_format, $t1), date($long, $t2));
回答by Ciges
I would use strtotimeANDstrftime. Is a much simpler way of doing it.
我会使用strtotimeANDstrftime。是一种更简单的方法。
By example, if a have a date string like "Oct 20 18:29:50 2001 GMT" and I want to get it in format day/month/year I could do:
例如，如果有一个像“Oct 20 18:29:50 2001 GMT”这样的日期字符串，我想以日/月/年的格式获取它，我可以这样做：
$mystring = "Oct 20 18:29:50 2001 GMT";
printf("Original string: %s\n", $mystring);
$newstring = strftime("%d/%m/%Y", strtotime($mystring));
printf("Data in format day/month/year is: %s\n", $newstring);
回答by Robus
As for the second one:
至于第二个：
$time1 = time();
$time2 = $time1 + 345600; // 4 days
if( date("j",$time1) != date("j",$time2) && date("FY",$time1) == date("FY",$time2) ){
echo date("j",$time1)." - ".date("j F Y",$time2);
}
Just make up more conditions
多补些条件
