在 PHP 中将特定字符串转换为时间
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Convert specific string to time in PHP
提问by Dênis Montone
I need to convert a string into date format, but it's returning a weird error. The string is this:
我需要将字符串转换为日期格式，但它返回一个奇怪的错误。字符串是这样的：
21 nov 2012
I used:
我用了：
$time = strtotime('d M Y', $string);
PHP returned the error:
PHP 返回错误：
Notice: A non well formed numeric value encountered in index.php on line 11
What am I missing here?
我在这里缺少什么？
回答by mpen
You're calling the function completely wrong. Just pass it
您完全错误地调用了该函数。只要通过它
$time = strtotime('21 nov 2012')
The 2nd argument is for passing in a timestamp that the new time is relative to. It defaults to time().
第二个参数用于传递新时间相对于的时间戳。它默认为time().
Edit:That will return a unix timestamp. If you want to then format it, pass your new timestamp to the datefunction.
编辑：这将返回一个 Unix 时间戳。如果您想格式化它，请将您的新时间戳传递给该date函数。
回答by Rubin Porwal
To convert a date string to a different format:
要将日期字符串转换为不同的格式：
<?php echo date('d M Y', strtotime($string));?>
strtotimeparses a string returns the UNIX timestamp represented. dateconverts a UNIX timestamp (or the current system time, if no timestamp is provided) into the specified format. So, to reformat a date string you need to pass it through strtotimeand then pass the returned UNIX timestamp as the second argument for the datefunction. The first argument to dateis a template for the format you want.
strtotime解析字符串返回表示的 UNIX 时间戳。 date将 UNIX 时间戳（或当前系统时间，如果未提供时间戳）转换为指定格式。因此，要重新格式化日期字符串，您需要传递它strtotime，然后将返回的 UNIX 时间戳作为date函数的第二个参数传递。的第一个参数date是您想要的格式的模板。
Click herefor more details about date format options.
单击此处了解有关日期格式选项的更多详细信息。
回答by Joakim Ling
For more complex string, use:
对于更复杂的字符串，请使用：
$datetime = DateTime::createFromFormat("d M Y H:i:s", $your_string_here);
$timestamp = $datetime->getTimestamp();
回答by Naftali aka Neal
You are using the wrong function, strtotimeonly return the amount of seconds since epoch, it does not format the date.
您使用了错误的函数，strtotime只返回自纪元以来的秒数，它没有格式化日期。
Try doing:
尝试做：
$time = date('d M Y', strtotime($string));
