PHP DateTime :: createFromFormat返回错误的日期 [英] PHP DateTime::createFromFormat returning the wrong date

问题描述

在尝试使用太平洋/奥克兰时区和格式字符串 F-Y运行 createFromFormat 时。返回的日期是10月1日，即使我提供了 September-2019。

When trying to run createFromFormat using the Pacific/Auckland timezone and the format string 'F-Y'. The date returned is the first of October even though I have supplied it with 'September-2019'.

我尝试在PHP 7.3.9和7.2.22上运行它

I have tried running it on PHP 7.3.9 and 7.2.22 in CLI and FPM, and online in a PHP sandbox.

<?php
echo DateTime::createFromFormat('F-Y', 'September-2019')                                                           
    ->setTimezone(new DateTimeZone('Pacific/Auckland'))
    ->format('Y-m-d');
// 2019-10-01

echo DateTime::createFromFormat('F-Y', 'September-2019')
    ->format('Y-m-d');
// 2019-09-01

在这两个示例中，返回日期都应具有已于2019-09-01。

In both of these examples the returned date should have been 2019-09-01. This wasn't happening yesterday.

推荐答案

之所以会出现这种情况，是因为您没有指定a的缺失部分输入到 DateTime :: createFromFormat ，它使用当前本地日期和时间的值。在奥克兰，这是10月31日，因此它尝试将日期定为2019年9月31日，即2019年10月1日。为避免此问题，请使用！在格式字符串的开头；取而代之的是将1970年1月1日00:00:00（Unix纪元）以来的值替换为未在时间值中指定的值：

The reason for this behaviour is that when you don't specify the missing parts of a date/time input to DateTime::createFromFormat, it uses the values from the current local date and time. In Auckland, that is October 31st and so it tries to make a date out of September 31 2019, which comes out as October 1 2019. To avoid this problem, use a ! at the start of the format string; this will instead substitute values from January 1 1970, 00:00:00 (the Unix Epoch) as required for those that are not specified in the time value:

echo DateTime::createFromFormat('!F-Y', 'September-2019')
    ->setTimeZone(new DateTimeZone('Pacific/Auckland'))
    ->format('Y-m-d');

输出：

2019-09-01

在3v4l.org上进行演示

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