PHP - 使用区域设置字符串解析日期时间

发布于 2024-11-28 02:12:04 字数 454 浏览 1 评论 0原文

我想解析诸如“Ayer，16:08”之类的日期时间，即西班牙语中的“昨天，16:08”。

我已经尝试过这个

$dateString = 'Ayer, 16:08';
setlocale(LC_ALL, 'es');
$time = strtotime($dateString);
echo date('d-m-Y H:i', $time);

，但它回显

01-01-1970 00:00

不过，如果我用英语字符串来做它，它工作得很好：

$dateString = 'Yesterday, 16:08';
$time = strtotime($dateString);
echo date('d-m-Y H:i', $time);

这是区域设置的问题吗？

谢谢

原文

I want to parse datetimes like 'Ayer, 16:08' which is 'Yesterday, 16:08' in spanish.

I have tried this

$dateString = 'Ayer, 16:08';
setlocale(LC_ALL, 'es');
$time = strtotime($dateString);
echo date('d-m-Y H:i', $time);

but it echoes

01-01-1970 00:00

Nevertheless, if I do it with english strings it works just fine:

$dateString = 'Yesterday, 16:08';
$time = strtotime($dateString);
echo date('d-m-Y H:i', $time);

Is it a problem with locale?

Thanks

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允世 2024-12-05 02:12:05

如今，有 IntlDateFormatter 用于此目的，请参阅此 stackoverflow 答案：https://stackoverflow.com/a/32265594/682317

复制到这里：

这就是答案：
$formatter = new IntlDateFormatter("en_US", IntlDateFormatter::SHORT, IntlDateFormatter::NONE);
$unixtime=$formatter->parse($date);
这是之前使用我的答案进行的测试。
<前><代码>\r\n";
$日期=“01/02/2015”； //1月2日
$formatter = new IntlDateFormatter("en_US", IntlDateFormatter::SHORT, IntlDateFormatter::NONE);
$unixtime=$formatter->parse($date);
$datetime=new DateTime();
$datetime->setTimestamp($unixtime);
echo $datetime->format('Ymd');
echo "
\r\n";
echo "IT 区域设置
\r\n";
$日期=“01/02/2015”； //2月1日
$formatter = new IntlDateFormatter("it_IT", IntlDateFormatter::SHORT, IntlDateFormatter::NONE);
$unixtime=$formatter->parse($date);
$datetime=new DateTime();
$datetime->setTimestamp($unixtime);
echo $datetime->format('Ymd');
echo "
\r\n";
不幸的是，我无法获得赏金...:-)

These days there is IntlDateFormatter for this purpose, see this stackoverflow answer: https://stackoverflow.com/a/32265594/682317

Copied here:

This is the answer:

$formatter = new IntlDateFormatter("en_US", IntlDateFormatter::SHORT, IntlDateFormatter::NONE);
$unixtime=$formatter->parse($date);

And this is the previous test working with my answer.

<?php
echo "EN locale<br>\r\n";
$date="01/02/2015"; //2th Jan
$formatter = new IntlDateFormatter("en_US", IntlDateFormatter::SHORT, IntlDateFormatter::NONE);
$unixtime=$formatter->parse($date);
$datetime=new DateTime();
$datetime->setTimestamp($unixtime);
echo $datetime->format('Y-m-d');
echo "<br>\r\n";

echo "IT locale<br>\r\n";
$date="01/02/2015"; //1th Feb
$formatter = new IntlDateFormatter("it_IT", IntlDateFormatter::SHORT, IntlDateFormatter::NONE);
$unixtime=$formatter->parse($date);
$datetime=new DateTime();
$datetime->setTimestamp($unixtime);
echo $datetime->format('Y-m-d');
echo "<br>\r\n";

Unfortunately I cannot earn my bounty... :-)

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方圜几里 2024-12-05 02:12:04

在约会之前你需要将其翻译成英语。

创建一个包含西班牙语单词的数组，另一个包含 PHP 识别的相应英语翻译的数组。然后只需使用 $dateString 运行 str_ireplace() 即可。

像这样的东西应该有效：

$spanish = array("spanish1", "spanish2", "spanish3");
$english = array("en_translation_of_spanish1", "en_translation_spanish2", "en_translation_of_spanish3");
$dateString = str_ireplace($spanish, $english, 'Ayer, 16:08');

You'll need to translate it into English before making the date.

Create an array with the Spanish words, and another with the corresponding English translations, as recognised by PHP. Then simply run str_ireplace() with $dateString.

Something like this should work:

$spanish = array("spanish1", "spanish2", "spanish3");
$english = array("en_translation_of_spanish1", "en_translation_spanish2", "en_translation_of_spanish3");
$dateString = str_ireplace($spanish, $english, 'Ayer, 16:08');

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