Question

我有这个代码。我想在字符串中提取2014年1月1日，但我一直在变为空

$paragraph = "today is 2014 January 1";

preg_match_all('/(\d{4}/) \b(?:Jan(?:uary)?|Feb(?:ruary)?|Mar(?:ch)?|Apr(?:il)?|May?|Jun(?:e)?|Jul(?:y)?|Aug(?:ust)?|Sept(?:ember)?|Oct(?:ober)?|Nov(?:ember)?|Dec(?:ember)?) (\d|\d{2}) ', $paragraph, $date);

var_dump($date);

但var_dump（$ date）返回null

Answer 1

尝试以下操作，您只需对正则表达式进行一些调整。

$paragraph = "today is 2014 January 1, and tomorrow will be 2015 March 12";

$date = Array();

// Find all dates
preg_match_all('@((?:\d{4}) (?:Jan(?:uary)?|Feb(?:ruary)?|Mar(?:ch)?|Apr(?:il)?|May?|Jun(?:e)?|Jul(?:y)?|Aug(?:ust)?|Sept(?:ember)?|Oct(?:ober)?|Nov(?:ember)?|Dec(?:ember)) (?:\d{1,2}))@', $paragraph, $date);

// Display results
echo '<pre>';
print_r($date[0]);
echo '</pre>';

// Iterate over the results
$total = count($date[0]);
for($i=0;$i<$total;$i++){
    // Use the following if you want to change the display format
    date_create_from_format('Y-M-d', $date[0][$i]);
    echo $date[0][$i].'<br />';
}

Answer 2

您可以使用DateTime::createFromFormat()作为示例，但是您需要转义每个字符，因为今天文本不代表日期格式。

<?php
$paragraph = "today is 2014 January 1";
$date = DateTime::createFromFormat('\t\o\d\a\y \i\s Y F j', $paragraph);
echo $date->format('Y F j'); //"prints" 2014 January 1

如何使用格式2014年1月1日提取日期？ PHP

2 个答案: