PHP正则表达式用于特定的文本模式

Question

在我的网站上，我插入了当年在主体中创建一个项目的年份，并将其替换为“六年前”（或时间长短）。

所以在我的内容中，我有：

我们从1998年开始从事业务，并于[2011年]之前制作了这种包装设计。

我正在尝试使用正则表达式将2011年放入变量中，以便稍后进行搜索和替换，但无法弄清楚。 每页只有一个实例。 我可以进行搜索和替换，这只是正则表达式，我从来没有动过头。

为了解决以下问题-年份是可变的，这就是为什么我要使用正则表达式。

例

$bodycontent = <p>We've been in business since 1998 
and produced this logo design [2002] years ago.</p>

要么

$bodycontent = <p>We've been in business since 1998 
and produced this website design [2016] years ago.</p>

因此，我将花括号中的年份放入正则表达式为$ then的变量中，然后从当前年份中减去该年份以得出$ age（由另一个函数转换为单词）

$bodycontent = str_replace("[".$then."]",$age,$bodycontent)

我试过了

preg_match("[\d{4}]",$bodycontent,$found); 

但它会返回第一个日期，而不是大括号中的那个。

Answer 1

使用函数preg_replace_callback() ：

$bodycontent = preg_replace_callback('~\[(\d{4})\]~', function($match) {
    return (date('Y') - $match[1]) . " ago";
}, $bodycontent);

演示

Answer 2

如果这是我的项目，那么我可能会利用一个查找字词数组进行preg_replace_callback()调用，并在括号中的日期“超出范围”时进行后备替换。

*请务必在适当的时候将year复数。
*我模式中的第二个]不需要转义，但是如果您认为它可以提高可读性，则可以添加它。
*我也在years ago匹配，因此替换文字在所有情况下都有意义； 您可能希望在原始输入文本中删除此尾随文本。

代码：（ 演示 ）

$bodycontent = "<p>We've been in business since 1998 and produced this logo design [1995] years ago.</p>\n";
$bodycontent .= "<p>We've been in business since 1998 and produced this website design [2018] years ago.</p>\n";
$bodycontent .= "<p>We've been in business since 1998 and produced this website design [2017] years ago.</p>\n";
$bodycontent .= "<p>We've been in business since 1998 and produced this website design [2016] years ago.</p>";

$lookup=['less than a','one','two','three','four','five','six','seven','eight','nine',
         'ten','eleven','twelve','thirteen','fourteen','fifteen','sixteen','seventeen','eighteen','nineteen'
];  // extend as needed

$bodycontent = preg_replace_callback('/\[(\d{4})] years ago/', function($m)use($lookup){
    $diff=date('Y')-$m[1];
    if(isset($lookup[$diff])){
        return $lookup[$diff].' year'.($diff>1?'s':'').' ago';
    }else{
        return "in {$m[1]}";  // out of lookup range, use backup text
    }
}, $bodycontent);

echo $bodycontent;

输出：

<p>We've been in business since 1998 and produced this logo design in 1995.</p>
<p>We've been in business since 1998 and produced this website design less than a year ago.</p>
<p>We've been in business since 1998 and produced this website design one year ago.</p>
<p>We've been in business since 1998 and produced this website design two years ago.</p>

Answer 3

假设此格式为[2002] years ，则可以使用此正则表达式作为替代方法：

\\[(\\d{4})\\] years

说明

\[      # Match [
(       # Capturing group
  \d{4} # Match 4 digits
)       # Close capturing group
\]      # Match ]
 years  # Match `whitespace`years

然后，您可以使用preg_match来匹配组1中的年份，以年为单位计算差异，并执行单数或复数格式。

例如：

$bodycontent = "<p>We've been in business since 1998 and produced this logo design [2002] years ago.</p>";

preg_match('/\[(\d{4})\] years/', $bodycontent, $matches);
$years = date('Y') - $matches[1];
$result = sprintf("%s year%s",
    $years,
    $years === 1 ? "": "s"
    );
$bodycontent =  str_replace($matches[0], $result, $bodycontent);

演示php输出