使用 preg_replace 在 PHP 中识别和替换字符串中的短编码变量对

Question

在有人问之前，我已经花了一个小时试图找到关于 SO 的解决方案，但似乎没有什么能完全达到我想要的......如果它在那里，我道歉！

所以...我正在尝试在一堆文本上运行 preg_replace 来替换成对的短代码。

示例文本：

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]Abc123[/img][cap]This is an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]321cbA[/img][cap]This is also an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.

我想把每张图片和标题放在一起并替换它们：

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
<img src="Abc123.jpg" alt="This is an image" />
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
<img src="321cbA.jpg" alt="This is also an image" />
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.

我需要的代码如下：

$text="Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]Abc123[/img][cap]This is an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]321cbA[/img][cap]This is also an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.";

$pattern = '/\[(img)\](.*?)\[\/\1\]?/';
// This doesn't work: $pattern = '/\[(img)\](.*?)\[\/\1\]\[(cap)\](.*?)\[\/\4\]?/';

$replacement = '<img src="$2.jpg" alt="Help" />';
// "Help" will be $5 when we get both values

echo preg_replace($pattern, $replacement, $text);

此代码获取一个值，但我无法在我的一生中获得获取这两个值的模式（上面的注释行是我所拥有的）。我已经尝试了几十个 SO 页面和其他网站，但似乎没有任何效果。毫无疑问，修正后的模式会很简单，但如果有人能让我摆脱痛苦，我将不胜感激！

感谢您的宝贵时间。

Answer 1

我已经像这样修改了你的代码：

$text="Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]Abc123[/img][cap]This is an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]321cbA[/img][cap]This is also an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.";

$pattern = '/\[img\](\w+)\[\/img\]\[cap\]([\w\s]+)\[\/cap\]/';
// This doesn't work: $pattern = '/\[(img)\](.*?)\[\/\1\]\[(cap)\](.*?)\[\/\4\]?/';

$replacement = '<img src="$1.jpg" alt="$2" />';
// "Help" will be $5 when we get both values

echo preg_replace($pattern, $replacement, $text);

我只更改了模式和替换字符串。

Answer 2

$str = <<<'EOD'
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]Abc123[/img][cap]This is an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
[img]321cbA[/img][cap]This is also an image[/cap]
Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt.
EOD;

$trans = ['[img]' => '<img src="', '[/img][cap]' => '" alt="', '[/cap]' => '" />'];

echo strtr($str, $trans);

demo

您代码中的最后一个引用应该是\3 而不是\4（但您真的需要捕获标签名称吗？）。