如何使 php 正则表达式 preg_replace 工作

Question

我必须使用 转换字符串preg_replace。下面是我需要转换的字符串

@[ Test Career 12](career:235)@[ Testing11](business:2)@[ Username](user:1)some text

我已经创建了代码来替换内容，但它不起作用。请检查下面的代码，

$Rtm = '@[ Test Career 12](career:235)@[ Testing11](business:2)@[ Username](user:1)some text';

if (preg_match("/@\[(.*?)\]\(user:(.*?)\)/", $Rtm, $match)) {

    $Rtm0 = preg_replace("/@\[(.*?)\]\(user:(.*?)\)/", '<a href="/en/main/profile_page_link/$2">$1</a>, ', $Rtm);

    $Rtm = rtrim($Rtm0, ', ');

}

if (preg_match("/@\[(.*?)\]\(business:(.*?)\)/", $Rtm, $match)) {

    $slug = "1";

    $Rtm01 = preg_replace("/@\[(.*?)\]\(business:(.*?)\)/", '<a href="/en/business/' . $slug . '/about">$1</a>, ', $Rtm);

    $Rtm = rtrim($Rtm01, ', ');

}

if (preg_match("/@\[(.*?)\]\(career:(.*?)\)/", $Rtm, $match)) {

    $slug = "2";

    $Rtm02 = preg_replace("/@\[(.*?)\]\(career:(.*?)\)/", '<a href="/en/main/' . $slug . '/about">$1</a>, ', $Rtm);

    $Rtm = rtrim($Rtm02, ', ');

}

echo $Rtm;

上述代码的输出是，

<a href="/en/main/profile_page_link/1"> Test Career 12](career:235)</a><a href="/en/business/1/about"> Testing11</a>, @[ Username, some text

但我需要的输出是，

<a href="/en/main/2/about"> Test Career 12</a>, <a href="/en/business/1/about"> Testing11</a>, <a href="/en/main/profile_page_link/1"> Username</a> some text

给定的字符串只是一个演示，顺序可能会改变，因为它是动态的。但结构是一样的。

如何获得所需的输出。我的编码有问题吗。

Answer 1

为此，我将preg_replace_callback捕获 中的文本[]、slug 类型（user、career或business）和值（对于 slug user）分组，并将它们传递到回调以形成 URL：$Rtm = '@[ Test Career 12](career:235)@[ Testing11](business:2)@[ Username](user:1)some text';$Rtm = preg_replace_callback('/@\[([^]]*)\]\(([a-z]+):([^)]*)\)/', function ($match) {    switch($match[2]) {        case 'user':            return "<a href=\"/en/main/profile_page_link/$match[3]\">$match[1]</a>";            break;        case 'business':            return "<a href=\"/en/business/1/about\">$match[1]</a>";            break;        case 'career':            return "<a href=\"/en/main/2/about\">$match[1]</a>";            break;        default:            return "";            break;    }}, $Rtm);echo $Rtm;输出（对于您的输入字符串）：<a href="/en/main/2/about"> Test Career 12</a><a href="/en/business/1/about"> Testing11</a><a href="/en/main/profile_page_link/1"> Username</a>some text3v4l.org 上的演示

Answer 2

使用@\[([^][]*?)\]而不是@\[(.*?)\]&nbsp;演示和解释。rtrim所有更换后仅执行一次。$Rtm = '@[ Test Career 12](career:235)@[ Testing11](business:2)@[ Username](user:1)some text';if (preg_match("/@\[([^][]*?)\]\(user:(.*?)\)/", $Rtm, $match)) {&nbsp; &nbsp; $Rtm = preg_replace("/@\[([^][]*?)\]\(user:(.*?)\)/", '<a href="/en/main/profile_page_link/$2">$1</a>, ', $Rtm);}if (preg_match("/@\[([^][]*?)\]\(business:(.*?)\)/", $Rtm, $match)) {&nbsp; &nbsp; $Rtm = preg_replace("/@\[([^][]*?)\]\(business:(.*?)\)/", '<a href="/en/business/$2/about">$1</a>, ', $Rtm);}if (preg_match("/@\[([^][]*?)\]\(career:(.*?)\)/", $Rtm, $match)) {&nbsp; &nbsp; $Rtm = preg_replace("/@\[([^][]*?)\]\(career:(.*?)\)/", '<a href="/en/main/$2/about">$1</a>, ', $Rtm);}$Rtm = rtrim($Rtm, ', ');echo $Rtm;