我必須使用 轉(zhuǎn)換字符串preg_replace。下面是我需要轉(zhuǎn)換的字符串@[ Test Career 12](career:235)@[ Testing11](business:2)@[ Username](user:1)some text我已經(jīng)創(chuàng)建了代碼來替換內(nèi)容，但它不起作用。請(qǐng)檢查下面的代碼，$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;上述代碼的輸出是，<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但我需要的輸出是，<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給定的字符串只是一個(gè)演示，順序可能會(huì)改變，因?yàn)樗莿?dòng)態(tài)的。但結(jié)構(gòu)是一樣的。如何獲得所需的輸出。我的編碼有問題嗎。