PHP编程技巧:如何实现3秒内跳转网页

发布日期:2025-11-28 19:50:27 阅读:0

<?php$targetUrl = "https://www.example.com/target-page";$delay = 3;echo &#34;页面将在 $delay 秒后跳转...";echo " let seconds = $delay; const countdown = setInterval(() => { seconds--; document.getElementById('countdown').textContent...

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