php - 错误: The terms you entered did not bring any results with twitter share on mobile only

Question

我有一个简单的 Facebook、Twitter 和 Pinterest 分享按钮

<a href="<?php echo "$base_url/share.php"; ?>" class="btn btn-default btn-facebook btn-icon btn-block" target="popup" 
onclick="window.open('<?php echo "$base_url/share.php?id=$id&share=fb&type=article"; ?>','popup','width=600,height=600'); return false;"><i class="fab fa-facebook"></i> <span class="post-sharing__label d-none d-sm-inline-block">Share on Facebook</span></a>

<a href="<?php echo "$base_url/share.php"; ?>" class="btn btn-default btn-twitter btn-icon btn-block" target="popup" 
onclick="window.open('<?php echo "$base_url/share.php?id=$id&share=tw&type=article"; ?>','popup','width=600,height=600'); return false;"><i class="fab fa-twitter"></i> <span class="post-sharing__label d-none d-sm-inline-block">Share on Twitter</span></a>

<a href="<?php echo "$base_url/share.php"; ?>" class="btn btn-default btn-twitter btn-icon btn-block" target="popup" 
onclick="window.open('<?php echo "$base_url/share.php?id=$id&share=tw&type=article"; ?>','popup','width=600,height=600'); return false;"><i class="fab fa-twitter"></i> <span class="post-sharing__label d-none d-sm-inline-block">Share on Twitter</span></a>

当用户单击共享时，它会转到名为 share.php 的 php 文件；该文件获取有关将要共享的页面的信息。

if ($share =="tw"){     

header("Location: http://twitter.com/share?text=$seo_description&url=$share_url");
die();
}

在桌面上，它可以正常工作，但在移动设备上，我收到以下错误

The terms you entered did not bring any results, please try again later.

Facebook 和 Pinterest 在移动设备上运行，我正在苦苦思索为什么我会得到这个。

我在手机上登录的 Twitter 帐户与我在电脑上登录的帐户相同。

我从 this question 获取了我的网址信息

Answer 1

我通过更改链接在移动设备上实现了此功能

if ($share =="tw"){     

header("Location: http://twitter.com/share?text=$seo_description&url=$share_url");
die();
}

到此

if ($share =="tw"){     

header("Location: https://twitter.com/intent/tweet?text=$seo_description&url=$share_url");
die();
}