我有我的网页的链接/按钮应该引导用户到一个CSV文件。 CSV文件是动态生成使用PHP，基于一些POST变量：

I have a link/button on my page that should lead the user to a csv file. The csv file is generated dynamically using PHP, based on some POST variables:

<?php
    header("Content-type: application/csv");
    header("Content-Disposition: attachment; filename=e".date('ymdHis').".csv");
    ...
?>

我

如何可以重定向到脚本，同时通过POST变量，而无需使用一种形式？

How can I redirect to the script, at the same time passing POST variables, without using a form?

解答我发现，我真的不喜欢：

Answers I found that I didn't really like:

推荐答案

您可以使用GET方法 - 用户将看不到GET字符串（丑陋的URL），它不会显示在新窗口。但是，如果你真的需要POST（虽然我不能想象为什么），你应该隐藏字段创建表单 - 这就是做你的任务的唯一方法（是的，我以前说这句话的测试方法很多）

You may use GET method - user won't see GET string (ugly URL), it won't be displayed in the new window. But if you really need POST (though i cannot even imagine why), you should create form with hidden fields - this is the only way to do your task (yes, i've tested many ways before saying this).