PHP发布方法无法从URL获取数据

本文介绍了PHP发布方法无法从URL获取数据的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个URL，该URL将用于粘贴并在浏览器中按Enter后获取帖子数据.我的链接是: vtrails.us/mixtape-builder/?song_urls=vtrails.us/wp-content/uploads/2015/12/mpthreetest.mp3&song_artist=ganja

I have an URL which is to be used for getting post data after pasting and pressing enter in the browser. My link is : vtrails.us/mixtape-builder/?song_urls=vtrails.us/wp-content/uploads/2015/12/mpthreetest.mp3&song_artist=ganja

<?php if(isset($_POST['song_urls']) && !empty($_POST['song_urls'])){ $song_url = $_POST['song_urls']; $song_artist = $_POST['song_artist']; } echo $song_url; echo $song_artist; ?>

但是我什么都没收到.那我现在该怎么办?

But i am not getting any of them.So what can i do now?

推荐答案

如果使用 $ _ GET 而不是 $ _ POST ，您将获得所需的结果:

If you use $_GET instead of $_POST you will get desired result:

这是代码

<?php if(isset($_POST['song_urls']) && !empty($_POST['song_urls'])){ $song_url = $_POST['song_urls']; $song_artist = $_POST['song_artist']; } echo $song_url; echo $song_artist; ?>