本文介绍了表单提交结果后用php代码打开新窗口的处理方法，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
这是我一直在编写的脚本，它应该集成用户并在打开时通过
here is the script that i been working on , it is supposed to integrate user and pass when opened
<?php
$name = $_POST['name']; // contain name of person
$pass = $_POST['pass']; // Email address of sender
$link = window.open(https://secure.brosix.com/webclient/?nid=4444&user=$name&pass=$pass&hideparams=1 'width=710,height=555,left=160,top=170');
echo $link;
?>
我这样做对吗，我想在用户将表单提交到 php 代码后打开一个弹出窗口，但我总是收到错误.
am i doing this right, i want to open a pop up windows after the user submits the form to the php code but i always get an error.
推荐答案
把你的代码改成这个
<?php
$name = $_POST['name']; // contain name of person
$pass = $_POST['pass']; // Email address of sender
$link = "<script>window.open('https://secure.brosix.com/webclient/? nid=4510&user=$name&pass=$pass&hideparams=1', 'width=710,height=555,left=160,top=170')</script>";
echo $link;
?>
补充说明
您应该考虑使用 fancybox，它可以使用 iframe 在弹出窗口中加载整个网页.还有其他选择，请随意探索！
You should consider using fancybox which can load webpages as a whole in a popup window using iframes. There are other options as well feel free to explore!
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