表单提交结果后，用php代码打开新窗口 [英] open new window with php code after form submits results

问题描述

这里是我一直在处理的脚本，它应该在打开时集成用户和传递

 < php 
 
 $ name = $ _POST ['name']; //包含人的名字
 $ pass = $ _POST ['pass']; //发件人的电子邮件地址
 $ link = window.open（https://secure.brosix.com/webclient/?nid=4444&user=$name&pass=$pass&hideparams=1'width = 710，高度= 555，左= 160，顶部= 170' ）; 
 
 echo $ link; 
 
？> 
  

我是这样做的，我想在用户提交表单之后打开一个弹出窗口PHP代码，但我总是得到一个错误。

解决方案

更改您的代码到这个

 <？php 
 
 $ name = $ _POST ['name']; //包含人的名字
 $ pass = $ _POST ['pass']; //发件人的电子邮件地址
 $ 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; 
 
？> 
  

其他注意

您应该考虑使用 fancybox ，它可以在使用iframe的弹出窗口中加载整个网页。还有其他的选择，以及随时探索！

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;

?>

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.

解决方案

Change your code to this

<?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;

?>

Additional Note

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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