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

open new window with php code after form submits results

这是我一直在编写的脚本，它应该在打开时集成用户和传递

<?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;
?>

我做得对吗？我想在用户将表单提交给php代码后打开一个弹出窗口，但我总是收到一个错误。

将代码更改为此

<?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;
?>

附加说明

您应该考虑使用fancybox，它可以使用iframes在弹出窗口中加载整个网页。还有其他选择，请随意探索！

您忘记在$link的值周围加引号和标记。

$link = "<script>window.open('"https://secure.brosix.com/webclient/?nid=4444&user=$name&pass=$pass&hideparams=1width=710,height=555,left=160,top=170''")</script>";

您不必使用php，只需创建具有特定id的提交按钮，然后告诉jquery在提交时触发新选项卡

<form id="itemreport_new" type="post" action="">
          <input id="submit2" type="submit" value="show"  target=_blank   />
</form>
    $(document).ready(function () {

        $('#submit2').click(function() {
                 $('#itemreport_new').attr('target','_blank');
        });
    });

<?php
echo "<h1>Hello, PHP!</h1>";
$name = $_POST['name']; // CONTAIN NAME OF PERSON
$pass = $_POST['pass']; // ANY DETAIL OF PERSON
$link = "<script>window.open('https://google.co.in')</script>";
echo $link;