如何根据要求打开多个浏览器窗口? (PHP) [英] How to open multiple browser windows on request? (PHP)
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问题描述
所以我在 PHP / HTML
页面上有一个表单。用户将其提交给相同的 PHP / HTML
页面。所以现在PHP页面将有 $ _ POST
数据。我想要在页面刷新的时候弹出一些popUp浏览器窗口,其中的URL将与用户相关 POST
请求。如 www.example.com/bal-bla-bla.php? id = $ _ POST ['StreamId']
解决方案
if(isset($ _ POST ['submit'])){
echo'< script> window.open(''。$ _ SERVER ['PHP_SELF']。'myplayer.php?stream_id = 。$ _ POST [ '流ID'] ', myplayer);< /脚本>';
$ b $ / code>
编辑:
您可以在打开窗口之前始终显示一条消息,建议用户接受这个新窗口!
var flag = confirm(这个窗口不是ADV!;-));
if(flag)
window.open('。$ _ SERVER ['PHP_SELF']。',myplayer);
So I have a form on PHP/HTML
page. User submitss it to that same PHP/HTML
page. So now PHP page I will have $_POST
data. I want to when page is refreshed opnt some popUp browser windows which url's will be relative to users POST
request. like www.example.com/bal-bla-bla.php? id=$_POST['StreamId']
解决方案
if ( isset($_POST['submit']) ) {
echo '<script>window.open ("'.$_SERVER['PHP_SELF'].'myplayer.php?stream_id='.$_POST['StreamId'].'","myplayer");</script>';
}
edited:
you can always display a message before open the window that advice the user to accept this new window!
var flag = confirm(" This window is not an ADV! ;-) ");
if (flag)
window.open("'.$_SERVER['PHP_SELF'].'","myplayer");
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