Question

我是php新手并试图弄明白，如何在保持其他网页打开的同时关闭一个打开的网页。例如，我想打开yahoo.com网页，如下面的代码所示。然后我想关闭这个yahoo.com页面并打开google.com页面。

<?php

//To open yahoo.com page
echo '<script type="text/javascript" language="javascript"> 
window.open("http://www.yahoo.com/"); 
</script>'; 

//How can I close the yahoo.com page and let the program keep running
echo "<script>window.close();</script>";
sleep(10);

//To open google.com page
echo '<script type="text/javascript" language="javascript"> 
window.open("http://www.google.com/"); 
</script>'; 

echo "Done";

?>

Answer 1

这是一个JavaScript问题。您从PHP回显脚本标记的事实对答案没有影响。

window.open的返回值是您创建的窗口的句柄。如果存储它，您可以稍后关闭它：

var w = window.open("http://www.yahoo.com/");  // Opens window with yahoo.com
w.close();                                     // Closes window with yahoo.com

jsFiddle for the above code snippet

请注意：（a）许多浏览器会阻止弹出窗口，（b）它会在打开后立即关闭窗口。如果您想睡在两者之间，请使用JavaScript的setTimeout方法：

var w = window.open("http://www.yahoo.com/");  // Opens window with yahoo.com
setTimeout(function() { w.close() }, 1000);    // Closes window with yahoo.com after waiting 1000 milliseconds

jsFiddle for the above code snippet

Answer 2

简单的例子

<!DOCTYPE html>
    <html>
    <head>
    <script>
    function openWind()
    {
    myWindow=window.open("http://www.yahoo.com/","","width=1000,height=500");
    }

    function closeWind()
    {
    myWindow.close();
    }
    </script>
    </head>
    <body>

    <input type="button" value="Open 'myWindow'" onclick="openWind()" />
    <input type="button" value="Close 'myWindow'" onclick="closeWind()" />

    </body>
    </html>

如何使用PHP关闭特定网页？

2 个答案: