从 JavaScript 弹出窗口传递 PHP 页面的值
我有一个用 JavaScript 编写的弹出按钮。当用户单击该按钮时,它会打开一个 PHP 弹出窗口。单击时,我想通过 GET 或 POST 方法将值传递到该 PHP 页面,这是我的弹出按钮代码:
<form input type="BUTTON" value="popup button" onClick="javascript:popup('actionpage.php')">
</form>
<SCRIPT LANGUAGE="JavaScript">
function popup(URL) {
day = new Date();
id = day.getTime();
eval("page" + id + " = window.open(URL, '" + id + "', 'toolbar=0,scrollbars=0,location=0,statusbar=0,menubar=0,resizable=0,width=880,height=300');");
}
</script>
这里,当用户单击“弹出按钮”时,我想将值传递到 actionpage.php。谢谢和问候,
I have a Popup button written in JavaScript. When user clicks that button, it opens a PHP popup. While clicking, i want to pass values through GET or POST method to that PHP page, Here is my code for Popup button:
<form input type="BUTTON" value="popup button" onClick="javascript:popup('actionpage.php')">
</form>
<SCRIPT LANGUAGE="JavaScript">
function popup(URL) {
day = new Date();
id = day.getTime();
eval("page" + id + " = window.open(URL, '" + id + "', 'toolbar=0,scrollbars=0,location=0,statusbar=0,menubar=0,resizable=0,width=880,height=300');");
}
</script>
Here i want to pass values to actionpage.php when user clicks "popup button". Thanks and Regards,
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onClick="javascript:popup('actionpage.php?foo=bar')"
onClick="javascript:popup('actionpage.php?foo=bar')"
您可以使用以下代码发送数据:
you can send data using this code:
希望这会对您有所帮助。
在PHP页面中只需使用
$_GET['id']
即可获取ID。Hope this will help you.
In PHP page just use
$_GET['id']
to obtain the ID.您可以在查询字符串上传递参数并在 PHP 页面的服务器端处理它们。
另请注意,使用
eval()
是不必要的,并且通常不受欢迎,并且您的FORM/INPUT
标记格式错误。我相信我已经在下面的示例中纠正了它。You can pass parameters on your querystring and process them server-side in your PHP page.
Also note that the use of
eval()
is unneeded and generally frowned upon and yourFORM/INPUT
tags are malformed. I believe I've corrected it in the following example.