Question

我的网页上有一个表单，其中'action'指定用于执行查询的php文件。我想按下按钮执行该查询并打开一个名为details.php的新网页。我怎样才能做到这一点？查询工作正常。在执行查询后，它应该打开details.php页面。我的输出打印代码如下所示。谢谢提前

if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
echo "The image has been uploaded, and your information has been added";  
}
else {
echo "Sorry, there was a problem uploading your file.";
}

我需要使用主页按钮将此结果打印在网页（页面尺寸与之前数据输入页面相同）

即使同一页面显示结果也没问题，但是应该清除表格中的数据，以便用户可以输入新数据

Answer 1

以最简单的形式，

if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
   $msg = "The image has been uploaded, and your information has been added";
   $size = <place the value here>
   $filename = <place file name here>
   header("Location: details.php?msg={$msg}&size={$size}&fn={$filename}&target={$target}");
   exit();
}
else {
echo "Sorry, there was a problem uploading your file.";
}

然后在details.php

$msg = $_GET['msg'];
$size = $_GET['size'];
$fn = $_GET['fn'];
$destination = $_GET['target'];

echo "<p>".$msg."</p>";
echo "<p>File size is: ".$size."</p>";
echo "<p>File name is: ".$fn."</p>";
echo "<p>File uploaded at :".$destination."</p>";
echo "<input type='button' value='Home' onclick='window.location.href=\"home.php\"' />";

PS：不要忘记清理所有数据

Answer 2

使用header()只会导航用户当前所在页面到您想要的页面。要创建具有所需尺寸的新弹出窗口，您必须使用JavaScript。

您可以使用AJAX处理请求，并让PHP向客户端JS返回一条消息，以触发新的弹出页面打开。

查询执行后PHP打开新页面

2 个答案: