PHP：如何在关闭弹出窗口时返回上一页

Question

I have a form submission. While i Click Submit button, will check for mandatory fields and if the fields are blank, I am opening a popup window using javascript to show error message. So once I close the popup I am not going back to previous page Instead it shows a white blank screen.

My code for verification and opening popup is :

<?php 

 if (isset($_POST['submit']))
    { 
if ($task == '' || $comments == '')
 {
 // generate error message
print "<script type='text/javascript'>";
print "window.open('error.php','new_window1','status=1,scrollbars=1,resizable=0,menu=no,width=320,height=220');";
print "</script>";  

}
else
{
$sql3= "INSERT INTO work (task, comments, assignee, type, priority, dataum1, dataum2, status) VALUES ('$task', '$comments', '$assignee', '$type', '$priority', '$dataum1', '$dataum2', '$status')";
mysqli_query($mysqli,$sql3) or die(mysqli_error($mysqli));
}
}
?> 

Answer 1

由于您使用的是JS弹出窗口，因此您可能希望使用JavaScript进行客户端强制检查（尽管也要进行服务器端检查），然后在未填写必要字段的情况下阻止表单发布出来。

Answer 2

Post the form data to the same page where you have the form. And add the above validation code to the top of that page. So when a validation error occurs, the user sees a popup with the error message, but he is on the same page where the form is.