Question

我正在显示数据库中的设备信息列表。每个设备都会有一个“显示警告”按钮。用户可以单击“显示警告”，我正在使用隐藏字段（来自隐藏字段的值来标识）来显示设备警告消息。数据库中的每个警告都有一个接受按钮（警告列表，每个警告都有接受按钮）。用户可以单击接受按钮，以使警告状态从未接受变为接受（并且页面将刷新）。问题在于刷新，因为它没有首页上的信息，因此php返回错误。

//page-1.. under each category (like electrical, mechanical, more than one types of devices
//device can be identified by using ID


echo '<tr><th>Category Name</th>';
echo '<td >' . $row['category']. '</td></tr>';

echo '<tr><th>Device ID</th>';
echo '<td >' . $row['ID']. '</td></tr>';

//form section. 

echo "<form action = alarms_list_display.php method =post>";
echo '<tr>';

echo "<td >"." <input type=submit    name=acceptID value=ShowAlarm"."  </td>";
echo "<td >"." <input type=hidden  name=hiddenID  value =". $row["ID"]."  </td>";
echo "</tr>";
echo "</form>";

//second page..    alarms_list_display.php

include ("DBconnect.php");
$conn= mysqli_connect( $dbhost, $dbuser, $dbpass, $db );

if(isset($_POST ['acceptID']))
    {$deviceID = $_POST['hiddenID'];

    }
     else {$deviceID='';}


//above code is problematic. as it is available only when i submit the first page.
//when updating the second page (and refreshing), $deviceID is not available anymore. 



//listing all warnings related to that device when user clicked the 'ShowAlarm' button.

if ($conn->connect_error) {die("Connection failed: " . $conn->connect_error);}
$sql = "SELECT * FROM dataTable WHERE ID='$deviceID' ";
$result = $conn->query($sql);

//form with buttons to accept warning
//skipped table heading parts
//each warning will have a serialNumber. Using to accept the warning and update the database

if ($result->num_rows > 0) {
         while($row = $result->fetch_assoc()) {
           echo "<form action = alarms_list_display_copy.php method =get>";


            echo "<td>" . $row["Date"]. "</td>";
            echo "<td>" . $row["Warning"]. "</td>";

            echo "<td>"." <input type=hidden  name=hidden value =" .$row["SerialNumber"]."  </td>";
            echo "<td>"." <input type=submit    name=accept value=AcceptAlarm id=button1 class = formclass"."  </td>";
            echo "</tr>";

           echo "</form>";
  } else { echo "0 results"; }



//code to update when clicking the AcceptAlarm button.

  if (isset($_GET['accept'])){
        $query= "SELECT WarningAccept FROM dataTable WHERE  SerialNumber= '$_GET[hidden]' ";
        $result = mysqli_query($conn, $query);
        while($row = $result->fetch_assoc()) {

        $updateQuery = " UPDATE dataTable SET WarningAccept=1 WHERE  SerialNumber= '$_GET[hidden]' ";
        mysqli_query($conn, $updateQuery);

     }

可以提交第一页。它将显示与该设备相对应的所有警告。但是当接受警告时，页面将刷新并且$ deviceID不再存在。因此更新后不会显示任何警告消息。如何改变它？？？任何帮助

Answer 1

您的second page仅在提交<form action = alarms_list_display.php method =post>";后才获得价值，这就是刷新第二页时会出错的原因，因为第二页会丢失$_POST数组，然后

解决方案

您可以将<form action = alarms_list_display.php method =post>"的操作设置到同一页面并将值存储在会话中，然后将其重定向到第二页，然后可以使用会话中的值

两种php形式。第二种形式从第一种形式获取数据。但刷新第二页会显示错误

1 个答案: