Question

此页面生成一个包含所有内容的表格我数据库中billing_reports表的数据。当你加倍单击某个项目，您可以在该字段中输入新值，然后按输入保存。问题是，当我按提交时，它只刷新页面，而不更新对数据库的更改。我该如何解决？

<?php
        include_once ("includes/config.php");
        include_once ("includes/functions.php");
        include_once ("includes/header.php");
    ?>
    <!--How this works, is that this page generates a table that containes all 
        the data on the billing_reports table in my database. When you double 
        click on an item, you can enter in a new value in the field, then press
        enter to save. The problem I am having, is that when I press submit it 
        just refreshes the page, when I want it to update the changes to the database. 
        How do I fix this? -->  
    <script type="text/javascript" src="http://code.jquery.com/jquery-1.8.3.min.js"></script>
    <script type="text/javascript" src="script.js"></script>
    <?php
     // $data = All the rows in billing reports
     $data = mysql_query("SELECT * FROM billing_reports") 
     or die(mysql_error());

     // This is printing the Javascript table listed above
    ?> <table class="editableTable"> <?php
     Print "<editableTable border cellpadding=4>"; 

    //This is defining $info as the action that seperates the $data pulled from the database
     while($info = mysql_fetch_array( $data )) 
    { 
     Print "<tr>"; 
     Print "<th>Name:</th> <td>".$info['cc_name'] . "</td> ";
     Print "<th>ID:</th> <td>".$info['br_id'] . "</td> ";
     Print "<th>Listing ID:</th> <td>".$info['listing_id'] . "</td> ";
     Print "<th>Payment:</th> <td>".$info['billing_amount'] . "</td> ";
     Print "<th>Status:</th> <td>".$info['status'] . " </td></tr>"; 
     } 
     Print "</table>";

    if (!$_POST['submit']) {
        $result = mysql_query($info);
        $person = mysql_fetch_array($result);
    }

    if(isset($_POST['submit'])) {
        $info = "UPDATE billing_reports SET cc_name='$_POST[cc_name]', br_id='$_POST[br_id]', listing_id='$_POST[listing_id]', billing_amount='$_POST[billing_amount]', status='$_POST[status]'"; 
        mysql_query($info) or die(mysql_error());

        echo"Agent has been modified!";

    }
    ?>

    <form action="" method="post" class="button">
        <li>Save Changes</br>
        <input type="submit" <name="submit" value="Modify"/></li>
    </form>

    <?php
    include_once ("includes/footer.php"); 
    ?>

Answer 1

您的输入中存在误导，导致输入本身不起作用，isset()提交永远不会是真的

<input type="submit" <name="submit" value="Modify"/></li>
                   //^ here you don't need <

另一方面，我不会建议您极易受mysql injections攻击，我宁愿切换到pdo或mysqli并使用预备语句

Answer 2

此行中有一个拼写错误，<之前name：

  <input type="submit" <name="submit" value="Modify"/></li>

通常建议使用带代码突出显示的编辑器。

如何编辑它以便在我编辑此页面上的单元格时将其保存到数据库的表中？

2 个答案: