php - 用户 ID 进入数据库，而不是通过电子邮件发送

Question

无论如何，我已经尝试了几个小时，这个表单假设要获取正在显示的我的电子邮件，并在提交时将包括电子邮件在内的详细信息插入到数据库中。

它不是获取我的“电子邮件”，而是从用户表中获取我的用户 ID(自动增量)并将其更新到反馈表中。

反馈.php

<?php   


$email =(isset($_SESSION['email']) ? $_SESSION['email'] : null);
$name='';$feedback=''; $topic=''; $details=''; 
$action =(isset($_POST['submit']) ? $_POST['submit'] : null);

if($action!=null) {                 

    $name =(isset($_POST['name']) ? $_POST['name'] : null);
    $feedback =(isset($_POST['feedback']) ? $_POST['feedback'] : null);
    $topic =(isset($_POST['topic']) ? $_POST['topic'] : null);
    $details =(isset($_POST['details']) ? $_POST['details'] : null)

    if($details==null) {
        echo "<br><p style='text-align:center;color:red'>Please fill up all text fields!</p>";
    }
    else {
        $query="insert into feedback values('','$email','$name','$feedback','$topic','$details','',null)";
        $result=mysql_query($query);


        echo "<br><p style='text-align:center;color:blue'>Successfully submit the feedback to system </p>";
        $feedback=''; $topic=''; $details=''; 
    }

    if( mysql_error()!="") {
        echo "<font style='text-align:center;color:red'>" . mysql_error() . "</font><br>"; 
    } 


}
?>


<form method="post" action="user_feedback.php">     
    <?php
    $email =(isset($_SESSION['email']) ? $_SESSION['email'] : null);
    $query="select * from user where id=$email";
    $result=mysql_query($query);
    $row = mysql_fetch_array($result);


    ?>
    <br>
    <table cellpadding="5">
            <tr><td style="width:150px">User Email</td><td><input type="text" name="email" value="<?php echo $row['email'] ?>" disabled style="width:200px;"  ></td></tr>
            <tr><td>User Name</td><td><input type="text" name="name" value="<?php echo $name ?>" style="width:300px;" ></td></tr>

    <tr><td>Feedback</td>
    <td>
        <select name="feedback" style="height:32px">            
        <option <?php if($feedback=="Inquiry") { echo 'selected'; } ?> >Inquiry</option>

        </select>
    </td>
    </tr>
    <tr><td>Feedback Topic</td><td><input type="text" name="topic" value="<?php echo $topic ?>" style="width:300px;" ></td></tr>
    <tr><td>Feedback Details</td><td><textarea type="comment" name="details" value="<?php echo $details ?>" style="width:500px;" ></textarea></td></tr>
    <tr><td></td><td><input type="submit" name="submit" value="Submit" class="tall"></td></tr>
    </table>


</form>

Answer 1

通过使用建议修复了这个问题，因为登录页面本身的 ID 和电子邮件之间的登录 session 是混合的。

<?php   

//echo("{$_SESSION['id']}"."<br />");

$email =(isset($_SESSION['id']) ? $_SESSION['id'] : null);  
$name='';$feedback=''; $topic=''; $details=''; 
$action =(isset($_POST['submit']) ? $_POST['submit'] : null);
if($action!=null) {                 

    $name =(isset($_POST['name']) ? $_POST['name'] : null);
    $feedback =(isset($_POST['feedback']) ? $_POST['feedback'] : null);
    $topic =(isset($_POST['topic']) ? $_POST['topic'] : null);
    $details =(isset($_POST['details']) ? $_POST['details'] : null);

    if($topic==null || $details==null) {
        echo "<br><p style='text-align:center;color:red'>Please fill up all text fields!</p>";
    }
    else {
        $query="insert into feedback values('','$email','$name','$feedback','$topic','$details','',null)";
        $result=mysql_query($query);


        echo "<br><p style='text-align:center;color:blue'>Successfully submit the feedback to system </p>";
        $feedback=''; $topic=''; $details=''; 
    }

    if( mysql_error()!="") {
        echo "<font style='text-align:center;color:red'>" . mysql_error() . "</font><br>"; 
    } 


}
?>  

<form method="post" action="user_feedback.php">     
    <?php
    $email =(isset($_SESSION['email']) ? $_SESSION['email'] : null);
    $query="select * from user where id=$email";
    $result=mysql_query($query);
    $row = mysql_fetch_array($result);


    ?>
    <br>
    <table cellpadding="5">
            <tr><td style="width:150px">User Email</td><td><input type="text" name="email" value="<?php echo $row['email'] ?>" disabled style="width:200px;"  ></td></tr>
            <tr><td>User Name</td><td><input type="text" name="name" value="<?php echo $name ?>" style="width:300px;" ></td></tr>

    <tr><td>Feedback</td>
    <td>
        <select name="feedback" style="height:32px">            
        <option <?php if($feedback=="Inquiry") { echo 'selected'; } ?> >Inquiry</option>

        </select>
    </td>
    </tr>
    <tr><td>Feedback Topic</td><td><input type="text" name="topic" value="<?php echo $topic ?>" style="width:300px;" ></td></tr>
    <tr><td>Feedback Details</td><td><textarea type="comment" name="details" value="<?php echo $details ?>" style="width:500px;" ></textarea></td></tr>
    <tr><td></td><td><input type="submit" name="submit" value="Submit" class="tall"></td></tr>
    </table>


</form>