Question

我遇到了一个简单的问题，但我不知道如何克服......一点帮助就会明显...... !!

代码 commentinsert.php

<?php
    if(isset($_POST['action']) && $_POST['action'] == 'Send'){
        sendfeedback();
    }

     function sendfeedback() {
        $name=$_POST['name'];   
        $email = $_POST['email'];
        $subject = $_POST['subject'];
        $feedback = $_POST['feedback'];

        $servername= "localhost";
        $username= "root";
        $password= "******";
        $dbname= "mydb";

        $conn = mysqli_connect($servername, $username, $password, $dbname);

        $sql= "INSERT INTO feedback_tbl (name,email,subject,feed_text) VALUES ('$name','$email','$subject','$feedback')";

        if(mysqli_query($conn, $sql))
        {
            echo "Feedback Successfully Sent";
        }
        else
        {
            echo "Feedback not sent";
        }
    }

?>

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>commentinsert</title>
<link href="css/style.css" rel="stylesheet" type="text/css" />
<script type="text/javascript" src="js/delete.js"></script>
</head>
<body>

<div class="bg">
  <div id="header">
  <div id="logo">
    <h1> xyz </h1><h4>Company</h4>

    </div>
    <div id="navlist">
      <ul>
        <li> <a href="/"> Home</a></li>
        <li> <a href="/gallery"> Gallery</a></li>
        <li> <a href="/legal"> Legal</a></li>
        <li> <a href="/contact" class="current"> Contact Us</a></li>
        <li> <a href="/about"> About Us</a></li>
        <li> <a href="/plan"> Plan</a></li>
      </ul>
    </div>

  </div>
</div>
<br>

<div class="bodywrapper">
<div id="conoverflw">

<div id="con_detail"> <p>asiubvisdvbidsubvauY </p>
  <p>iuivsduvbidubvuaovidosvni<br> asfv<br>abvckivdai<br>Email : xyz@gmail.com</p>   
</div>

<div id="feedback_form">

<div class="contact_email">

<form action ="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post"  id="emailForm" >

<label for="contact_name"> &nbsp;Enter your Name: </label>
<input type="text" name="name" id="contact_name" required size="30" class="inputbox" value="">
<label id="contact_emailmsg" for="contact_email"> &nbsp;E-mail address: </label>
<input type="text" id="contact_email" name="email" size="30" required value="" class="inputbox"  maxlength="100">
<label for="contact_subject"> &nbsp;Feedback Subject: </label>
<input type="text" name="subject" required  size="30" class="inputbox" value="">
<label id="contact_textmsg" for="contact_text"> &nbsp;Enter your Feedback: </label>
<textarea cols="50" rows="10" name="feedback" id="contact_text"  required style="margin: 4px; width: 370px; height: 150px;"></textarea>
<br>
<input class="" name="action"  type="submit" value="Send">
<input type="submit" name="action" value="Cancel">



</form></div>

</div>
</div>
</div> 
<div class="bg_footer">
<div id="footer">
<div class="sec1">
<ul>
<li id="1"><a href="http://localhost/about">&gt; Post it on Youtube</a></li>
<li id="2"><a href="http://localhost/about">&gt; Post it on your Facebook page</a></li>
<li id="3"><a href="http://localhost/about"> &gt; Optimize it for search on Youtube</a></li>
</ul>
</div>
<div class="sec1">
<ul>
<li id="1"><a href="http://localhost/about">&gt; Post it on Youtube</a></li>
<li id="2"><a href="http://localhost/about">&gt; Post it on your Facebook page</a></li>
<li id="3"><a href="http://localhost/about"> &gt; Optimize it for search on Youtube</a></li>
</ul>
</div>
<div class="sec3">
<h1>our company</h1><p>© 2016 <span> l </span> PRIVACY POLICY</p>

</div>
</div></div>
</body>
</html>

在另一页displaycomment.php
中

<?php
ini_set('display_errors', 1);

error_reporting(~0);

$serverName = "localhost";

$userName = "root";

$userPassword = "******";

$dbName = "mydb";

$conn= mysqli_connect($serverName,$userName,$userPassword,$dbName);

$sql="SELECT * FROM feedback_tbl";

$query = mysqli_query($conn,$sql);
if(isset($_GET['delete_id']))
        {
        $sql = "DELETE FROM feedback_tbl WHERE feed_id=".$_GET['delete_id'];
        $conn->query($sql);
        header("Location: http://localhost/displaycomment.php");
}

    while($row=mysqli_fetch_array($query,MYSQLI_ASSOC))
    {
    ?>

<html>
<head><title>displaycomment</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

</head>
<body>
 <script type="text/javascript" src="js/delete.js"></script>
<hr align="left" width="250px">
<div>

<p><?php echo(($row['name']))?>  &nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp </p>
<p><?php echo(($row['subject']))?></p>
<p><?php echo(($row['feed_text']))?></p>
<a href="javascript:delete_id(<?php echo $row["feed_id"]; ?>)"> x </a>

</div>
<hr align="left" width="250px">
<?php
}
?>
</body>
</html>

这两个单独的页面工作正常..但我想在单页中创建。即我想在commentinsert.php中显示插入的注释

Answer 1

忽略样式代码和不适：

<?php
if (isset($_POST['action']) && $_POST['action'] == 'Send')
    sendfeedback();

function sendfeedback()
{
    $name=$_POST['name'];   
    $email = $_POST['email'];
    $subject = $_POST['subject'];
    $feedback = $_POST['feedback'];

    $servername= "localhost";
    $username= "root";
    $password= "******";
    $dbname= "mydb";

    $conn = mysqli_connect($servername, $username, $password, $dbname);

    $sql= "INSERT INTO feedback_tbl (name, email, subject, feed_text) VALUES ('$name', '$email', '$subject', '$feedback')";

    if (mysqli_query($conn, $sql))
    {
        echo "Feedback Successfully Sent";
        $sql = "SELECT * FROM feedback_tbl";
        $query = mysqli_query($conn, $sql);
        while ($row = mysqli_fetch_array($query,MYSQLI_ASSOC))
        {
            ?>
            <hr align="left" width="250px">
            <div>

            <p><?php echo(($row['name']))?>  &nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp </p>
            <p><?php echo(($row['subject']))?></p>
            <p><?php echo(($row['feed_text']))?></p>
            <a href="javascript:delete_id(<?php echo $row["feed_id"]; ?>)"> x </a>

            </div>
            <hr align="left" width="250px">
            <?
        }
    }
    else
    {
        echo "Feedback not sent";
    }
}
?>

我还建议在此功能上方移动<html>，<head>，</head>，<body>标记并删除重复的标记。

Answer 2

注意：考虑转义用户输入并使用预准备语句。使用SQL注入，用户可以从“反馈”表中删除所有数据。

在单个网页中显示插入表单和插入的数据

2 个答案: