Question

这是我的主页代码：

<html>
<head>
    <title>PHP test</title>
    <link href="style.css" rel="stylesheet" type="text/css">
</head>
<body>


<div id="formDiv">  
    <form action="testsql.php" method="post">
        Email: <input type="text" name="email"><br>
        <textarea name="comment" rows="10" cols="40">Your comment here.</textarea>
        <input type="submit" value="submit">
    </form>


<div id="show">         
<?php

    $con=mysqli_connect("","root","","my_db");
    // Check connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $result = mysqli_query($con,"SELECT * FROM Comments");

    echo "<table border='1'>
    <tr>
    <th>Email</th>
    <th>Comment</th>
    </tr>";

    while($row = mysqli_fetch_array($result))
    {
        echo "<tr>";
        echo "<td>" . $row['Email'] . "</td>";
        echo "<td>" . $row['comment'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";

    mysqli_close($con);
?>
</div>
</div>
</body>
</html>

这是testsql.php文件的代码：

<?php
    $con=mysqli_connect("","root","","my_db");

if($_SERVER["REQUEST_METHOD"] == "POST")
{ 

header("Location: testphp.php");
if(empty($_POST["comment"]))
    {echo "You have to write something. <br>";}

else
    {
      if (mysqli_connect_errno())
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }

        if(empty($_POST["email"]))
        {
            $sql="INSERT INTO Comments (Email, comment)
            VALUES
            ('$_POST[email]', '$_POST[comment]')";
            echo "1 record added.";
            if (!mysqli_query($con,$sql))
                {
                    die('Error: ' . mysqli_error($con));
                }

        }
        else{
            if (!preg_match("/([\w\-]+\@[\w\-]+\.[\w\-]+)/",$_POST["email"]))
            {
                echo "Invalid email format <br>"; 
            }
        else{
            echo "Your email is: {$_POST["email"]} <br>";
            $sql="INSERT INTO Comments (Email, comment)
            VALUES
            ('$_POST[email]', '$_POST[comment]')";
            echo "1 record added.";
            if (!mysqli_query($con,$sql))
                {
                    die('Error: ' . mysqli_error($con));
                }

            }
        }
    }
   }
 ?>

我想要做的是从表单中获取值并将它们放在表中。单击“提交”按钮后，应该更新该表。当我有header(Location: "testphp.php");时，它会工作，但如果用户以错误的方式写邮件或不写评论，它就不会显示错误消息。

当我不包含标题时，代码正常工作，但我被重定向到 testsql.php ，我必须返回 testphp.php 才能看到更新表。我知道我可以使用javascript，但我们的教授告诉我们网站应该完全可用，关闭javascript 。有什么想法吗？

Answer 1

您需要使用JavaScript的 Ajax 方法。请参阅JQuery的DataTable插件，以便轻松完成。

使用同一页面上的表单中的值更新表

1 个答案: