PHP表单不插入数据

This is not going online anytime soon. I am just trying to learn PHP/MySQL. This is my first attempt with inserting to a database. I cannot figure out why the data won't insert. This is the HTML form:

    <form action="testdbserverside.php" method="post">
    Name:<br />
    <input type="text" id="ContactName" name="Name" /><br />
    E-mail:<br />
    <input type="text" id="ContactEmail" name="Email" /><br />
    Comment:<br />
    <input type="text" id="ContactComment" name="Comment" size="50" />
    <br /><br />
    <input type="submit" id="submit" name="submit" value="Send">
    <input type="reset" id ="reset" value="Reset">
    </form>

This is the PHP:

    <?php
     $ContactName = $_POST["ContactName"];
     $ContactEmail = $_POST["ContactEmail"];
     $ContactComment = $_POST["ContactComment"];

     $sql_connection = mysqli_connect("localhost:8889","root","root","derek_website_tmp");

     if (mysqli_connect_errno())
        {
         echo "failed to connect" . mysqli_connect_error();
        }

     $sql = "INSERT INTO MyContacts (

                ContactName,
                ContactEmail,
                ContactComment,
                ContactDateCreated
            )
            VALUES (
              '$ContactName',
              '$ContactEmail',
              '$ContactComment',
              NOW()
        )";

     if (!mysqli_query($sql_connection,$sql))
        {
        die('Error : ' . mysqli_error($sql_connection));
        }

     mysql_close ($sql_connection);

    ?>

And this is the database:

    mysql> USE derek_website_tmp;
    Database changed
    mysql> CREATE TABLE MyContacts  (
        -> ContactID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
        -> ContactName VARCHAR(100),
        -> ContactEmail VARCHAR(100),
        -> ContactComment VARCHAR(200),
        -> ContactDateCreated DATETIME
        -> );
    Query OK, 0 rows affected (0.19 sec)

Any help would be greatly appreciated. I realize this is not sanitizing anything and is not worthy to go live. I'm just working on getting it to insert for now.

Thank you,

$ContactName = $_POST["ContactName"]; 
$ContactEmail = $_POST["ContactEmail"]; 
$ContactComment = $_POST["ContactComment"];

Should be:

$ContactName = $_POST["Name"]; 
$ContactEmail = $_POST["Email"]; 
$ContactComment = $_POST["Comment"];

The name is your parameter not the id

And your query should be:

$sql = "INSERT INTO MyContacts (ContactName, ContactEmail, ContactComment, ContactDateCreated ) VALUES ( "'.$ContactName.'", "'.$ContactEmail.'", "'.$ContactComment.'", NOW() )";

You tried to read values from "ID" instead of "NAME" attribute.

The form can be modified like this (ID part is optional in terms of form submission):

Name:<br />
<input type="text" id="ContactName" name="ContactName" /><br />
E-mail:<br />
<input type="text" id="ContactEmail" name="ContactEmail" /><br />
Comment:<br />
<input type="text" id="ContactComment" name="ContactComment" size="50" />

All the rest of your code should work fine.

FYI, You can use print_r() or var_dump() to print the contents of given variable.

<?
// this will show what really is coming in through the $_POST
print_r($_POST);
?>

In your PHP page, you are gathering the value by the id, it won't work like that you have to gather the value by text field name like

$ContactName = $_POST["Name"];
$ContactEmail = $_POST["Email"];
$ContactComment = $_POST["Comment"];  

I think this will work fine.