PHP form not inserting data

Question

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,

(1) What isn't working? (2) Please don't use those insertions like "$variable" for queries they are very fragile to SQL injection attacks... — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Sep 17, 2013 at 2:31
I am not getting any errors. The only thing being inserted is ContactID and the DateCreated. — user2785839
– user2785839, Commented Sep 17, 2013 at 2:33
In future cases please add error_reporting(-1); ini_set('display_errors', 'On'); during development work. — Ja͢ck
– Ja͢ck, Commented Sep 17, 2013 at 2:42

Niket Malik · Accepted Answer · 2013-09-17 02:32:41Z

1

$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() )";

answered Sep 17, 2013 at 2:32

Niket Malik

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Comments

Kita · Accepted Answer · 2013-09-17 02:41:34Z

0

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);
?>

edited Sep 17, 2013 at 2:41

answered Sep 17, 2013 at 2:36

Kita

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Comments

justjiten · Accepted Answer · 2013-09-17 02:41:57Z

0

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.

answered Sep 17, 2013 at 2:41

justjiten

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