How to store a forms dropdown list selected value into mysql database

Question

Have a form with a dropdown list. How can I get the value the user selects into my data base once a user submits the form? Can't get the code below to store the user selected value? Thanks for your help.

FORM

<form action="xxx.php" class="well" id="xxx" name"xxx" method="post">


<select name="extrafield5">
 <option value="NOW" selected="selceted">Submit order now</option>
 <option value="REVIEW">Submit my order for review</option>
</select>


</form>

PHP FILE

<?php

define('DB_NAME', 'xxx');
define('DB_USER', 'xxx');
define('DB_PASSWORD', 'xxx');
define('DB_HOST', 'xxx');

$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);

if(!$connection){
die('Database connection failed: ' . mysqli_connect_error());
}

$db_selected = mysqli_select_db($connection, DB_NAME);

if(!$db_selected){
die('Can\'t use ' .DB_NAME . ' : ' . mysqli_connect_error());
}

echo 'Connected successfully';


if (isset($_POST['extrafield5'])){
    $extrafield5 = $_POST['extrafield5'];
}

else {$extrafield5 = '';}




$sql = "INSERT INTO seguin_orders (extrafield5) 
        VALUES ('$extrafield5')";

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

DATABASE

http://oi60.tinypic.com/9ppc0i.jpg

You should include more of the code. Is the <select> in a form that is posting to that PHP file? If you var_dump($extrafield5); exit; after the IF statement, what is the output? Do you have an established DB connection to run the INSERT with? I recommend PDO php.net/manual/en/pdo.connections.php — Mikel Bitson
– Mikel Bitson, Commented Dec 3, 2014 at 3:38
you should get the option value as the value of extrafield5. Are you getting a null value? also check if you have set the valid form method=post or else use if(isset($_REQUEST['extrafield5'])). Post you mysql connection code as well if possible — Aditya
– Aditya, Commented Dec 3, 2014 at 3:38
you'll have to connect to it first, and use either mysqli or PDO API, then create an insert statement, also consider using prepared statements — Kevin
– Kevin, Commented Dec 3, 2014 at 3:38

user4246994 · Accepted Answer · 2014-12-03 04:53:23Z

2

change your form as below.

<form action="xxx.php" class="well" id="xxx" name"xxx" method="post">


<select name="extrafield5">
<option value="NOW" >Submit order now</option>
<option value="REVIEW">Submit my order for review</option>
</select>


</form>

answered Dec 3, 2014 at 4:53

user4246994

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