Inserting Data Into Mysql Using Php

Ive been having difficulties trying to load form data into my database. Im trying to input theatre info using the following php script.

Solution 1:

try this

<?phprequire('connect.php');

         if (isset($_POST['theatre_name']) && isset($_POST['website'])){
$theatre_name = $_POST['theatre_name'];
$phone_number = $_POST['phone_number'];
$website = $_POST['website'];
$num_screens = $_POST['num_screens'];
$address = $_POST['address'];
$city = $_POST['city'];

//**change code to below**$queryd = "INSERT INTO `Theatres` (theatre_name, phone_number, website, num_screens, address, city) VALUES ('{$theatre_name}', '{$phone_number}', '{$website}', '{$num_screens}', '{$address}', '{$city}')";
$result = mysql_query($queryd);
if($result){
    $msg = "Theatre created.";
}

} ?>

link single quoted

Note: Unlike the double-quoted and heredoc syntaxes, variables and escape sequences for special characters will not be expanded when they occur in single quoted strings.

Solution 2:

I once had the same issue. Your query variable should look like this:

$queryd = "INSERT INTO `Theatres` (theatre_name, phone_number, website, 
                                 num_screens, address, city)
                VALUES ('".$theatre_name."', '".$phone_number."', '".$website."',
                        '".$num_screens."', '".$address."', '".$city."')";

Explanation: In your original query, you would have just inserted literally $theatre_name, not the variables value. In order to get around this, you have to close the string, with ", concatenate the variable to the preceding string, with . , and then re open the string.

Also, I don't know what version of PHP you are using, but you should be using mysqli_query(). mysql_query() is depreciated as of PHP v5.5. PHP manual entry.

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