Parse error: syntax error, unexpected T_VARIABLE

calsteve9790

I keep getting this error message. After going through the code many times I am stumped. If there is anyone who can help I would appreciate it.

<?php
$link = mysql_connect('mydatabase.com', 'mydatabase', 'password'); 
if (!$link) { 
    die('Could not connect: ' . mysql_error()); 
} 
echo 'Connected successfully'; 
mysql_select_db(mydatabase) or die(mysql_error()); 

// Get values from form
$user_type=$_POST['user_type'];
$ven_name=$_POST['ven_name'];
$address=$_POST['address'];
$email_addy=$_POST['email_addy'];
$phone=$_POST['phone'];
$class=$_POST['class'];

// Insert data into mysql
$sql="INSERT INTO vendor_games(user_type, ven_name, address, email_addy, phone, class) VALUES('$user_type', '$ven_name', '$address', '$email_addy', '$phone', '$class')"

$result = mysql_query($sql);

// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='/index.html'>Back to Bike Show Page</a>";
}

else {
echo "ERROR";
}

// close connection
mysql_close();
?>

djjjozsi

close the string with a ; in the 18th line.

$sql="INSERT INTO vendor_games(user_type, ven_name, address, email_addy, phone, class) VALUES('$user_type', '$ven_name', '$address', '$email_addy', '$phone', '$class')";

how to make it better?

if(!empty($_POST))
{
$to_database=array_map("mysql_real_escape_string" , $_POST);
extract($to_database , EXTR_PREFIX_SAME, "x_");

$sql="INSERT INTO vendor_games(`user_type`, `ven_name`, `address`, `email_addy`, `phone`, `class`) VALUES('$x_user_type', '$x_ven_name', '$x_address', '$x_email_addy', '$x_phone', '$x_class')";
unset($to_database);
/* The rest of your code comes here..*/
}

calsteve9790

Thanks, I'll give it a try...

Paavero

On which line is the error?

the_lynx123

you are missing as semicolon at the end of the query

bradgrafelman

Also note that your code is vulnerable to SQL injection attacks. User-supplied data should never be placed directly into a SQL query. Instead, it must first be sanitized with a function such as [man]mysql_real_escape_string/man.

EDIT: Threads moved. Please do not post the same topic multiple times!

calsteve9790

djjjozsi, thanks for the help. I used your suggestion and found I may have been using a reserved word. I am now getting a "Array NULL ERROR" I have inserted the code below.

<form action="includes/action.php" method="post">
<input type="hidden" value="show" name="user_type">
Name: <input type="text" name="ven_name"><br>
Address, City, State, Zip: <input type="text" name="address"><br>
Email: <input type="text" name="email_addy"><br>
Phone: <input type="text" name="phone"><br>
Pick a Class from the Menu Below:<br>
<select name="bike_kind[]">
<option value="full">Full Custom - Any make or year custom motorcycle with major changes such as Engine,Frame, Front End, Fabricated Bikes.</option>
<option value="mild">Mild Custom - Any make or year custom with original Engine, Frame Front End.</option>
<option value="manufactured">Manufactured Custom - Any Custom bike built by a franchised company, Harley, Bourget, Iron Horse, etc.</option>
<option value="street">Street Stock - Any factory stock street bike.</option>
<option value="performance">Street Performance - Any Performance Oriented street bike. Must be licensed.</option>
<option value="classic">Classic - Any make motorcycle, 1960 or later. Minor changes allowed.</option>
<option value="antique">Antique - Any make motorcycle, pre 1960. Near stock.</option>
<option value="touring">Touring - Any make or year touring motorcycle. Must have hard bags and/or windshield. Changes allowed.</option>
<option value="trike">Trike - Any three wheeled vehicle.</option>
<input type="submit" name="submit" value="Submit">
</form>

<?php
$link = mysql_connect('website.com', 'database', 'password'); 
if (!$link) { 
    die('Could not connect: ' . mysql_error()); 
} 
echo 'Connected successfully'; 
mysql_select_db(database) or die(mysql_error()); 

// Get values from form
$user_type=$_POST['user_type'];
$ven_name=$_POST['ven_name'];
$address=$_POST['address'];
$email_addy=$_POST['email_addy'];
$phone=$_POST['phone'];
$bike_kind=$_POST['$bike_kind'];

// Insert data into mysql
if(!empty($_POST))
{
$to_database=array_map("mysql_real_escape_string" , $_POST);
extract($to_database , EXTR_PREFIX_SAME, "x_");

$sql="INSERT INTO vendor_games(`user_type`, `ven_name`, `address`, `email_addy`, `phone`, `bike_kind`) VALUES('$x_user_type', '$x_ven_name', '$x_address', '$x_email_addy', '$x_phone', '$x_bike_kind')";
unset($to_database);
}

$result = mysql_query($sql);

// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='/bike_show.html'>Back to Bike Show Page</a>";
}

else {
echo "ERROR";
}

// close connection
mysql_close();
?>

djjjozsi

bike_kind has set as array with your form.

<select name="bike_kind[]">

you cant insert an array into a single string directly.

in my example there is a mistake, the extract function takes an _ already between my prefix and the original index names, so you should use it like this:

extract($to_database, EXTR_PREFIX_ALL, "x");

calsteve9790

So can I use the "bike_kind" as just a $_POST variable instead of an array?

djjjozsi

if you need to allow more then one item from that selection, you should use another database model to be able to store its values not in a single field.

<select name="bike_kind">

as a select name,

and you can reach its value in your select as: $x_bike_kind