I get this error when I try and include a variable from an external php file: Unexpected T_VAR on line 1
Here is my external php code:
<?php var dbStr = 'host::,username::,password::,database'; ?>
Here is the PHP file that is including it:
if($_POST['admincreate'] == "ok")
{
include('db.php');
$info = explode("::,", dbStr);
$con=mysqli_connect($info[0],$info[1],$info[2],$info[3]);
mysqli_query($con,"INSERT INTO admins (id, user, pass)
VALUES ('" . $_POST['user'] . "', '" . $_POST['pass'] . "',0)");
}
It works if I change the code to
if($_POST['admincreate'] == "ok")
{
$dbStr = 'host::,username::,password::,database';
$info = explode("::,", $dbStr);
$con=mysqli_connect($info[0],$info[1],$info[2],$info[3]);
mysqli_query($con,"INSERT INTO admins (id, user, pass)
VALUES ('" . $_POST['user'] . "', '" . $_POST['pass'] . "',0)");
}
But I need the included file.
If I change the included file to
<?php $dbStr = 'host::,username::,password::,database'; ?>
I get this error: Parse error: syntax error, unexpected '=' in .../db.php on line 1
And I find that the external php file has changed to
<?php = 'host::,username::,password::,database'; ?>
How do I include this string properly without it giving me an error?
P.S. The external php is generated by the main php file use fopen and fwrite, also the actual values of host, username, password, and database have been censored just because I feel better that way.
Thanks in advance, -p0iz0n