Parse Error?

+4 Brandon Bennett · November 5, 2014
I'm having some trouble and i'm not sure what a parse error is. 

I keep getting this error: 

"Parse error: syntax error, unexpected 'return' (T_RETURN) in /functions/users.php on line 14" and I'm not sure what i'm missing.

Here is the code i'm using:

<?php
function user_data($user_id){ 
$data = array();

}

function logged_in() {
return (isset($_SESSION['user_id'])) ? true : false;
}

function user_exists() {
$username = sanitize($username);
$query = mysql_query("SELECT COUNT ('user_id') FROM `users` WHERE `username` = '$username'")
return (mysql_result($query, 0) ==1) ? true : false;
}

function user_active() {
$username = sanitize($username);
$query = mysql_query("SELECT COUNT ('user_id') FROM `users` WHERE `username` = '$username' AND `active` = 1")
return (mysql_result($query, 0) ===1) ? true : false;
}

function user_id_from_username($username) {
$username = sanitize($username);
return mysql_result(mysql_query("SELECT `user_id` FROM `users` WHERE `username` = '$username'"), 0 'user_id');
}

function login($username, $password) {
$user_id = user_id_from_username($username);

$username = sanitize($username);
$password = md5($password);

return (mysql_result(mysql_query("SELECT COUNT('user_id') FROM `users` WHERE `username` = '$username' AND `password` = '$password'"), 0) ==1) ? $user_id : false;
}
?>

Please let me know if there is anything that i can do to fix this. 

Thank you!

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0 Brandon Bennett · November 22, 2014
what do i do if i have a separate "connect.php"?
0 Brandon Bennett · November 22, 2014
And what does this mean? i did the edits and got this

"Warning: mysql_query() expects parameter 1 to be string, resource given in /core/functions/users.php on line 13"
0 Ron Butcher · November 23, 2014
you can include the connect.php file with an include or require statement. That would have all of your $link information.
require 'connect.php';  

connect.php:
$link = mysqli_connect("host", "login_name", "login_password", "database");

The only real catch is that when you include this file at the top of your PHP page, you will not be able to use the $link variable in your functions without adding this line of code in each function that needs to run a query:
function example()
{
global $link; // This allows a variable that was defined outside the function
// Rest of function code i.e. $query = mysqli_query($link, "SELECT * FROM myTable");
}


On your error, it looks like you are mixing mysql and mysqli.  The mysql_query only expects a string that is the actual query, the mysqli_query expects two paramaters.  The first is the connect information, the second is the query.
Make sure that your query statement is as follows:
$query = mysqli_query($link, $qryString);  //  Do not forget the 'i' in mysqli
0 Ayodele Joseph · October 6, 2015
Hello guyz.... I have the same similar problem but have corrected it using this corrections seen here. The problem am having now is that my $data is not submitted to the my database......Plz, i need way forward.
0 OLayemii Garuba · October 19, 2015
gotten a solution yet?
0 Teslim Salako · October 19, 2015
paste your code so we can see whats wrong

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