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PHP : Parse error: syntax error, unexpected '"', expecting ')' in your code on line 28


nouaman

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please im seeking help through your helpful webite

 

im using a code snipet from a website regarding user logon session so i got the error :

 

Parse error: syntax error, unexpected '"', expecting ')' in your code on line 28

 

i check on php code checker and got the same error

 

below is the code in question please point me to the error to be fixed as im unable to locate it even if im considering the error line and had a look at my code but eas unable to resolve this issue

 

also i found in google somethingn related to php version but im not sur so FYI im using version 5.4,16 ofo PHP

 

thank you in advance

 

code :

 

------------------------------------------

 

/* Snippet #1 */

<?php
// login_check.php
define("server", "localhost");
define("user", "root");
define("password", "");
define("name", "test");
$connection;
$this->connection = mysql_connect(localhost,root,'') or die(mysql_error());
mysql_select_db(name, $this->connection) or die(mysql_error());
function is_logged_in () {
if (!($_SESSION["id"]) || ($_SESSION["id"] == "") || ($_SESSION["id"] == 0)) {
Header("Location: ./login.php");
exit();
}
}
function clean_input($input) {
$clean = array("",'<','>','`',':',';','/','(',')','{','}','[',']');
//$with = array();
return str_ireplace($clean,'', $input);
}
function login_check ($forms) {
$error = "";
$username = clean_input($forms["username"]);
$password = clean_input($forms["password"]);
if (trim($username) == "") $error .= "<li>Your username is empty.</li>";
if (trim($password) == "") $error .= "<li>Your password is empty.</li>";
/* from here, do your sql query to query the database to search for existing record with correct username and password */
$query = "SELECT password, username FROM users WHERE username = '".mysql_real_escape_string($username)."' AND password = '".mysql_real_escape_string($password)."'";
$result = mysql_query($query, $this->connection);
if(!$result || (mysql_numrows($result) < 1)) {
$error = "Invalid username or password";
}else
{
$error = "";
}
if (trim($error)!="") return $error;
}
function login ($forms) {
$username = clean_input($forms["username"]);
$password = clean_input($forms["password"]);
/* do your sql query again, but now returning the id of member */
$query = "SELECT member_id FROM users WHERE username = '".mysql_real_escape_string($username)."' AND password = '".mysql_real_escape_string($password)."'";
$result = mysql_query($query, $this->connection);
$result = mysql_query($query, $this->connection);
if(!$result || (mysql_numrows($result) < 1)) {
$id = 0;
}else
{
$id = $result;
}
return $id;
}
?>

 

 

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On this line you have an escaped double-quote:

$clean = array("",'<','>','`',':',';','/','(',')','{','}','[',']');

If you need to put a backslash in a string you need to escape it with another backslash:

$clean = array("",'<','>','`',':',';','/','(',')','{','}','[',']');
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Hi there

 

thank you for your reply it was helpful fixing that problem but now i got another error related to this line :

 

$this->connection = mysql_connect('localhost','root','') or die(mysql_error());

 

and the error was :Fatal error: Using $this when not in object context in C:wampwwwphploginsnipetlogin_check.php on line 10

 

any suggestion would be appreciated

 

thanks in advance

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hi there

 

thanks for your reply first , actually as i mentioned in my first post i m using this code snipet so i copied/pasted the code to see if it s working and then try to customized for my need so i got stack with these errors.so. i guess that the code snipet should not use it as you mentioned

 

in this case could you please suggest me what to use instead or point me to what can be done and i will dig on it on my own that will be very appreciated

 

thanks again for taking time to read my posts

 

Note : im beginer in PHP so when im stack on something i use google to find out solution or this site w3schools or hints from this forum :)

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