Parse error: syntax error, unexpected '$password' (T_VARIABLE) Line 4

Question

I was trying to create a simple login in html using phpmyadmin database and xampp server but then this error shows *Parse error: syntax error, unexpected '$password' (T_VARIABLE) *

Here's my code.

<?php

$username = $_POST['user']
//*here's the Erro// $password = $_POST['pass']

$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);

mysql_connect("localhost", "root", "");
mysql_select_db("Login");

$result = mysql_query("select * from users where username = '$username' and password = '$password'") or die("Failed to query Database".mysql_error());

$row = mysql_fecth_array($result);

if ($row['username'] == $username && $row['password'] == $password ){
    echo "Login Succesful! Welcome" =.$row['username'];
}else {
    echo "Login Failed"
}

?>

i put `;` at the end of both `$username = $_POST['user']` and `$password = $_POST['pass']` and there's another error, says "Parse error: syntax error, unexpected '=', expecting ',' or ';' in C:\xampp\htdocs\Logintry1\bs-advance-admin\advance-admin\loginpro.php on line 23" — NiewBiee2020, Feb 26 '17 at 06:19
Pease don't use `mysql` functions. they have deprecated. use [mysqli](http://php.net/manual/en/book.mysqli.php) or [PDO](http://php.net/manual/en/book.pdo.php) — gaurav, Feb 26 '17 at 06:19
thank you for the ref's :) sorry i'm newbie in html and mysql — NiewBiee2020, Feb 26 '17 at 06:33

Anthony Rutledge · Answer 1 · 2017-02-26T06:25:31.223

2

You forgot the semi-colon.

$username = $_POST['user'];

You also forgot it here.

echo "Login Failed";

edited Feb 26 '17 at 06:25

answered Feb 26 '17 at 06:19

Anthony Rutledge

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B. Desai · Accepted Answer · 2017-02-26T06:30:55.247

You have multiple errors:

<?php

$username = $_POST['user']; //<-------put ';'
//*here's the Erro// $password = $_POST['pass']


$username = stripcslashes($username);
$password = stripcslashes($password);
$username = mysql_real_escape_string($username);
$password = mysql_real_escape_string($password);



mysql_connect("localhost", "root", "");
mysql_select_db("Login");


$result = mysql_query("select * from users where username = '$username' and password = '$password'") or die("Failed to query Database".mysql_error());

$row = mysql_fecth_array($result);

if ($row['username'] == $username && $row['password'] == $password ){
echo "Login Succesful! Welcome =".$row['username']; //<-------change this line
}else {
echo "Login Failed";//<-------put ';'

}

?>

what about `echo "Login Failed"`? – Sean Feb 26 '17 at 06:30 — Sean, Feb 26 '17 at 06:30
yes. thanks for it. Updated – B. Desai Feb 26 '17 at 06:31 — B. Desai, Feb 26 '17 at 06:31

Parse error: syntax error, unexpected '$password' (T_VARIABLE) Line 4

2 Answers2