I've got a "Parse error: syntax error, unexpected T_VARIABLE in C:\wamp\www\php\vali"

[Mandyrjs]

hi im new to php and i want to code a admin login page where i want to match the values from the login form with the values of the database table.....when i run the login page after entering the user name and the password i've got a Parse error: syntax error, unexpected T_VARIABLE in C:\wamp\www\php \validate.php on line 16. Can anybody please help me to slove this matter.

validate.php

<?php 
session_start();
$username= $_POST['adminusername'];
$password= $_POST['adminpass'];
	
$conn= mysql_connect('localhost','root','');
mysql_select_db('onlinestore',$conn);

$query='SELECT * FROM admin WHERE adminusername='$username' and adminpassword='$password'';
$result=mysql_query($query);
if(mysql_num_rows($result)==1)
{
header("location:menu.php");
exit();
}
header("location:adminlogin.php");
exit();
?>
</body>

[Dormilich]

wrap your query in double quotes. single quoted strings are not parsed and in your case cause the error.

[Mandyrjs]

Hey thanks but when i wrap it with double quotes it should go to the menu.php when giving correct username & password right???? but it act as an incorrect input........ca n u help me???

[Dormilich]

what does the corrected code look like?

[Mandyrjs]

<?php 

$username= $_POST['adminusername'];
$password= $_POST['adminpass'];
	
$conn= mysql_connect('localhost','root','');
mysql_select_db('onlinestore',$conn);

$query="SELECT * FROM `admin` WHERE adminusername='$username' and adminpassword='$adminpass';
$result=mysql_query($query);
if(mysql_num_rows($result)==1)
{
header("location:menu.php");
exit();
}
header("location:adminlogin.php");
exit();
?>

now it says "Parse error: syntax error, unexpected T_STRING in C:\wamp\www\php \validate.php on line 20"
which is ,header("locati on:menu.php");

[Dormilich]

if you open a string, you should also close it (i.e. missing closing string delimiter).