Warning: mysql_error() expects parameter 1 to be resource, string given - php

Question

<?php
session_start();
$username = "root";
$password = "password";
$database = "meipolytechnic";
mysql_connect('localhost', $username,$password);
#mysql_select_db($database) or die(mysql_error());
$username=$_SESSION['MM_Username'];
$query = "SELECT rollno FROM users where username = '".$username."'";
$result = mysql_query($query) or die(mysql_error());
$num = mysql_num_rows($result);
mysql_close();
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
echo ($rows['rollno']);
?>
i want to retrieve only the logged in users roll no from users table in database
when i run this code
and log in as foo
i get the following stuff
Unknown column 'foo' in 'where clause'

Answer 1

There should be session_start() at the top of the page
query need to change as
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."' ";
EDIT
Please try something before posting a question here. Please google or go through www.w3school.com for clearing this kind of issues. Make a good knowledge about arrays and mysql connection. And mysql_query function won't work latest PHP version.
Please try following code.
$result = mysql_query($query) or die(mysql_error());
$rows = array();
while($r = mysql_fetch_row($result))
{
$rows[] = $r[0];
}
print_r($rows);

Answer 2

To use an array inside a string you need to put a curly bracket before it and after it
so
$query = "SELECT rollno FROM users where username = {$_SESSION['MM_Username']}";
or
$query = "SELECT rollno FROM users where username = ".$_SESSION['MM_Username'];

Answer 3

First of all start session using start_session()
then change your query:
$query = "SELECT rollno FROM users where username = ".$_SESSION['MM_Username'];
then change:
$num = mysql_num_rows($result); instead of $num = mysql_numrows($result);

Answer 4

Try this query
$query = "SELECT rollno FROM users where username = ".$_SESSION[MM_Username]." ";
And start session on same page.

Answer 5

You can use
$username=$_SESSION[MM_Username];
$query = "SELECT rollno FROM users where username = '".$username."'";
and start the session by using session_start()
and you have used two closing tags omit one make it like below
echo ($rows['rollno']);
?>

Answer 6

This type of error occur when query goes false
May be becouse You have not start session, becouse if you dont have session_start(), then nothing will come in session variable... just try as
$query = "SELECT rollno FROM users where username = '".$_SESSION[MM_Username]."'";
may this help you

Answer 7

You must need to use session_start() before using $_SESSION variable in code.
So put below code at start,
session_start();
Then do some modification in query like,
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";

Answer 8

first start your session
session_start();
and Change your query like this...
$query = "SELECT rollno FROM users where username = '".$_SESSION['MM_Username']."'";