Problem in executing sql query

[bandy]

Hi there,
I am trying to execute a query on database but it's giving error as:
Warning: mysql_fetch_row (): supplied argument is not a valid MySQL result resource in C:\wamp\www\201 00519\HonestFol lowers.com\desi gn\datagraph.ph p on line 21.
and also shows Resource id #46.
The connection is successful, database is also selected.
What is the problem? can U help me on this,
help will be greatly appreciated.
Here is my code:

if($con = mysql_connect('localhost', 'Rushi', 'bhanage'))
	echo "connection successful!!<br>";
else
    die (mysql_error());
	 
 $db = mysql_select_db("rohit25_honestfollowers", $con);
 	echo "Database successfully selected!!<br>";
$sql = mysql_query("select * from usersearchmaster") or die (mysql_error());
echo $sql;
 $dishf = array();
  while($res = mysql_fetch_row($exec))
  {
  	echo $dishf[] = $res[DishoestFollower]."<br>";
  }

[code green]

You are using the wrong variable for the result resource

$sql = mysql_query("select * from usersearchmaster") or die (mysql_error()); 
echo $sql; 
 $dishf = array(); 
  while($res = mysql_fetch_row($exec))

$exec instead of $sql.

[bandy]

You are using the wrong variable for the result resource

$sql = mysql_query("select * from usersearchmaster") or die (mysql_error()); 
echo $sql; 
 $dishf = array(); 
  while($res = mysql_fetch_row($exec))

$exec instead of $sql.

Hi there,
extremely sorry, it happened while cleaning code, but I tried it with:

    $res = mysql_fetch_row($sql);

but no change in error.

[aung1986]

DishoestFollowe r is the right field name in the table.
first you test output data with"echo $res[DishoestFollowe r]."<br>";"
and then you should add to array.

[bandy]

DishoestFollowe r is the right field name in the table.
first you test output data with"echo $res[DishoestFollowe r]."<br>";"
and then you should add to array.

I got my problem solved. Actually problem was with
mysql_fetch_row ($sql);
when I went with array i.e:
mysql_fetch_arr ay($sql)
I got expectd result.

Thank you all.