Validation for display message "no result found"

Few Suggestions:

• Stop using mysql_* extension, its deprecated and close in PHP 7, use mysqli_* or PDO.
• try to debug print_r($_POST) if need.
• You must need to use IF for ELSE condition.
• When you are using HTML Form input directly in your Query Statement, you need to learn mysqli::real_escape_string

Here is the complete example of your code by using MYSQLi Object Oriented:

Example:

<?php    
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$yourInput = $conn->real_escape_string($_POST['compName']);
$sql = "SELECT * FROM user WHERE username='$yourInput'";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        //your stuff
    }    
} 
else 
{
    echo "no record found";
}
$conn->close();

?>

Please try this

include("incls/connector.php"); 

$result = mysql_query("SELECT * FROM user WHERE username='$_POST[compName]'");

$num_rows = mysql_num_rows($result); //Check for total number of rows

if($num_rows){
 while($rowval = mysql_fetch_array($result))
    {
        $complainer = $rowval['username'];
        $department = $rowval['department'];
        $phonenumber = $rowval['phone_number'];
    }   
  }
else { echo 'No Results were found'; }

mysql_close();

please use if for else statement

if( $rowval ) {
} else {
   echo "No Rows Found";
}

Get the count of records firstly to display the records. Also add the code snippet in a try catch block to track the exceptions without displaying them on UI.

try {
$result = mysql_query("SELECT * FROM user WHERE username='$_POST[compName]'");

$count = mysql_num_rows($result); //get the count of rows
if($count > 0)
  {
        while($rowval = mysql_fetch_array($result))
       {
        $complainer = $rowval['username'];
        $department = $rowval['department'];
        $phonenumber = $rowval['phone_number'];
       }
   }    
   else
   {
        echo 'No Results were found';
   }    
    mysql_close();  // Missing semicolon in your code
} catch (e){
   print_r("Error while getting the results.");
}
?>

There is a syntax error in your code. Else has no if

if(count($result)==0){
   echo 'No Results were found';
}

Moreover, checkout sql injection you are passing in line parameter which can lead to sql injection