Parse error: syntax error, unexpected '$result' (T_VARIABLE) in my code

Question

What is going wrong with this code and I get this message about unexpected '$result' ??? I can't get rid of it. I searched a lot about this issue but didn't find a solution.

<?php

//error_reporting(0);

if(isset($_REQUEST['v']) && !empty($_REQUEST['v'])){

    $conx = mysqli_connect("localhost", "root", "mypass", "country_db");

    if (!$conx) {
        echo "Error connecting database!";
        exit;
    }

    $search = $_REQUEST['v'];

    $query = "SELECT `country_name` FROM `apps_countries` WHERE `country_name` LIKE '$search%'";

    $result = mysqli_query($conx, $query); // here is the error

    if (mysqli_num_rows($result) > 0) {

        while ($d = mysqli_fetch_assoc($result)) {
            echo $d['country_name'];
        }

    }

    else{
        echo "Nothing matched with your query";
    }

}

else{
    echo "No data received from the server";
}

?>

You have a BOM at the end of your query line. So just go at the end of the line and press delete once. — Rizier123, Apr 25 '16 at 05:52
change query to: "SELECT `country_name` FROM `apps_countries` WHERE `country_name` LIKE %".$search."%"; — Dhara Parmar, Apr 25 '16 at 05:53
Thank u both for this reply. Rizier123 cause of you, I made it. Thank u very much. — Danis35, Apr 25 '16 at 12:36

Parse error: syntax error, unexpected '$result' (T_VARIABLE) in my code

0 Answers0