Jquery token-input not working when populating data from database

Question

I am trying to populate data from database using jquery token-input plugin http://loopj.com/jquery-tokeninput/ .but nothing shows up. I wish to know where i am going wrong.

I ACCEPT ANY ALTERNATIVE APPROACH

Every example i search does not seem to work.

My index page

<script type="text/javascript" 
        src="https://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js">
</script>
<script type="text/javascript" src="src/jquery.tokeninput.js"></script>

<link rel="stylesheet" href="styles/token-input.css" type="text/css" />
<link rel="stylesheet" href="styles/token-input-facebook.css" type="text/css" />
<div>
<input type="text" id="tagcool"   name="blah" />
<input type="button" value="Submit" />       
<script type="text/javascript">
$(document).ready(function() {                                  
   $("#tagcool").tokenInput("phpsearch.php", {
   theme: "facebook",
   preventDuplicates: true,              
});

});
</script>

And my PHP page

    <?php
$link = mysqli_connect("localhost","root","","dbname") or die("Couldn't make connection.");

$look = $_GET['q']; 
$arr = array();
$rs = mysqli_query($link,"SELECT * FROM `country` WHERE name like '%".$look."%'");
# Collect the results
while($obj = mysqli_fetch_object($rs)) {
    $arr[] = $obj;
}
# JSON-encode the response
$json_response = json_encode($arr);
# Optionally: Wrap the response in a callback function for JSONP cross-domain support
if($_GET["callback"]) {
    $json_response = $_GET["callback"] . "(" . $json_response . ")";
}
# Return the response
echo $json_response;

?>

Example of my db

first remove extra comma in plugin options(not an answer) & explain more what exactly is not working, what is there on console & more debug info — vinayakj
– vinayakj, Commented Jul 17, 2016 at 14:28
It doesn't return any error or result. just says searching... — Omari Victor Omosa
– Omari Victor Omosa, Commented Jul 17, 2016 at 14:32
whats in the network tab, what does ajax status shows? can you create a jsfiddle? — vinayakj
– vinayakj, Commented Jul 17, 2016 at 14:33
Select only needed field SELECT id, name FROM country and the jQuery version is outdated, at least use 1.7+ ! — Ismail RBOUH
– Ismail RBOUH, Commented Jul 17, 2016 at 14:49

Daksh B · Accepted Answer · 2016-07-23 11:28:39Z

This works. You need to change a bit of the coding below according to your setup.

HTML

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="styles/token-input-facebook.css" type="text/css" />
</head>
<body>
<form name='jqueryAutocompleteForm' id='jqueryAutocompleteForm'>
    <label for='query'>Search:</label>
    <input type='text' name='q' class='form-control input-lg' id='query' placeholder='Please Start Typing'>
    <input type='submit' value='Submit' class='btn btn-info'>
</form>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script> 
<script type="text/javascript" src="src/jquery.tokeninput.js"></script> 
<script type="text/javascript">
        $(document).ready(function() {
            $("#query").tokenInput("php.php", {
                theme: "facebook"
            });
        });
        </script>
</body>
</html>

Note: Be sure the path to your "jquery.tokeninput.js" and the "php to json", in my case "php.php" and yours "phpsearch.php" is correct.

The php.php

<?php

    //open connection to mysql db
    $connection = mysqli_connect("localhost","root","","tags") or die("Error " . mysqli_error($connection));


    $s = $_GET["q"];
    //fetch table rows from mysql db
    $sql = "SELECT name from mytable WHERE name LIKE '%".$s."%'";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray = array();
    while($row =mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }
    echo json_encode($emparray);

    //close the db connection
    mysqli_close($connection);


?>

Replace the select query, "$sql = "SELECT name from mytable WHERE name LIKE '%".$s."%'";" with yours.

In case you need to know, my database has just 1 table with 2 columns 1.id, 2, name.

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