使用PHP将数据库中的URL转换为Javascript

I've added a table in my mySQL database which has two variables: ID and a URL.

ID has the value 1 and
URL has http://www.examplesite.com

I want to get "www examplesite com" from the database into a PHP file.

The PHP file should then send this string to a javascript file which will then open that URL.

So far I've been using getJSON with little success.
I'm new to PHP and Java and would really appreciate some help!

I want something like this in my .js file

$.getJSON('getlink.php', {'link'}, function(e) {    
alert('Result from PHP: ' + e.result);    
});


window.open(linkVariable'_blank')    

I would like linkVariable to be www examplesite com

The javascript is linked to another php file which has a clickable element for the window.open.

How can I get the getlink.php and the .js file to communicate with each other?

EDIT:

My getlink.php would look something like this without any echo. The connection to mySQL is already written.

function get_links($url_link) {
$sql = "SELECT `name` FROM `variables` WHERE ID=?";
$res = $this->db->query($sql, array($url_link))->row_array();
return $res['URL'];

it would be better if you posted what code is in getlink.php.

I'd suggest you proceed like this. In getlink.php query the database to return the correct row.

There is different ways to do that, depending on what framework or library (PDO,mysql_,) you are using.

Then return the result as json

In the same file, call the function you just defined. It should look to something like this

<?php

function get_links($url_link) {
$sql = "SELECT `name` FROM `variables` WHERE ID=?";
$res = $this->db->query($sql, array($url_link))->row_array();
return $res;
}

$result = get_links(1); //whatever the id is
return json_encode($result);
?>

In your javascrit you can do something like this

$.get( 
     "getlink.php",
      function(data) {
         alert(data.URL);
       }
   );

//My Idea
Read URL from the database and give it to the variable $link code below

 $result=mysqli_query($dbconnection, "SELECT * FROM url"); //You can specify conditions
       while($row=mysqli_fetch_array($result)){
        $link=$row["URL"]; //We have our url from database here
    }

//Link between JAVASCRIPT AND PHP Put this code in a php to take $link and give it to getlink() function as a parameter

<script>
 getlink("<?php echo $link?>"); //A java script function from your JS file to get url
</script>