PHP Database ajax not updating

Question

A while ago i made a search function with ajax and php. You could fill in a textbox with text and it would try to find a match among all countries stored in the database. Now i am refining the code and making it PDO, but i broke something and i cant find out what.

this is my plain HTML

<head>
    <title>Ajax</title>
    <link href="style/style.css" rel="stylesheet" type="text/css" />
    <link rel="stylesheet" type="text/css" />
    <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
    <script type="text/javascript" src="scripts/Javascript.js"></script>
</head>

<body>
    <div id="main">
            <h1 class="title">Enter your country please</h1>

        <input type="text" id="search" autocomplete="off" onchange="">
            <h4 id="results-text">Showing results for: <b id="search-string">Array</b></h4>

        <ul id="results"></ul>
    </div>
</body>

here is my Jquery and javascript. note i have not changed anything to the HTML nor javascript so it can not by a type error.

$(document).ready(function() {
alert('asdf');

function search() {
    var query_value = $('input#search').val();
    $('b#search-string').html(query_value);

    if(query_value !== ''){
        $.ajax({
            type: "POST",
            url: "search.php",
            data: { query: query_value },
            cache: false,
            success: function(html){
                $("ul#results").html(html);
            }
        });
    }

    return false;
}

$("input#search").live("keyup", function(e) {
    clearTimeout($.data(this, 'timer'));
    var search_string = $(this).val();

    if (search_string == '') {
        $("ul#results").fadeOut();
        $('h4#results-text').fadeOut();
    }

    else {
        $("ul#results").fadeIn();
        $('h4#results-text').fadeIn();
        $(this).data('timer', setTimeout(search, 100));
        };
    });
});

And here is my Search.PHP

<?php
class SearchEngine{

    private $html;

    public function __construct($conn){

        $this->html = '<li class="result">
                            <h3>NameReplace</h3>
                            <a target="_blank" href="ULRReplace"></a>
                        </li>';

        if (isset($_POST["query"])) {
            $search_string = mysql_real_escape_string($search_string);
        }

        else{
            $search_string = 'b';
        }

        if (strlen($search_string) >= 1 && $search_string !== ' ') {

            $query = 'SELECT * FROM country WHERE name LIKE "%' . $search_string . '%"';
            $result = $conn->prepare($query);
            $result->execute();
            $result_array = $result->fetchAll();

                foreach ($result_array as $result) {
                    $display_name = preg_replace("/" . $search_string . "/i", "<b>" . $search_string . "</b>", $result['name']);
                    $display_url = 'sadf';

                    $output = str_replace('NameReplace', $display_name, $this->html);
                    $output = str_replace('ULRReplace', $display_url, $output);
                    echo($output);
                }
        }
    }

    }
?>

The problems:

the Post query is never created, for this i made a isset so for now when there is no Post Query created. It will create a Post Query with value "B".
I think the page never gets updated, but i cant be 100% sure since the post never gets created so there is never a update to the query. And for some reason the results are not placed in there correct spot.

Any help will be much appreciated. Please be gentle i am new to Ajax and i rather want to understand than have the solution. Thank you

if you var_dump($_POST); - is anything displayed? What happens if you try to manually navigate to the PHP Page that updated the SQL? — Dutchie432
– Dutchie432, Commented Jun 29, 2014 at 10:48
I can not var_Dump the post because it never gets created. I tested the javascript.js and the functions within are called but aren't doing what they are supposed to do. and what do you mean with manual navigate to PHP page? — Rien
– Rien, Commented Jun 29, 2014 at 11:02
I asked this question 10 minutes again and it was solved. but now i have another problem. that it returns the HTML twice. But your answer has a few other flaws mentioning my code so i will look into it. — Rien
– Rien, Commented Jul 7, 2014 at 22:06

meda · Accepted Answer · 2014-06-29 21:58:14Z

Are you really posting your search string to a PHP class?

That's not how it works, you need to create an instance of that class and use it.

Create a separate php file for that class, and included it in search.php

You can have a simple class looking like the following:

search_engine.php

<?php
class SearchEngine{

    private $conn;

    function __construct($conn){
        $this->conn = $conn;
    }

    function get_search_results($search_string){
        $query = 'SELECT * FROM country WHERE name LIKE :search';
        $result = $this->conn->prepare($query);
        $result->execute(array(':search'=>'%'.$search_string.'%'));
        $result_array = $result->fetchAll();

        return $result_array;   
    }
}
?>

search.php

if (isset($_POST["query"])) {
    $search_string = $_POST["query"];
    if(strlen($search_string) >= 1 && $search_string !== ' ') {
        include 'search_engine.php'
        $engine = new SearchEngine($conn);
        $results = $engine->get_search_results($search_string);
        foreach($results as $result){
          //do something
        }
    }
}

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