How to fill a template file with data from a table - php

Question

I am looking for some initial direction on this one because I cannot seem to find my way with it. Let me explain... I'm creating a music website and having a search bar. It filters information as the user types. I don't want to make a separate .php file for each song on the website. (Eg: song1.php, song2.php, etc...). There should be one PHP template file, that outputs the webpage for ALL songs. With my code, when I try searching with the search bar, it opens the template file as expected but it fills the file with information of only the first row from the mysql table. This is the form, its in the index page:
<script type = "text/javascript "src = "jquery.js">
<form class="navbar-form navbar-left" >
<div class="form-group">
<input type="text" class="form-control" id="search" placeholder="Search for songs, artists" autocomplete="off">
<div id = "searchresults"> </div>
Then there's the search.js file having two tasks, that is to check if a result has been clicked and also if the user has pressed a key. Its like this:
$('#search').keyup(function()
{
var searchterm = $ ('#search').val();
if (searchterm != '')
{
$.post('search.php', {searchterm:searchterm},
function(data)
{
$('#searchresults').html(data);
});
}
else{
$('#searchresults').html('');
}
});
$('#mylink').click(function(){
var wanted = $('#mylink').val();
$.post('/web/ztemplate.php', {wanted:wanted});
});
I think it's the one having an error but I can't figure out where it is. The template file has this php code :
$search = $_POST['wanted'];
$find = mysql_query("SELECT * FROM search WHERE title LIKE '%$search%'");
$row = mysql_fetch_assoc($find); $title = $row["title"];
There's a search.php file which queries the database to provide information for the instant search. It looks like this :
$search = mysql_real_escape_string(trim($_POST['searchterm']));
if ($search == '' && ' '){
echo 'No results found';
}
else {
$find_videos = mysql_query("SELECT * FROM search WHERE keywords LIKE '%$search%'");
$count = mysql_num_rows($find_videos);
if ($count ==0){
echo 'No Results found for '.$search;
}
else {
while($row = mysql_fetch_assoc($find_videos)){
$title = $row["title"];
$link = $row["link"];
echo "<a href = '$link'><h5 id = 'mylink'> $title <h5> </a> <hr /> ";
}
}
}
Any help is much appreciated.

Answer 1

why you not send an AJAX??
$('#search').keyup(function(){
var searchterm = $ ('#search').val();
if (searchterm != ''){
$.ajax({
type: 'POST',
data: {
"searchterm": searchterm,
"_token":"{{ csrf_token() }}"
},
url: "{{URL::asset('yourPHP')}}",
success: function(response){
$('#searchresults').html(response);
}
});
}else{
$('#searchresults').html('');
}
});
PHP
function searchSong($search =''){
$search = (trim($_POST['searchterm']));
$out='';
if ($search == '' && ' '){
echo 'No results found';
}else {
$sql = "SELECT * FROM search WHERE title LIKE '%$search%'";
$data= DB::select($sql);
$count = count($data);
if ($count ==0){
echo 'No Results found for '.$search;
}else {
foreach ($data as $key => $row) {
$title = $row["title"];
$link = $row["link"];
$out .= "<a href = '$link'><h5 id = 'mylink'> $title </h5> </a> <hr /> ";
}
}
}
return $out;
}