How to define a var in PHP coming from jquery - php

Question

So i already asked some questions here and fixed my problems,but i think they are growing instead of going down,lol.My problem now is i get "Undefined index" error.I will leave picture with the error here.I tried to add variables to define my fsearch.Tried things like
$fsearch = mysqli_escape_string($connect, $_POST['fsearch']);
And more simple like.
$fsearch = $_POST['fsearch'];
But it doesn't work.Here is code.
jQuery(document).ready(function ($) {
$("#food_search").keyup(function(event){
var search_term =$(this).val();
$.ajax({
type:"POST",
url:"/Food-Search",
data:{fsearch:search_term},
success:function(res){
$("#food_search_result").html(res);
},
error: function (xhr, ajaxOptions, thrownError) {
alert(xhr.status);
alert(xhr.responseText);
alert(thrownError);
}
});
});
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<!----------------------------------------------------------------
HTML
----------------------------------------------------------------->
<form method="POST">
<p>Търсене на храни: <input type="text" name="fsearch" id="food_search"></p>
</form>
<!----------------------------------------------------------------
PHP
-----------------------------------------------------------------><?php
$hostname = "localhost";
$username = "name";
$password = "password";
$databaseName = "dbName";
$connect = mysqli_connect($hostname, $username, $password, $databaseName);
if (!empty($_POST['fsearch'])) {
$fsearch = mysqli_escape_string($connect, $_POST['fsearch']);
$req = $connect->prepare('SELECT * FROM food_data_bg WHERE title LIKE "%".$fsearch."%"');
$req->execute(array(
'fsearch'=>'%'.$_POST['fsearch'].'%'
));
if($req->rowCount()==0){
echo 'Не бяха намерени резултати!';
}
else{
while($foodsearch=$req->fetch()){
?>
<div class="search-result">
    <img src="<?php echo $data['fimage']; ?>" class="fimage"/>
    <span class="result-title"><?php echo $data['title'];?></span><br>
    <span class="calories-total"><?php echo $data['calories total'];?></span><br>
</div>
<?php
}
}
}
?>
The code is used for searching,it must get the value of #food-search (input field) on every keyup and send it to PHP,which must check for matches in database and print it.(I think i got 1 more mistake by printing result in PHP page and returning it to java to print it in #food_search_result div)

Answer 1

Hope this helps.
PHP Code
$hostname = "localhost";
$username = "root";
$password = "";
$databaseName = "stackoverflow";
if (!empty($_POST['fsearch'])) {
$fsearch = $_POST['fsearch'];
$mysqli = new mysqli($hostname , $username, $password, $databaseName);
if($mysqli->connect_error)
{
die("$mysqli->connect_errno: $mysqli->connect_error");
}
$query = "SELECT * FROM food_data_bg where title LIKE ?";
$stmt = $mysqli->stmt_init();
if(!$stmt->prepare($query))
{
print "Failed to prepare statement\n";
}
else
{
$searchVar = "%" . $fsearch . "%";
$stmt->bind_param("s", $searchVar);
$stmt->execute();
$result = $stmt->get_result();
if(!$result->num_rows){
echo 'Не бяха намерени резултати!';
}
while ($row = $result->fetch_array())
{
?>
<div class="search-result" style='clear:both; border:1px solid red;'>
<img src="<?php //echo $row['fimage']; ?>" class="fimage"/>
<span class="result-title"><?php echo $row['title'];?></span><br>
<span class="calories-total"><?php echo $row['calories total'];?></span><br>
</div>
<?php
}
}
$stmt->close();
$mysqli->close();
}
HTML Code
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
jQuery(document).ready(function ($) {
$("#food_search").keyup(function(event){
var search_term =$(this).val();
console.log(search_term);
$.ajax({
type:"POST",
url:"Food-Search.php",
data:{fsearch:search_term},
success:function(res){
$("#food_search_result").html(res);
console.log(res);
},
error: function (xhr, ajaxOptions, thrownError) {
alert(xhr.status);
alert(xhr.responseText);
alert(thrownError);
}
});
});
});
</script>
<!----------------------------------------------------------------
HTML
----------------------------------------------------------------->
<form method="POST">
<p>Търсене на храни: <input type="text" name="fsearch" id="food_search"></p>
</form>
<!----------------------------------------------------------------
PHP
----------------------------------------------------------------->
<div style="border:#FF0000;clear:both; padding:10px;min-height:300px;" id="food_search_result"></div>