PHP - MySql SELECT WHERE value = string JSON - php

Question

through a cURL connection, I can pick up data, from Json files, placed on a remote server. In particular, the codes of some products, which thanks to a foreach
foreach($data['results'] as $key=>$val){
$codici_hotel = $val['hotel_code'];
echo $codici_hotel.",";
}
I can see on video:
1074d0,19f726,1072ba,107104,183444,112438,15d8ab,1b326e,19d885,189b95,1071bf,107155,193e61,10aab2,138752,18dd7d,19d7f9,117b0d,1071b8,1398c4,107039,110851,107124,110669
Now I need to use that string to run a select on a local database, such as:
$sql = "SELECT * FROM hotels WHERE code = ('$codici_hotel')";
What is the correct sql string?
Thanks for your help
CODE UPDATE USED
$codici_hotel_arr = array();
foreach($data['results'] as $key=>$val){
$codici_hotel_arr[] = $val['hotel_code'];
}
$codici_hotel = "'".implode(",",$codici_hotel_arr)."'";
$conn2 = new mysqli($servername, $username, $password, $dbname);
if ($conn2->connect_error) {
die("Connection failed: " . $conn2->connect_error);
}
$sql2 = "SELECT name FROM hotels WHERE code IN ('$codici_hotel')";
$result2 = $conn2->query($sql2);
if ($result2->num_rows > 0) {
// output data of each row
while($row2 = $result2->fetch_assoc()) {
$nome_hotel = $row2["name"] ;
}
} else {
echo "0 results";
}
$conn2->close();
echo $nome_hotel;

Answer 1

You have to convert your all codes in string enclosed with '. Then use IN clause of mysql. change your code as below
$codici_hotel_arr = array();
foreach($data['results'] as $key=>$val){
$codici_hotel_arr[] = $val['hotel_code'];
}
$codici_hotel = "'".implode(",",$codici_hotel_arr)."'";
$sql = "SELECT * FROM hotels WHERE code IN ($codici_hotel)";