PHP MySQL WHERE column-value is in $_POST - php

Question

I'm trying to get all the rows which contain a particular text. However, when I execute the query, no rows are returned. I'm retrieving the text from a post request which looks like this "Krachttraining,Spinning" (= 2 values). I think my code fails on the following part (if I leave this out, the query returns some rows): AND CONCAT('%', sport.name, '%') LIKE $sports.
FYI. I know you can perform SQL injection on this, this will be fixed later.
<?php
$servername = "SECRET";
$username = "SECRET";
$dbpassword = "SECRET";
$dbname = "SECRET";
$lat = $_POST['lat'];
$lng = $_POST['lng'];
$sports = $_POST['sports'];
echo $sports; //Echo's: Krachttraining,Spinning.
// Create connection.
$conn = new mysqli($servername, $username, $dbpassword, $dbname);
// Check connection.
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT gym.id FROM gym, sport, gym_sport WHERE lat BETWEEN '$lat'-1 AND '$lat'+1 AND lng BETWEEN '$lng'-1 AND '$lng'+1 AND gym.id = gym_sport.gym_id AND sport.id = gym_sport.sport_id AND CONCAT('%', sport.name, '%') LIKE $sports";
$result = $conn->query($sql);
$output = array();
if ($result->num_rows > 0) {
// output data of each row.
while($row = $result->fetch_assoc()) {
$id = $row["id"];
array_push($output, $gym);
}
//Brackets because of GSON's parser.
echo "[" . json_encode($output) . "]";
}
$conn->close();
?>
EDIT: Changed SQL statement to:
$sql = "SELECT * FROM gym, sport, gym_sport WHERE lat BETWEEN '$lat'-1 AND '$lat'+1 AND lng BETWEEN '$lng'-1 AND '$lng'+1 AND gym.id = gym_sport.gym_id AND sport.id = gym_sport.sport_id AND sport.name LIKE '%".$sports."%'";
Still getting 0 rows returned.
EDIT 2: I ran the following code in my phpMyAdmin environment, and it returned 0 rows.
Select * FROM sport WHERE name LIKE '%Krachttraining,Spinning%';
However when I'm running the following code, it returns 1 row:
Select * FROM sport WHERE name LIKE '%Krachttraining%';
I don't really get it what I'm doing wrong, any thoughts?

Answer 1

I think you want to use the IN statement.
This will check if any word in the array matches. For instance:
Select * FROM sport WHERE name IN ('Spinning', 'Krachttraining'); Will return every row which has the name Spinning or Krachttraining.

Answer 2

Just use:
SELECT .FROM...WHERE AND sport.name LIKE '%".$sports."%'";
After question editing
After you changed the question, I suggest to take a look at this answer to better understand what you should to do: https://stackoverflow.com/a/3902567/1076753
Anyway I think that you have to learn a bit about the like command: http://www.mysqltutorial.org/mysql-like/

Answer 3

Change the sql to :
$sql = "SELECT gym.id FROM gym, sport, gym_sport WHERE lat BETWEEN '$lat'-1 AND '$lat'+1 AND lng BETWEEN '$lng'-1 AND '$lng'+1 AND gym.id = gym_sport.gym_id AND sport.id = gym_sport.sport_id AND sport.name LIKE '%".$sports%."'";