PHP and Mysql not echoing my variable

Question

Not sure if I have coded this incorrectly or if it is just because I have a very large database that I am trying to pull from.

I'm not receiving any errors but there is just nothing being displayed on the page. Is having a large database an issue which might prevent me from being able to find my result. I know that the postcode is within the database as I have found it in PHPMYADMIN SQL tab.

<?php

$location = $user['location'];

$postcode = DB::query("SELECT * FROM postcodes WHERE Postcode LIKE '%" . $location . "%'");

$longitude = $postcode['Longitude'];

echo $longitude;

?>

I'm getting the $user['location'] from a query which is already loaded in and echos on the page shows the postcode on the page already.

I'm very new to PHP and MYSQL so am trying to learn but when it isn't giving me any errors it makes it very hard to search for what I am looking for.

Thank you very much - Jonny Dommett

EDIT - Table Schema

id          int(11)         NO      PRI     NULL    auto_increment  
Postcode    varchar(8)      NO              NULL        
Latitude    decimal(9,6)    NO              NULL        
Longitude   decimal(9,6)    NO              NULL        

EDIT - PDO

    <?php
  class DB{

    private static function connect(){
      $pdo = new PDO('mysql:host=localhost;dbname=vapoural_wsc;charset=utf8','testing','testing123');
      $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
      return $pdo;

    }
    public static function query($query, $params = array()){
      $statement = self::connect()->prepare($query);
      $statement->execute($params);


   if (explode(' ', $query)[0] == 'SELECT') {
  $data = $statement-> fetchAll();
  return $data;
}

    }
  }

 ?>

EDIT - VAR DUMP

var_dump($postcode);

produced a result of

array(0) { }

EDIT - SQL WITHIN PHPMYADMIN

EDIT - The ISSUE = FIXED

The Issue was from within my database on where I had entered my data. Make sure to check your columns by doing VARDUMP to see what that results first.

Answer 1

Your error was in because of your typo in the variables $longitude/$logitude and in the array key Logitude.

Here is a working code, tested on my machine.

<?php
class DB{
    private static function connect(){
      $pdo = new PDO('mysql:host=localhost;dbname=vapoural_wsc;charset=utf8','testing','testing123');
      $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
      return $pdo;

    }
    public static function query($query, $params = array()){
      $statement = self::connect()->prepare($query);
      $statement->execute($params);
      if (explode(' ', $query)[0] == 'SELECT') {
          $data = $statement-> fetchAll();
          return $data;
      }
    }
}


$location = '4200';
$query = "SELECT * FROM postcodes WHERE Postcode LIKE ?";
var_dump(DB::query($query, ['%'.$location.'%']));

Output :

array(1) {
[0]=>
array(8) {
  ["id"]=>
  string(1) "1"
  [0]=>
  string(1) "1"
  ["Postcode"]=>
  string(5) "42000"
  [1]=>
  string(5) "42000"
  ["Latitude"]=>
  string(4) "2344"
  [2]=>
  string(4) "2344"
  ["Longitude"]=>
  string(4) "2334"
  [3]=>
  string(4) "2334"
  }
}

The output is an example of random datas that I've inserted in my database.

Important note : this code use a prepared statement. You are using a variable in your query, so please use a prepared statement to avoid SQL injection.

In this code I've set $location to a static value to be sure the provided code works.