Use POST method to display data from database PHP - php

Question

I am using this SQL query in a link to retrieve data from database
<div class="nav-laptop">Laptop
and display it using
$sql = $_REQUEST['upit'];
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<div class='proizvodi'>";
// output data of each row
$result->data_seek(0);
while($row = $result->fetch_assoc()) {
echo "<div class='row'>";
foreach($row as $key => $value){
echo "<div class='" . $key . "'>" . $value . "</div>";
}
echo "</div>";
echo "<hr />";
}
echo "</div>";
}
else {
echo "<div class='search-query-none'><img src='index/no result.png' width='754' height='198' /></div>";
}
I realized this is very vulnerable and that I should use POST method to hide parameters from URL. I tried reading online forums, but I found nothing that would help me to convert this to POST way of retrieving data.
So, how do I use POST method to achieve the same result as I am achieving right now using GET?

Answer 1

This will give you a general idea on how to do this.
HTML form:
<form method="post" action="your_handler.php">
<input type = "text" name = "search_query">
<input type = "submit" name = "submit" value = "Search">
</form>
SQL/PHP and assuming a successful connection using the MySQLi API.
$conn = mysqli_connect("your_host", "user", "password", "db");
if (!$conn) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}
if(isset($_POST['submit'])){
if(!empty($_POST['search_query'])){
$search_query = mysqli_real_escape_string($conn, $_POST['search_query']);
$result = mysqli_query($conn, "SELECT * FROM TABLE WHERE col = '$search_query' ");
if(!$result) { echo "Error: " . mysqli_error($conn); }
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
// perform what you want here
// and check for errors on your query
}
}
}
}
You can substitute SELECT * with the said columns also.
Ideally, a prepared statement is nice to work with.
http://www.php.net/manual/en/mysqli.quickstart.prepared-statements.php
http://php.net/pdo.prepared-statements (if you want to look into PDO).
Sidenote: Do not intermix different MySQL APIs such as mysqli_ with PDO. They just don't mix together.
Check for errors also against your query:
http://php.net/manual/en/mysqli.error.php
Add or die(mysqli_error($conn)) to mysqli_query().
Add error reporting to the top of your file(s) which will help find errors.
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
// Then the rest of your code
Sidenote: Displaying errors should only be done in staging, and never production.
Plus, make sure that no whitespace gets introduced into your input, otherwise your query may fail.
Use trim() against the input.

Answer 2

You don't need to use POST for a SELECT query. You can, but it's really better suited for INSERT / UPDATE / DELETE, things that actually change your data. A possible advantage to using a link like that for search results is that it can be saved, bookmarked, emailed, etc., where a form submission cannot. But you are right that putting your entire query into a link like that definitely is extremely vulnerable.
Instead of passing the entire query through the link, you can just pass the parameters, like this:
Laptop
Then in your display code you can use a prepared statement and safely bind the parameter:
$kategorija = $_GET['kategorija'];
$sql = 'SELECT Slika, Naziv, Opis, Cijena FROM Proizvodi
WHERE Kategorija=? ORDER BY Proizvodac';
$stmt = $conn->prepare($sql);
$stmt->bind_param("s", $kategorija);
$stmt->execute();
// etc.