User clicks button:
<input type="image" src="img/like.png" alt="Like" value="Like">
<input type="hidden" name="IP" value="<?php $_SERVER ["REMOTE_ADDR"] ?>">
Save into database (errors are in comments):
<?php
$servername = "localhost";
$username = "username";
$password = "password";
if (!empty($_POST)) {
$connection = mysqli_connect ($servername, $username, $password);
$statement = mysqli_prepare ($connection, "INSERT INTO Like (User, PageId) VALUES (?, ?)");
mysqli_stmt_bind_param ($statement, "si", $_POST[IP], $_GET[id]);
//mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, boolean
mysqli_stmt_execute ($statement);
//mysqli_stmt_execute() expects parameter 1 to be mysqli_stmt, boolean
exit;
}
?>
Display like amount:
<?php
$connection = mysqli_connect ($servername, $username, $password);
$statement = mysqli_prepare ($connection, "SELECT * FROM Like WHERE PageId=$_GET[id];");
?>
Everything looks right to me, but I'm new and have a hard time learning PHP/SQL.
Can O' Spam :
The error is found by looking at your connection, You don't connect to a database, what you have is:\n\n$connection = mysqli_connect ($servername, $username, $password);\n\n\nWhere what you need is:\n\n$database = 'my_db';\n$connection = mysqli_connect ($servername, $username, $password, $database);\n",
2016-01-18T09:42:26