How to get variables from URL and insert into database - php

Question

So I am trying to get the variables from the URL (http://example.com/pb.php?id=123&affiliate=abd123&lp1=dun.com&lp2=dun2.com&lp3=dun3.com) and Ive tried this code but I receive this error
Prepare failed: (1136) Column count doesn't match value count at row 1
Fatal error: Call to a member function bind_param() on boolean in /home/recondes/public_html/postback.php on line 25
and also
<?php
define("MYSQL_HOST", "localhost");
define("MYSQL_PORT", "3306");
define("MYSQL_DB", "db");
define("MYSQL_TABLE", "tbl");
define("MYSQL_USER", "user");
define("MYSQL_PASS", "pass");
$mysqli = new mysqli(MYSQL_HOST, MYSQL_USER, MYSQL_PASS, MYSQL_DB);
if ($mysqli->connect_errno)
{
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
$id = $_GET['id'];
$affiliate = $_GET['affiliate'];
$lp1 = $_GET['lp1'];
$lp2 = $_GET['lp2'];
$lp3 = $_GET['lp3'];
if (!($stmt = $mysqli->prepare("INSERT INTO ".MYSQL_TABLE." VALUES (id, affiliate, lp1, lp2, lp3);")))
{
echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
$stmt->bind_param('dds', $id, $affiliate, $lp1, $lp2, $lp3 );
if (!$stmt->execute())
{
echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
else
{
printf("%d Row updated, added ".$id." to ".$affiliate." .\n", mysqli_stmt_affected_rows($stmt));
}
?>

Answer 1

Your query doesn't list the columns to be inserted into, so it expects you to supply values for all the table columns. You haven't shown the table schema, but it doesn't have only 5 columns.
You're also missing the placeholders that will get filled in by bind_param(). I suspect the values you listed in VALUES() were intended to be the table columns. So try:
if (!($stmt = $mysqli->prepare("INSERT INTO ".MYSQL_TABLE." (id, affiliate, lp1, lp2, lp3) VALUES (?, ?, ?, ?, ?)")))
Also, in your call to bind_param, the string that specifies the datatypes needs to have as many letters as there are parameters. So it should be:
$stmt->bind_param('dssss', $id, $affiliate, $lp1, $lp2, $lp3 );
Finally, when you get an error in one step, and you print the error message, you should stop this script rather than going on to the next step. It makes no sense to use the prepared statement if prepare() fails.