mysql database search remote host or page name [duplicate] - php

Question

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 years ago.
The purpose of my lab is to have the user search through a database called "pageVistis" from what they enter in the field. They can search through the remote host or page name. I have Been having trouble fixing an error that i am recieving saying "Notice: Undefined variable: searchtype in C:\xampp\htdocs\pageVisitsLab\DataBC.php on line 23
Fatal error: Uncaught Error: Call to a member function bind_param() on boolean in C:\xampp\htdocs\pageVisitsLab\DataBC.php:25 Stack trace: #0 {main} thrown in C:\xampp\htdocs\pageVisitsLab\DataBC.php on line 25". Also the code is form a file called "DataBC.php". If any extra information is needed please let me know and i will edit this post and include it thank you.
include "interface.php";
if(isset($_POST['submit'])) {
$sName = $_POST['sN'];
$sHost = trim($_POST['sH']);
if(!$sName || ! $sHost) {
echo "<p> Please enter a search </p>";
exit;
}
/*switch ($sName) {
case 'sName':
case 'sHost':
break;
default:
echo "<p> Invalid search type and please try again </p>";
}
*/
$query = "SELECT page_name, visit_date, previous_page, request_method, remote_host, remote_port FROM visitInfo
WHERE $searchtype= ?";
$stmt = $databse -> prepare($query);
$stmt-> bind_param('s',$sName);
$stmt-> execute();
$stmt->store_result();
$stmt->bind_result($pageName,$visitDate,$previouPage,$requestMethod,$remoteHost,$remotePort);
echo "<p> The number of items found: .$stmt->num_rows</p>";
while ($stmt->fetch()) {
echo "<p><strong>page Name: ".$pageName."</strong></p>";
echo "Visit Date: ".$visitDate."<br>";
echo "Previous Page: ".$previouPage. "<br>";
echo " Request Method: ".$requestMethod. "<br>";
echo "Remote Host: ".$remoteHost. "<br>";
echo "Remote Port: ".$remotePort. "<br>";
}
$stmt->free_result();
$databse->close();
}
?>

Answer 1

I think your problem is in your query, is $searchtype a variable you define in php? maybe in that interface.php.
If you want to use a php variable in a mysql query then that is not the correct way of doing it.