Getting error when trying to fetch data from MySQL using PHP - php

Question

When trying to fetch data from a Mysql database using PHP, the following code gets a message:
Getting error:
undefined variable result.
<?php
require_once 'login.php';
$conn = new mysqli($hn, $un, $pw, $db);
if ($conn->connect_error) die($conn->connect_error);
echo <<<_END
<form action="fetchdata.php" method="post"><pre>
Enter Country <input type="text" name="field">
<input type="submit" value="Display Records">
</pre></form>
_END;
if (isset($_POST['field'])) {
$field=$_post($conn,'field');
$query="SELECT * FROM customers WHERE Country = '$field'";
$result=$conn->query($query);
if (!$result) die($conn->error);
}
$rows = $result->num_rows;

Answer 1

Instead of
$field=$_post($conn,'field');
Maybe you mean
$field=$_POST['field'];
Additionally you use $result at the end, even when it is not defined:
$rows = $result->num_rows;
Also you never output any data and print it. You just store it in variables.
In any case: What you are doing there by writing form data directly into a query string, is dangerous. I recommend you to use PDO together with Named Parameters. Also maybe read up about SQL injections.
Here is another stackoverflow question with a nice answer, regarding SQL injections. It includes both PDO and mysqli: https://stackoverflow.com/a/60496/6637731

Answer 2

Without telling how request to this code is done, one guess is that isset($_POST['field']) returns false, hence variable $result is never defined but you use it anyway below in $result->num_rows.

Answer 3

You made a mistake in the below line. It should be
$field=$_POST['field'];
not
$field=$_post($conn,'field');