Passing value of variable to next page using post method - php

Question

In a code below I want to pass variable $user_id to next page using post method suppose my next page is followers.php . The whole code is link itself i want to put post method in it so that on clicking the link, variable is also passed to follower.php . Don't want to use sessions.
<a href='follower.php' style='text-decoration:none;'>
<h2>Followers</h2>
<?php
$checkfollowers = "SELECT * FROM follow_user WHERE user_id='$user_id'";
$resultfollowers = mysqli_query($con,$checkfollowers);
echo mysqli_num_rows($resultfollowers );
?></a>
I guess code for post method is correct:
<form method="get" action="follower.php">
<input type="hidden" name="varname" value="user_id">
<input type="submit">
</form>
So how to put this post method inside the other code so on clicking, follower.php is opened and variable is also passed to this page.

Answer 1

You are using GET method. Change it to
<form method="POST" action="follower.php">
<input type="hidden" name="user_id" value="user_id">
<input type="submit">
</form>
Also change the name to user_id of the hidden field. Now receive as
<?php
$user_id = $_POST['user_id'];
$checkfollowers = "SELECT * FROM follow_user WHERE user_id='$user_id'";
$resultfollowers = mysqli_query($con,$checkfollowers);
echo mysqli_num_rows($resultfollowers );
?>
Note: Above query is vulnerable to SQLI. Better to use Prepared statements.

Answer 2

You are talking about post method but in your code you are using the method GET. You might want to consider to change this into POST. This way the stuff filled out in the form will not enter the URL. (bit more secure). After this you could call the global variable $_POST['user_id']

Answer 3

A link cannot POST data, only GET.
use:
<a href='follower.php?user_id=1' style='text-decoration:none;'>
and in follower.php file:
$user_id = $_GET['user_id'];