Fatal error: Function name must be a string in line 12 - php

Question

Edited: This is the error I got.
$categoryName = $_POST['category_name'] ;
$categoryDesc = $_POST['category_desc'] ;
$sql = "INSERT INTO category (category_title, category_desc) VALUES ('$categoryName','$categoryDesc')";
if (mysqli_query($con,$sql))
{
echo 'Inserted successfully';
}
else
{
echo 'Inserted Failed';
}
mysqli_close($con);
?>
The error I got is Fatal error: Function name must be a string in line 12
line 12 : $categoryName = $_POST['category_name'] ;
[EDITED]
dbconnect.php (I am not sure that I am right or wrong because I am using virtual host, that's why my servername is my virtual host name)
<?php
$servername = "wp-one";
$username = "root";
$password = "";
$dbName = "personality_test";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbName);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>
html code
<div class="col-lg-6">
<form role="form" action="../controller/AddCategory&Question.php?create=true" method="post">
<div class="form-group">
<label>Category Name</label>
<input class="form-control" type="text" placeholder="CategoryName" name="category_name" />
</div>
<div class="form-group">
<label>Category Description</label>
<input class="form-control" type="text" placeholder="CategoryDesc" name="category_desc" />
</div>
<div class="form-group input_fields_wrap">
<button class="add_field_button btn btn-default" style="margin-bottom:10px;">Add New Question</button>
<div class="form-group">
<div class="row">
<div class="col-lg-2"><input type="text" placeholder="Number" class="form-control" name="criteria[]"></div>
<div class="col-lg-5"><input type="text" placeholder="Question" class="form-control" name="grade[]"></div>
</div>
</div>
</div>
<button type="submit" class="btn btn-default">Submit</button>
<button type="reset" class="btn btn-default">Reset</button>
Please help me, Thank You so much

Answer 1

$conn = new mysqli(...);
so you must use $conn instead of $con
if (mysqli_query($conn,$sql))
{
echo 'Inserted successfully';
}
else
{
echo 'Inserted Failed';
}
mysqli_close($conn);

Answer 2

whats your data type for category_title in database?
Your Error in connection
// Create connection
$conn = new mysqli($servername, $username, $password, $dbName);
you use $conn and in insert file you use $can instead of $conn...