Php not inserting data - php

Question

my code is not inserting any data on my php, im using a form that will display values but my code in update is not working. please help,
here is my code in php :
if (isset($_POST['update'])) {
$landowner_id = $_POST['landowner_id'];
$firstname = $_POST['firstname'];
$middlename = $_POST['middlename'];
$lastname = $_POST['lastname'];
$municipality = $_POST['municipality'];
$barangay = $_POST['barnagay'];
$areacovered = $_POST['areacovered'];
$sex = $_POST['sex'];
mysqli_query($db, "UPDATE info SET firstname='$firstname', middlename='$middlename', lastnamename='$lastname', municipality='$municipality', barangay='$barangay', areacovered='$areacovered', sex='$sex' WHERE landowner_id=$landowner_id");
$_SESSION['message'] = "Address updated!";
}
here is my html
<div class="form-wrapper">
<input type="number" id = "check" name="firstname" placeholder="First Name" class="input-field" value="<?php echo $landowner_id;?>" required>
</div>
<div class="form-wrapper">
<input type="text" id = "check" name="firstname" placeholder="First Name" class="input-field" value="<?php echo $firstname;?>" required>
</div>
<div class="form-wrapper">
<input type="text" name="middlename" placeholder="Middle Name" class="input-field" value="<?php echo $middlename;?>">
</div>
<div class="form-wrapper">
<input type="text" name="lastname" placeholder="Last Name" class="input-field" value="<?php echo $lastname;?>" required>
</div>
<div class="form-wrapper">
<input type="text" name="municipality" placeholder="Municipality" class="input-field" value="<?php echo $municipality;?>" required>
</div>
<div class="form-wrapper">
<input type="text" name="barangay" placeholder="Barangay" class="input-field" value="<?php echo $barangay;?>" >
</div>
<div class="form-wrapper">
<input type="text" id = "check" name="areacovered" placeholder="Area Covered" class="input-field" value="<?php echo $areacovered;?>" required>
</div>
<div class="form-wrapper">
<input type="text" id = "check" name="sex" placeholder="Sex" class="input-field" value="<?php echo $sex;?>" required>
<br>
<button class="btn" type="submit" name="update" >Update</button>
</div>

Answer 1

I don't see any "form" tag. Are you missing to wrap your "form-wrapper" into a tag? Something like this:
<form action="" method="post">
<div class="form-wrapper">
<input type="number" id="check" name="firstname" placeholder="First Name" class="input-field" value="<?php echo $landowner_id; ?>" required>
</div>
<!-- Other inputs -->
<div class="form-wrapper">
<input type="text" id="check" name="sex" placeholder="Sex" class="input-field" value="<?php echo $sex; ?>" required>
<br>
<button class="btn" type="submit" name="update">Update</button>
</div>
</form>
Other important things to consider:
Never EVER send to the DB plain inputs coming from the outside without cleaning them! Otherwise, you will be open to SQL injection. Use prepare-statements to solve this issue.
Instead of mysqli_query I recommend you to use PDO. You can prepare statement super easy. Here you can see an example of usage: https://stackoverflow.com/a/60988740/3454593