PHP form only shows if user_id is in the database - php

Question

I am working on a PHP form where the user can either add information or update previous information.
When Building I entered some information manually into the database to get things working. Now I am find out that the form will not show up for new users because there is no user_id in the database yet.(I tested and manually added the user_ids and the the form then showed up.)
I believe its because my form is wrapped in a while loop.
How can I change my form(INPUTS) to populate if there is no data for the user yet?
<?php
$user_id = $_SESSION['User']['logged_in'];
$businessProfileQuery = $db->query("SELECT * FROM sellerBusinessProfile
WHERE user_id = $user_id;") or die(mysqli_error($db));
while( $row = $businessProfileQuery->fetch_assoc() ) {
$savedBusinessName = $row['businessName'];
$$savedBusinessAddress = $row['businessAddress'];
?>
<form>
<div class="form-group row">
<div class="col-lg-12">
<p class="content-header">BUSINESS ADDRESS</p>
<p class="seller-profile-text"></p>
<input name="businessName" type="name" class="form-control
seller-input" value="<?php echo $savedBusinessName; ?>">
<input name="businessAddress" type="name" class="form-control"
value="<?php echo $savedBusinessAddress; ?>" >
</div>
</form>
<?php
}
?>

Answer 1

You can count number of rows.
If number of rows less then or equal 0 then there is no data set for this user.
If number of rows greater then 0 then data already available for this user.
<?php
$user_id = $_SESSION['User']['logged_in'];
$businessProfileQuery = $db->query("SELECT * FROM sellerBusinessProfile
WHERE user_id = $user_id;") or die(mysqli_error($db));
if ($businessProfileQuery->num_rows <= 0) {
?>
<form>
<div class="form-group row">
<div class="col-lg-12">
<p class="content-header">BUSINESS ADDRESS</p>
<p class="seller-profile-text"></p>
<input name="businessName`enter code here`" type="name" class="form-control
seller-input" value="">
<input name="businessAddress" type="name" class="form-control"
value="" >
</div>
</form>
<?php
}
?>

Answer 2

Hope to help you!
<?php
$user_id = $_SESSION['User']['logged_in'];
$businessProfileQuery = $db->query("SELECT * FROM sellerBusinessProfile
WHERE user_id = $user_id;") or die(mysqli_error($db));
$savedBusinessName = "";
$savedBusinessAddress = "";
while( $row = $businessProfileQuery->fetch_assoc() ) {
$savedBusinessName = $row['businessName'];
$savedBusinessAddress = $row['businessAddress'];
}
?>
<form>
<div class="form-group row">
<div class="col-lg-12">
<p class="content-header">BUSINESS ADDRESS</p>
<p class="seller-profile-text"></p>
<input name="businessName" type="name" class="form-control
seller-input" value="<?php echo $savedBusinessName; ?>">
<input name="businessAddress" type="name" class="form-control"
value="<?php echo $savedBusinessAddress; ?>" >
</div>
</form>