How to dislpay all animals that are status = 1 in PHP - php

Question

I'm making an adoption website for a project.
I'm trying to display all animals whose status = 1 but the animals whose status = 0 keeps appearing in the category.
Is something wrong with my code? suggestions and corrections are welcome.
Thank you!
function.php
function get_animals($cat_id='', $animal_id='')
{
global $con;
$query = "SELECT * FROM animals WHERE status= 1";
if($cat_id!='')
{
$query = "SELECT * FROM animals WHERE category_name='$cat_id'";
}
if ($animal_id!='')
{
$query = "SELECT * FROM animals WHERE id=$animal_id";
}
return $result = mysqli_query($con,$query);
}
This is my animals.php file
<?php
$cat_id = '';
if(isset($_GET['id']))
{
$cat_id = mysqli_real_escape_string($con,$_GET['id']);
}
$particular_animal = get_animals($cat_id);
?>
<!-- End Navigation -->
<!-- Product Grid -->
<div class="container mt-5 ">
<div class="row">
<?php
if(mysqli_num_rows($particular_animal))
{
while($row = mysqli_fetch_assoc($particular_animal))
{
?>
<div class="col-md-4 product-grid">
<div class="row">
<div class="image border border-info bg-light">
<a href="animal_details.php? a_id=<?php echo $row ['id'] ?>">
<img src="admin/image/<?php echo $row['img']?>" class="w-100" alt="">
</a>
<h4 class="text-center mt-2 text-info font-weight-bold"><?php echo $row ['name'] ?></h4>
<p class="text-center mt-2"><?php echo $row ['gender'] ?></p>
</div>
</div>
</div>
<?php
}
}
else
{
echo "record not here";
}
?>
</div>
</div>

Answer 1

You're overwriting your $query (the one with the status) when $cat_id or $animal_id is set. Instead, add to the WHERE clause of your first query:
$query = "SELECT * FROM animals WHERE status= 1";
if($cat_id != '') {
$query .= " AND category_name='$cat_id'";
}
if ($animal_id != '') {
$query .= " AND id=$animal_id";
}

Answer 2

If your function gets a category id or an animal id, then your query is overriden and your status criteria is lost. This is a 1-command solution:
function get_animals($cat_id='', $animal_id='')
{
return mysqli_query($con,"SELECT * FROM animals WHERE status= 1" .
($cat_id ? " AND category_name = {$cat_id} " : "").
($animal_id ? " AND animal_id = {$animal_id} " : ""));
}