Why does my echo not work given no results found?

Question

I am performing a routine to check if books exist in a database given a range. I want to echo if no books are found. I have the following:

$search = mysqli_query($con,$query);

while(list($book_id, $title, $authors, $description, $price) = mysqli_fetch_row($search)){

if(!empty($book_id)) {

echo "Book ID: " . $book_id . "<br/>";
echo "Book Title: " . $title . "<br/>";
echo "Authors: " . $authors . "<br/>";
echo "Description: " . $description . "<br/>";
echo "Price: " . $price . "<br/>";
echo "<br/>";

}

if(empty($book_id)){
    echo "Fail";
    }
}

If no books are found nothing is printed. The echo does not work? How come?

Thanks

Answer 1

Because if no records are returned, you won't enter in the while at all, as $book_is will contain the value false and while(false)... you know

In this situation you may use mysqli_num_rows to check if there are rows found

Answer 2

Your no results echo is inside the while loop which will never get called

while (list($book_id, $title, $authors, $description, $price) = mysqli_fetch_row($search)) {

  if (!empty($book_id)) {
    echo "Book ID: " . $book_id . "<br/>";
    echo "Book Title: " . $title . "<br/>";
    echo "Authors: " . $authors . "<br/>";
    echo "Description: " . $description . "<br/>";
    echo "Price: " . $price . "<br/>";
    echo "<br/>";
  }

}

if (empty($book_id)) {
  echo "Fail";
}

Move it outside like the code above

Answer 3

It is because your while loop has not been accessed at all. Try to echo something out of the if conditions, and see if it get's displayed.