1

I am extremely new to programming (started my first app 4 days ago) and am having an issue showing MySQL data on the HTML page. The app idea is a simple random restaurant generator. I am manually inputting 50 or so restaurants to put together a quick prototype. I used the code below (copied and tweaked from a YouTube video) to randomly generate a restaurant from the dinner "din" category (It is not a column in MySQL, just a keyword in the rows to hint this should show if dinner is selected over breakfast) within the Nashville location DB.

I am getting the message

"Warning: mysql_num_rows() expects parameter 1 to be to mysql_result"

. What do I need to add to resolve this?

Any help would be greatly appreciated! 

<?php
   $sql = "SELECT * FROM nashville WHERE name IN din ORDER BY rand() LIMIT 1";
   $result = mysqli_query($conn, $sql);
   $resultCheck = mysqli_num_rows($result);

   if ($resultCheck > 0){
        while ($row = mysqli_fetch_assoc($result)){
            echo $row['name'] . "<br>";
        }
   }
    ?>
Lizesh Shakya
  • 2,482
  • 2
  • 18
  • 42
AAndrews
  • 9
  • 2
  • If your error really is `Warning: mysql_num_rows() expects parameter 1 to be to mysql_result` than you have shown us the wrong code. Having a table for every city seems like a bad DB design. Have you run your query in your db, I don't you'll get back what you think. – user3783243 Jul 03 '19 at 03:10

0 Answers0