mysql_num_rows（）期望参数1是第17行的资源[重复]

I know this question has been asked loads of times, and I have looked at the answers but I cant get my code to work correctly. The code I have tried is displayed below.

<?php

include 'connection.php';

$connection=mysqli_connect("localhost","root","","c3392262");

if (isset($_POST['search'])) {

  $search = $_POST['search'];

  if (!empty($search)) {

    $query = "SELECT name FROM products WHERE name LIKE '%$search%'";
    $query_run = mysqli_query($connection, $query);
    if (mysql_num_rows($query_run)>=1) {

      echo 'found';

    } else {

      echo 'No results found';
    }
  }
}

?>

<form name = "searchbar" method = "POST" action = "searchBar.php">
<input name = "search" type = "text" size = "40" maxlength = "50" >

<input type="submit" value="Submit">
</form>

</div>

You are mixing mysql_ and mysqli_ functions. Don't do that, they are incompatible. Stick with mysqli_.

You mixed mysql & mysqli. Use mysqli_num_rows instead

if (mysqli_num_rows($query_run) >= 1) {
    echo 'found';
} else {
    echo 'No results found';
}

1. Make sure you use mysqli_num_rows and not mysql_num_rows
2. You need to prepare the query
3. You can just check if greater than 0

<?php

include 'connection.php';

$connection = mysqli_connect("localhost", "root", "", "c3392262");

if (isset($_POST['search'])) {
    $search = $_POST['search'];
    if (!empty($search)) {
        $query = "SELECT name FROM products WHERE name LIKE ?";
        $query_run = mysqli_prepare($connection, $query);
        mysqli_stmt_bind_param($query_run , 's', '%'.$search.'%');
        mysqli_stmt_execute($query_run );
        if (mysqli_num_rows($query_run) > 0) {
            echo 'found';
        } else {
            echo 'No results found';
        }
    }
}
?>