Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result

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PHP

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

        # Save-button was clicked
        $vname = $_POST['vendorname'];

        // connect to mysql
        $link =  mysqli_connect("localhost", "root", "", "magna-billing");
        // mysql search query
        $query = "SELECT  `vendoraddress`, `vendorcontact`, 'vendorgst', 'vendorstateid', 'vendorcode' FROM `vendordetail` WHERE `vendorname` = $vname ";
        $result = mysqli_query($link, $query) ;
        if(mysqli_num_rows($result) > 0)
        {
        while ($row = mysqli_fetch_array($result))
        {

          $vadd = $row['vendoraddress'];
          $vcont = $row['vendorcontact'];
          $vgst = $row['vendorgst'];
          $vstate = $row['vendorstateid'];
          $vcode = $row['vendorcode'];

}

}

mysqli_free_result($result);
mysqli_close($link);

else {
  echo "Undefined";

}

}

 else {
  $vadd = "";
  $vcont = "";
  $vgst = "";
  $vstate = "";
  $vcode = "";
}




?>

What I have tried:

I am trying to show the database value in the text box after clicking on search button, Here I stuck somewhere. Getting Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result error. Thanks in advance.