我想通过提交一个按钮在两个表的相同数据库中插入相同的值

<table>
<form action="" method="post">
<tr>
<td> <input type="text" name="name" id="name" /> </td>
<td> <input type="submit" name="submit" id="submit" value="submit" /> </td>
</tr>
</form>
</table>   



$conn = mysqli_connect('localhost','root','','dbname');

if(isset($_POST["submit"]))
{

    $name= $name1 = $_POST["name"];
    $query1= "insert into course (name) values ('$name')";
    $result1 = mysqli_query($conn,$query1);

if (mysqli_num_rows($result1))
{
    echo "run";
}else{
    echo "not";
}


$query2= "insert into user (name) values ('$name1')";
$result2 = mysqli_query($conn,$query2);

if (mysqli_num_rows($result2) )
    echo "run";
}else{
    echo "not";
}

}

我收到警告： -

警告：mysqli_query（）：无法在第33行的D：\ Xampp \ xampp \ htdocs \ join_query \ insert.php中获取mysqli警告：mysqli_num_rows（）期望参数1为mysqli_result，在D：\ Xampp \中给出null第35行的xampp \ htdocs \ join_query \ insert.php不警告：mysqli_query（）：无法在第46行的D：\ Xampp \ xampp \ htdocs \ join_query \ insert.php中获取mysqli警告：mysqli_num_rows（）需要参数1是mysqli_result，在第48行的D：\ Xampp \ xampp \ htdocs \ join_query \ insert.php中给出null

php sql mysqli