Return results from Select * query using prepared statements

Question

I have the following test script to adapt all my current programs to using prepared statements ... can't find the right syntax/structure to read the result rows:

    $userid = "admin";
    $stmt = mysqli_stmt_init($link);
    if (mysqli_stmt_prepare($stmt, 'SELECT * FROM user_info WHERE userid = ?')) {
        mysqli_stmt_bind_param($stmt, "s", $userid);
        mysqli_stmt_execute($stmt);
        $result = mysqli_stmt_store_result($stmt);
        $number_of_rows = mysqli_stmt_num_rows($stmt);
        echo "Number of Rows: $number_of_rows<br />";
        $result = mysqli_stmt_get_result($stmt);
        for($i=0;$i<$number_of_rows;$i++){  
            $row = mysqli_fetch_array($result, MYSQLI_ASSOC);
            echo $row["id"];
        }
        mysqli_stmt_close($stmt);
    }
    else{
        // Catch a database error here
        die("Could not query database.");
    }

How do I reference the result correctly (using procedural)?

Answer 1

Try this:

$userid = "admin";
$stmt = mysqli_stmt_init($link);
if (mysqli_stmt_prepare($stmt, 'SELECT * FROM user_info WHERE userid = ?')) {
    mysqli_stmt_bind_param($stmt, "s", $userid);
    mysqli_stmt_execute($stmt);
    $result = mysqli_stmt_store_result($stmt);
    $number_of_rows = mysqli_stmt_num_rows($stmt);
    echo "Number of Rows: $number_of_rows<br />";
    $result = mysqli_stmt_get_result($stmt);

    while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC))  
        echo $row["id"];
    }

    mysqli_stmt_close($stmt);
}
else{
    // Catch a database error here
    die("Could not query database.");
}

Look at php manual for get-result function: http://php.net/manual/en/mysqli-stmt.get-result.php (Example #2 Procedural style)