Question

在下面的代码中，我已经将功能从MYSQL升级到MYSQLI，我在之前的帖子中得到了一些帮助，他们说应该做些什么......但现在问题该怎么做..

问题是在id之前设置return，否则无论我登录哪个帐户，我仍然会收到用户ID = 1。

function user_id_from_username($username) {
    $username = sanitize($username);
    global $db_connect;

    $result = $db_connect->query("SELECT(id) FROM members WHERE username = '$username'");
    if (false === $result) {
        return false;
    }
    return ($result->num_rows == 1) ? id : false;

我尝试按照旧的MYSQL代码进行操作，但之后却出错了。我的旧代码：

    function user_id_from_username($username){
    $username = sanitize ($username);
    return mysql_result(mysql_query("SELECT(id) FROM members WHERE username = '$username'"), 0, 'id');
}

我将不胜感激任何帮助！

Answer 1

您是否尝试过fetch_array或mysqli_fetch_array？

$row = $result->fetch_array(MYSQLI_ASSOC);
// or
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);

// and returning
return $row['id'] ? $row['id'] : false;

Answer 2

我没试过，但这应该可行吗？我还把$ db_connect作为参数，因为它更安全，但这取决于你。

function user_id_from_username($username) {
    $username = sanitize($username);
    global $db_connect;
    $result = $db_connect->query("SELECT id FROM members WHERE username = '".$db_connect->real_escape_string($username)."'");
    if (!$result) {
        return false;
    }
    if($result->num_rows == 1){
        $row = $result->fetch_assoc();
        $output = $row['id'];
    }
    else{
        $output = "Username does not exist in table or is a duplicate.";
    }
    return $output;
}

Answer 3

return ($result->num_rows == 1) ? id : false;

应该是

if(mysqli_num_rows($result) ===1 ){
    $row = mysqli_fetch_assoc($result);
    return intval($row['id']);
}
return false;

在返回功能之前输入ID

3 个答案: