Question

在尝试登录或从我的Android应用程序注册时，我在PHP中不断收到错误未定义索引。我尝试更改PHP版本，但它没有解决问题。这是我在error_log中得到的错误：

class user
{

public function login()
{
    $data = [];
    require $_SERVER['DOCUMENT_ROOT'].'/socialnetwork-api/Config/db.php';
    $DB = new DB();$db=$DB->connection;
    $username =htmlentities($_POST['username'],ENT_QUOTES,"UTF-8");
    $password =htmlentities($_POST['password'],ENT_QUOTES,"UTF-8");
    $sql = "SELECT * FROM `tbl_users` WHERE `username`='$username' AND `password`='$password'";
    $result = $db->query($sql);
    $result = $result->fetch();
    if($result != null)
    {
        $data["result"]=$result['id'];
    }
    else
    {
        $data["result"]="0";
    }
    echo json_encode($data);
}

public function signup()
{
    $data = [];
    require $_SERVER['DOCUMENT_ROOT'].'/socialnetwork-api/Config/db.php';
    $DB = new DB();$db=$DB->connection;
    $fullname = htmlentities($_POST['full_name'],ENT_QUOTES,"UTF-8");
    $username = htmlentities($_POST['username'],ENT_QUOTES,"UTF-8");
    $email = $_POST['email'];
    $password = htmlentities($_POST['password'],ENT_QUOTES,"UTF-8");
    $sql = "SELECT * FROM `tbl_users` WHERE `username`='$username'";
    $result = $db->query($sql);
    $result = $result->fetch();
    if($result != null)
    {
        $data["result"] = "username";
    }
    else
    {
        $sql = "SELECT * FROM `tbl_users` WHERE `email`='$email'";
        $result = $db->query($sql);
        $result = $result->fetch();
        if($result != null)
        {
            $data["result"] = "email";
        }
        else
        {
            $sql = "INSERT INTO `tbl_users` (`full_name`,`username`,`email`,`password`) VALUES ('$fullname','$username','$email','$password')";
            $result = $db->prepare($sql);
            $result = $result->execute();

            if($result)
            {
                $data["result"]="1";
            }
            else
            {
                $data["result"]="0";
            }
        }
    }
    echo json_encode($data);
}

以下是我的PHP代码：

{{1}}

我希望有人帮助我，因为我对PHP不太熟悉，而且我在大学的项目中需要这个。

Answer 1

未定义的索引错误对应于您在班级中使用WITH dataset AS (SELECT 'T1' AS child_table, 'T0' AS parent_table FROM DUAL UNION ALL SELECT 'X1' AS child_table, 'T0' AS parent_table FROM DUAL UNION ALL SELECT 'X2' AS child_table, 'T0' AS parent_table FROM DUAL UNION ALL SELECT 'T2' AS child_table, 'T1' AS parent_table FROM DUAL UNION ALL SELECT 'Y1' AS child_table, 'T1' AS parent_table FROM DUAL UNION ALL SELECT 'Y2' AS child_table, 'X1' AS parent_table FROM DUAL UNION ALL SELECT 'T3' AS child_table, 'T2' AS parent_table FROM DUAL UNION ALL SELECT 'T4' AS child_table, 'T3' AS parent_table FROM DUAL), findset (parent_table, child_table) AS (SELECT parent_table, child_table FROM dataset WHERE child_table = 'T4' UNION ALL SELECT dataset.parent_table, dataset.child_table FROM findset INNER JOIN dataset ON dataset.child_table = findset.parent_table) SELECT * FROM findset;。这意味着当您调用这些类函数时，不会设置这些值。

通常，如果您要使用$_POST，您将拥有一个调用PHP脚本的表单，并且每个表单输入都对应一个$_POST值。

例如，如果您的表单包含输入，例如：

$_POST

然后，当用户提交表单时，您可以使用<input type="text" name="username" value="">获取值。

但是，如果您没有提交表单，那么您将没有任何$_POST['username']值。因此未定义的索引错误。

另一种方法是将您需要的值传递给类方法。然后，您需要确保在调用方法时传递这些值。

示例：

$_POST

然后您可以传递数据：

public function login($username = '', $password = '')
{
    $data = [];
    require $_SERVER['DOCUMENT_ROOT'].'/socialnetwork-api/Config/db.php';
    $DB = new DB();$db=$DB->connection;
    $username =htmlentities($username,ENT_QUOTES,"UTF-8");
    $password =htmlentities($password,ENT_QUOTES,"UTF-8");
    $sql = "SELECT * FROM `tbl_users` WHERE `username`='$username' AND `password`='$password'";
    $result = $db->query($sql);
    $result = $result->fetch();
    if($result != null)
    {
        $data["result"]=$result['id'];
    }
    else
    {
        $data["result"]="0";
    }
    echo json_encode($data);
}

然后您可以传递$class = new user(); $class->login($username, $password);和$username的值。

从数据库插入或选择数据时PHP出错

1 个答案: