Question

我正在使用Laravel框架。

我有2个表（用户和人员）。我想检查user_email和user_password是否在数据库中，如果是，我想登录。

表用户

CREATE TABLE IF NOT EXISTS `festival_aid`.`users` (
  `user_id` BIGINT NOT NULL AUTO_INCREMENT,
  `user_username` VARCHAR(45) NOT NULL,
  `user_email` VARCHAR(45) NOT NULL,
  `user_password` CHAR(64) NOT NULL,
  `user_salt` CHAR(32) NOT NULL,
  `user_created` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `user_modified` TIMESTAMP NULL,
  `user_deleted` TIMESTAMP NULL,
  `user_lastlogin` TIMESTAMP NULL,
  `user_locked` TIMESTAMP NULL,
  `user_token` VARCHAR(128) NULL,
  `user_confirmed` TIMESTAMP NULL,
  PRIMARY KEY (`user_id`, `person_id`),
  UNIQUE INDEX `user_email_UNIQUE` (`user_email` ASC),
  INDEX `fk_users_persons1_idx` (`person_id` ASC),
  CONSTRAINT `fk_users_persons1`
    FOREIGN KEY (`person_id`)
    REFERENCES `festival_aid`.`persons` (`person_id`)
    ON DELETE CASCADE
    ON UPDATE NO ACTION)
ENGINE = InnoDB;

表人

CREATE TABLE IF NOT EXISTS `festival_aid`.`persons` (
  `person_id` BIGINT NOT NULL AUTO_INCREMENT,
  `person_firstname` VARCHAR(45) NULL,
  `person_surname` VARCHAR(45) NULL,
  `person_created` TIMESTAMP NOT NULL,
  `person_modified` TIMESTAMP NULL,
  `person_deleted` TIMESTAMP NULL,
  PRIMARY KEY (`person_id`))
ENGINE = InnoDB;

用户迁移

Schema::table('users', function(Blueprint $table)
        {
            $table->increments('user_id');
            $table->string('user_email');
            $table->string('user_password', 64);
            $table->timestamp('user_created');
            $table->timestamp('user_modified');
            $table->timestamp('user_deleted');
            $table->timestamp('user_lastlogin');
            $table->timestamp('user_locked');

            $table->foreign('person_id')
                ->references('id')->on('persons')
                ->onDelete('cascade');
        });

人员迁移

public function up()
    {
        Schema::table('persons', function(Blueprint $table)
        {
            $table->increments('person_id');

            $table->string('person_firstname');
            $table->string('person_surname');
        });
    }

模型用户

class User extends Eloquent  {

    protected $primaryKey = 'user_id';

    protected $guarded = array();

    public static $rules = array();

    protected $table = 'users';

    public function getAuthIdentifier()
    {
        return $this->getKey();
    }

    public function getAuthPassword()
    {
        return $this->password;
    }

    public function getReminderEmail()
    {
        return $this->user_email;
    }

    public function getPasswordAttribute()
    {
        return $this->user_password;
    }

    public function persons()
    {
        return $this->hasOne('Person', 'person_id', 'user_id');
    }
}

模特人

class Person extends Eloquent {

    protected $table = 'persons';

    protected $primaryKey = 'person_id';

    protected $guarded = array();

    public static $rules = array();

    public function users()
    {
        return $this->belongsTo('User', 'user_id', 'person_id');
    }

    public $timestamps = false;
}

的LoginController

public function showLogin()
{
    return View::make('login');
}

public function doLogin()
{
    $rules = array(
    'user_email'    => 'required|email', 
    'user_password' => 'required'
);

$validator = Validator::make(Input::all(), $rules);

if ($validator->fails()) {
    return Redirect::to('login')
        ->withErrors($validator) 
        ->withInput(Input::except('user_password')); 
} else {

    $userdata = array(
        'user_email'    => Input::get('user_email'),
        'password'  => Input::get('user_password')
    );

    if (Auth::attempt($userdata)) {
        echo 'SUCCESS!';
    } else {
        echo 'FAIL!';
    }
}

登录视图

@extends('layouts.master')
@section('content')
{{ Form::open(array('url' => 'login')) }}
<h1>Login</h1>
<p>
    {{ $errors->first('user_email') }}
    {{ $errors->first('user_password') }}
</p>
<p>
    {{ Form::label('email', 'Email Address') }}
    {{ Form::text('user_email', Input::old('user_email'), array('placeholder' => 'email')) }}
</p>
<p>
    {{ Form::label('password', 'Password') }}
    {{ Form::password('user_password') }}
</p>
<p>{{ Form::submit('Submit!') }}</p>
{{ Form::close() }}
@stop

登录路线

Route::get('login', array('uses' => 'LoginController@showLogin'));

Route::post('login', array('uses' => 'LoginController@doLogin'));

auth.php文件

return array(
    'driver' => 'eloquent',

    'model' => 'User',

    'table' => 'users',

    'reminder' => array(

        'email' => 'emails.auth.reminder',

        'table' => 'password_reminders',

        'expire' => 60,
    ),
);

问题在于它总是回应“失败”，我希望看到改变成功，有人知道我做错了什么吗？是因为存在一种关系：人和用户之间的一对一关系不起作用？

Answer 1

您的用户模型必须至少实现UserInterface

<?php

use Illuminate\Auth\UserInterface;
use Illuminate\Auth\Reminders\RemindableInterface;

class User extends Eloquent implements UserInterface, RemindableInterface {
 ...
}

您的getAuthPassword（）必须返回正确的密码：

public function getAuthPassword()
{
    return $this->user_password;
}

Laravel将在凭证数组中查找密码密钥：

$userdata = array(
    'user_email'    => Input::get('user_email'),
    'password'     => Input::get('user_password')
);

if (Auth::attempt($userdata)) {
    echo 'SUCCESS!';
} else {
    echo 'FAIL!';
}

以下是Illuminate\Auth\DatabaseUserProvider.php中的相关代码：

public function validateCredentials(UserInterface $user, array $credentials)
{
    $plain = $credentials['password'];

    return $this->hasher->check($plain, $user->getAuthPassword());
}

在您的用户模型上创建一个getter，因此Guard可以使用password属性获取密码：

public function getPasswordAttribute()
{
   return $this->user_password;
}

您的密码字段应至少为60个字符：

`user_password` CHAR(60) NOT NULL,

Laravel - 登录保持呼应FAIL

1 个答案: