试图使facebook身份验证工作

时间:2019-01-15 17:52:41

标签: facebook firebase cakephp

每次我单击带Facebook的登录名时,都会弹出一个弹出窗口,它告诉我URL已被阻止,我在Facebook应用程序的有效OAuth重定向URI中添加了本地主机,但仍然无法正常运行

<!-- File: src/Template/Users/login.ctp -->
<div class="form-signin">
<?= $this->Flash->render() ?>
<?= $this->Form->create() ?>
    <h2><legend><?= __('Login')?></legend></h2>
    <?= $this->Form->control('username',
                             array('class' => 'form-control',
                                  'placeholder' => 'Enter your     username')) ?>
       <?= $this->Form->control('password',
                             array('class' => 'form-control',
                                  'placeholder' => 'Enter your    password')) ?>
<br>

<?= $this->Form->button(__('Login'), array('class' => 'btn btn-lg     btn-btn-secondary btn-block')); ?>
  <?= $this->Form->end() ?>

<hr>

<button type="button" name="button" onclick="fblogin()" class = "btn     btn-primary btn-lg btn-block">Login with Facebook</button>

</div>

<script src="https://www.gstatic.com/firebasejs/5.7.3/firebase.js">.   </script>
<script>
  // Initialize Firebase
 var config = {
apiKey: "AIzaSyBCPLPyLTlkWyVs3XK4LpgE3ASEkf1Zf9w",
authDomain: "sendit-e8103.firebaseapp.com",
databaseURL: "https://sendit-e8103.firebaseio.com",
projectId: "sendit-e8103",
storageBucket: "sendit-e8103.appspot.com",
messagingSenderId: "150641419524"
  };
   firebase.initializeApp(config);

    var provider = new firebase.auth.FacebookAuthProvider();
    function fblogin() {
        firebase.auth().signInWithPopup(provider)
        .then(function(result){
            var token = result.credential.accessToken;
            var user = result.user;

            var userName = user.displayName;
            var userEmail = user.email;

            var x = "<form method='POST' action='/ShareIT/users/fblogin'><input type='hidden' name='email' value='" + userEmail + "'></input><input type='hidden' name='name' value='" + userName + "'></input></form>";
                var y = $(x);
                $('body').append(y);
                $(y).submit();

            console.log(token);
            console.log(user);

        }).catch(function(error) {
            console.log(error.code);
            console.log(error.message);
        })
    }

</script>

https://ibb.co/nsmKZVd

https://ibb.co/99MVgv1

https://ibb.co/r5thcYF

0 个答案:

没有答案