Ionic2 离子2中的登录验证

我试图在我的应用程序中做登录功能，但我得到了这个 错误“响应状态为0 URL:Null”，我希望有人能帮助我解决此问题

授权服务

    let apiUrl='http://localhost:8888/PeopleService/auth/lognin.php

login(credentials) {
    return new Promise((resolve, reject) => {
        let headers = new Headers();
        headers.append('Content-Type', 'application/json');

        this.http.post(apiUrl+'login', JSON.stringify(credentials), {headers: headers})
          .subscribe(res => {
            resolve(res.json());
          }, (err) => {
            reject(err);
          });
    });
  }

login.ts

     loginData = { PHONENUM:'', PASSWORD:'' };

doLogin() {
    this.showLoader();
    this.authService.login(this.loginData).then((result) => {
      this.loading.dismiss();
      this.data = result;
      this.navCtrl.setRoot(HomePage);
    }, (err) => {
      this.loading.dismiss();
      this.presentToast(err);
    });
  }

 showLoader(){
    this.loading = this.loadingCtrl.create({
        content: 'Authenticating...'
    });

    this.loading.present();
  }

登录html

    <ion-content padding>
  <h2>Please, Login</h2>
  <form (submit)="doLogin()" name="login" id="login">
    <ion-item>
      <ion-label stacked>Username</ion-label>
      <ion-input [(ngModel)]="loginData.PHONENUM" name="PHONENUM" type="text" placeholder="Username" ></ion-input>
    </ion-item>
    <ion-item>
      <ion-label stacked>Password</ion-label>
      <ion-input [(ngModel)]="loginData.PASSWORD" name="PASSWORD" type="password" placeholder="Password"></ion-input>
    </ion-item>
    <button ion-button block type="submit">
      Login
    </button>
  </form>
</ion-content>

---

看起来url没有解析。您正在发布到

http://localhost:8888/PeopleService/auth/lognin.phplogin

我将URL添加到这个.http.pos'thttp://localhost:8888/PeopleService/auth/login.php，JSON.stringify（凭证），{headers:headers}），但仍然是空的

    if(isset(($_POST['login'])){
   $password=$_POST['PASSWORD'];
   $phone=$_POST['PHONENUM'];

     $password = $conn->real_escape_string($password);
     $phone = $conn->real_escape_string($phone);
        $arr = array();


 $sql = "SELECT EMPID FROM employee WHERE PASSWORD=$password AND PHONENUM=$phone";
        $select_users = $conn->query($sql);
        if(!$select_users){
            die('Query Failed' . mysqli_error($conn));
        }
        while($row = $select_users->fetch_assoc()) {
               $arr= $row


        } 

echo json_encode($arr);