Laravel Follow Unfollow 系统不适用于 Flutter 前端

问：

我正在使用 Laravel 作为我的 flutter 移动应用程序的后端，但对关注和取消关注系统有一些问题。

这些是我的 Laravel 代码：

用户模型：

class User extends Authenticatable
{
    use HasApiTokens, HasFactory, Notifiable;
    protected $fillable = [
        'name',
        'phonenumber',
        'image',
        'password',
        'usermoney',
        'vipuser',
        'userphonenumber',
        'job',
        'star',
        'codemelli',
        'bio',
    ];

    protected $hidden = [
        'password',
        'remember_token',
    ];

    protected $casts = [
        'phonenumber_verified_at' => 'datetime',
        'password' => 'hashed',
    ];

    public function followings()
    {
        return $this->belongsToMany(User::class, 'follows', 'follower_id', 'following_id');
    }
    public function followers()
    {
        return $this->belongsToMany(User::class, 'follows', 'following_id', 'follower_id');
    }
    public function is_following($id)
    {
        return $this->followings()->where('following_id', $id)->exists();
    }
}

这是我的 AuthController 的夏天：

public function is_following(User $follower, User $following)
{
    return response()->json([
        'status' => $follower->is_following($following->id),
            'message' => 'followed.',
        ], 200);
    
}

这是我的 API 路由：

Route::get('/status/{follower}/{following}', [AuthController::class, 'is_following']);

这是我为该函数进行的迁移类：

return new class extends Migration
{
    public function up(): void
    {
        Schema::create('follows', function (Blueprint $table) {
            $table->unsignedBigInteger('follower_id');
            $table->unsignedBigInteger('following_id');
            $table->unique(['following_id', 'follower_id']);
            $table->foreign('follower_id')
                ->references('id')
                ->on('users')
                ->cascadeOnDelete();
            $table->foreign('following_id')
                ->references('id')
                ->on('users')
                ->cascadeOnDelete();
            $table->timestamp('created_at');
        });
    }

    public function down(): void
    {
        Schema::dropIfExists('follows');
    }
};

现在我将分享我的前端颤振代码：

服务部分：

Future<ApiResponse> followunfollow(User follower, User following) async {
  ApiResponse apiResponse = ApiResponse();
  try {
    String token = await getToken();
    final response = await http
        .get(Uri.parse('$userURL/is_following/$follower/$following'), headers: {
      'Accept': 'application/json',
      'Authorization': 'Bearer $token'
    });

    switch (response.statusCode) {
      case 200:
        apiResponse.data = jsonDecode(response.body)['message'];

        break;
      case 401:
        apiResponse.error = unauthorized;
        break;
      default:
        apiResponse.error = somethingWentWrong;
        break;
    }
  } catch (e) {
    apiResponse.error = serverError;
  }
  return apiResponse;
}

在我的代码中，我想使用它的地方在这里： （用户，用户是用户类型的相同实例，我检查过它们是否正常）

   void fUNf() async {
    ApiResponse response = await followunfollow(user, user);
    if (response.error == null) {
      setState(() {});
    } else if (response.error == unauthorized) {
      logout().then((value) => {
            Navigator.of(context).pushAndRemoveUntil(
                MaterialPageRoute(builder: (context) => Login()),
                (route) => false)
          });
    } else {
      ScaffoldMessenger.of(context)
          .showSnackBar(SnackBar(content: Text('${response.error}')));
    }
  }

最后，我创建按钮，按下它会像：

  onTap: () {
              fUNf();
            },

当我点击按钮时出现的错误是在我的 flutter 服务部分定义的：

   apiResponse.error = somethingWentWrong; //this is a part of services part in my flutter code

无论如何，我每次都会遇到这个错误！