Question

我使用FirebasaRecyclerAdapter来显示我在Firebase控制台中收到的帖子列表。我想获得每个帖子引用，以便能够删除它们。我已经可以获得网址：

 public class PostAdapter extends FirebaseRecyclerAdapter<Post, PostViewHolder> {

    private static final String TAG = Post.class.getSimpleName();
    private Context context;

    public PostAdapter(Class<Post> modelClass, int modelLayout, Class<PostViewHolder> viewHolderClass, DatabaseReference ref, Context context) {
        super(modelClass, modelLayout, viewHolderClass, ref);
        this.context = context;
    }
    @Override
    protected void populateViewHolder(PostViewHolder viewHolder, Post model, int position) {
         String postReference = this.getRef(position).toString();
         viewHolder.bindPost(model, position, postReference);
    }

    @Override
    public int getItemViewType(int position) {
        return super.getItemViewType(position);
    }
}

我得到字符串postReference并将其发送到我的viewHolder。这个viewHolder是我用来在feed中显示帖子的那个。当用户点击帖子时，他可以将帖子详细信息放在另一个片段中（我发送postReference字符串：

Bundle bundle = new Bundle();
bundle.putSerializable("post", post);
bundle.putString("postReference", postReferece);

PostDetailFragment postDetailFragment = new PostDetailFragment();
postDetailFragment.setArguments(bundle);

FragmentTransaction fragmentTransaction = ((AppCompatActivity) context).getSupportFragmentManager().beginTransaction();
fragmentTransaction.setCustomAnimations(R.anim.slide_left_enter, R.anim.slide_left_exit, R.anim.slide_right_enter, R.anim.slide_right_exit);
fragmentTransaction.addToBackStack(null);
fragmentTransaction.replace(R.id.feed_container, postDetailFragment, "PostDetailFragment").commit();

在PostDetailFragment中，我想在点击删除按钮时删除该帖子，所以：

private View.OnClickListener deleteListener = new View.OnClickListener() {
    @Override
    public void onClick(View v) {
        String postReference = bundle.getString("postReference");
        FirebaseDatabase.getInstance().getReference(postReference).removeValue();
        ((FeedActivity)context).getSupportFragmentManager().popBackStack();
    }
};

我的问题是Firebase不接受我的＆＃34; postReference＆＃34;字符串，它说：

Invalid Firebase Database path: https://(myapp).firebaseio.com/users/(userkey)/posts/KIJ999. Firebase Database paths must not contain '.', '#', '$', '[', or ']'

对此有何想法？

Answer 1

所以，我终于解决了这个问题。删除我的帖子时我需要使用帖子“路径”，这不是我到目前为止的URL。为了获得路径而不是URL，我做了：

@Override
    protected void populateViewHolder(PostViewHolder viewHolder, Post model, int position) {

    String postUrl = this.getRef(position).toString();
    String postRoot = this.getRef(position).getRoot().toString();
    String postReference = postUrl.replace(postRoot,"");

    viewHolder.bindPost(model, position, postReference);
}

通过这样做，我删除了网址的“https://(myapp).firebaseio.com”部分（这是Firebase不接受的部分）并获取我要删除的帖子的路径。

Android FirebaseRecyclerAdapter获取对象引用并将其删除

1 个答案: