我正在开发一个应用程序 Firebase
作为 BaaS
.
问题是我越来越 com.firebase.client.FirebaseException: Failed to bounce to type at com.firebase.client.DataSnapshot.getValue(DataSnapshot.java:183) at com..xyz.MainActivity$2.onDataChange(MainActivity.java:119)
即使给出了正确的火基参考。stacktrace中另一个值得注意的行是 Caused by: com.fasterxml.jackson.databind.JsonMappingException: Conflicting setter definitions for property "colorFilter": android.widget.ImageView#setColorFilter(1 params) vs android.widget.ImageView#setColorFilter(1 params)
.
这里是 Firebase
参考文献: Firebase mFirebaseRef = new Firebase("https://appname.firebaseio.com/users");
下面是我的代码:
mFirebaseRef.addValueEventListener(new ValueEventListener() {
@Override
public void onDataChange(DataSnapshot snapshot) {
System.out.println("There are " + snapshot.getChildrenCount() + " blog posts");
for (DataSnapshot postSnapshot : snapshot.getChildren()) {
//error on this line
UsersDataFromFirebase post = postSnapshot.getValue(UsersDataFromFirebase.class);
View nameView = navigationView.getHeaderView(0);
userName = (TextView) nameView.findViewById(R.id.userName);
userName.setText(post.getUserName());
System.out.println(post.getUserName());
System.out.println(post.getUserEmail());
// View emailView = navigationView.getHeaderView(0);
// userEmail = (TextView) emailView.findViewById(R.id.userEmail);
// userEmail.setText();
userImageUrl = (URL) authData.getProviderData().get("profileImageUrl");
String userImageUrlString = userImageUrl.toString();
URL url = null;
try {
url = new URL(userImageUrlString);
} catch (MalformedURLException e) {
e.printStackTrace();
}
try {
image = BitmapFactory.decodeStream(url.openConnection().getInputStream());
} catch (IOException e) {
e.printStackTrace();
}
View imageView = navigationView.getHeaderView(0);
userImage = (ImageView) imageView.findViewById(R.id.userImage);
userImage.setImageBitmap(image);
}
}
@Override
public void onCancelled(FirebaseError firebaseError) {
System.out.println("The read failed: " + firebaseError.getMessage());
Toast.makeText(getBaseContext(), firebaseError.getMessage(), Toast.LENGTH_LONG).show();
}
});
} else {
android.app.AlertDialog.Builder builder = new android.app.AlertDialog.Builder(MainActivity.this);
builder.setTitle("No internet connection!");
builder.setMessage("Please connect to the internet.");
builder.setPositiveButton("Open Settings", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialogInterface, int i) {
Intent intent = new Intent(Settings.ACTION_SETTINGS);
startActivity(intent);
//progressBar.setVisibility(View.INVISIBLE);
}
});
builder.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialogInterface, int i) {
//progressBar.setVisibility(View.INVISIBLE);
}
});
builder.show();
}
这里是 UsersDataFromFirebase.java
文件代码:
public class UsersDataFromFirebase {
private String userName;
private String userEmail;
private ImageView userImage;
public UsersDataFromFirebase() {
// empty default constructor, necessary for Firebase to be able to deserialize blog posts
}
public String getUserName() {
return userName;
}
public String getUserEmail() {
return userEmail;
}
public ImageView getUserImage() {
return userImage;
}
}
作为 Firebase
对我来说是很新鲜的事情,我不知道这里出了什么问题。
请告诉我。
提前谢谢!
1条答案
按热度按时间euoag5mw1#
您在java类中包含了一个imageview,但jackson/firebase不知道如何将数据序列化到它。
解决方案是告诉firebase/jackson在从json读取/写入对象时忽略imageview。
请注意,我对这个问题的回答中包含了这一点:当我将firebase中的json转换为java对象时,为什么会出现“无法跳转到type”?