我有一个对象层次结构，其中包含父ID。我像这样解析json对象时，将parentId添加到子对象中。
library(data.table)
df[, paste0(rep(c("mean.", "sum."), each = 2), vars) :=
c(lapply(.SD, mean), lapply(.SD, sum)), by = .(city), .SDcols = vars]
df
# rain temp skip windw_sz city ord mean.rain mean.temp sum.rain sum.temp
# 1: 5 1 0 2 a 2 5 3 25 15
# 2: 5 2 1 1 a 5 5 3 25 15
# 3: 5 3 2 2 a 3 5 3 25 15
# 4: 5 4 2 1 a 4 5 3 25 15
# 5: 5 5 2 2 a 1 5 3 25 15
# 6: 5 6 0 1 b 2 5 8 25 40
# 7: 5 7 2 2 b 5 5 8 25 40
# 8: 5 8 1 2 b 3 5 8 25 40
# 9: 5 9 2 1 b 4 5 8 25 40
#10: 5 10 2 2 b 1 5 8 25 40
如果我有孙子孙女，有没有想到如何拔掉祖先？
数据在模拟袋装（Ancestries.json）上
作为示例，使用以下json和 mFusedLocationClient.getLastLocation()
.addOnSuccessListener(this, new OnSuccessListener<Location>() {
@Override
public void onSuccess(final Location location) {
// Got last known location. In some rare situations this can be null.
if (location != null) {
// Logic to handle location object
Double latittude = location.getLatitude();
Double longitude = location.getLongitude();
String uid = FirebaseAuth.getInstance().getCurrentUser().getUid();
DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference("Users");
DatabaseReference update = rootRef.child(uid);
GeoFire geoFire=new GeoFire(rootRef);
geoFire.setLocation("latandlong", new GeoLocation(location.getLatitude(), location.getLongitude()), new GeoFire.CompletionListener() {
@Override
public void onComplete(String key, DatabaseError error) {
if (error != null) {
System.err.println("There was an error saving the location to GeoFire: " + error);
} else {
System.out.println("Location saved on server successfully!");
}
}
});
geoFire.getLocation("latandlong", new LocationCallback() {
@Override
public void onLocationResult(String key, GeoLocation location) {
if (location != null) {
System.out.println(String.format("The location for key %s is [%f,%f]", key, location.latitude, location.longitude));
// save longitude and latitude to db
String uid = FirebaseAuth.getInstance().getCurrentUser().getUid();
DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference();
DatabaseReference update = rootRef.child("Users").child(uid);
Double longi = location.longitude;
Double lat = location.latitude;
update.child("latandlong").setValue(location.latitude+ "," +location.longitude);
//update.child("longitude").setValue(longi);
//update.child("latitude").setValue(lat);
} else {
System.out.println(String.format("There is no location for key %s in GeoFire", key));
}
}
@Override
public void onCancelled(DatabaseError databaseError) {
System.err.println("There was an error getting the GeoFire location: " + databaseError);
}
});
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_news_feed);
setRequestedOrientation(ActivityInfo.SCREEN_ORIENTATION_NOSENSOR);
menuItem=(MenuItem)findViewById(R.id.item_sign_in);
mFusedLocationClient = LocationServices.getFusedLocationProviderClient(this);
recycler=findViewById(R.id.recyclerView);
recycler.setLayoutManager(new LinearLayoutManager(this));
final DatabaseReference rootRef = FirebaseDatabase.getInstance().getReference();
mFusedLocationClient.getLastLocation().addOnSuccessListener(new OnSuccessListener<Location>() {
@Override
public void onSuccess(Location location) {
Double lat=location.getLatitude();
Double longi=location.getLongitude();
firebaseDatabase=FirebaseDatabase.getInstance().getReference("Users");
GeoFire geoFire=new GeoFire(firebaseDatabase.child("latandlong"));
GeoQuery geoQuery=geoFire.queryAtLocation(new GeoLocation(lat,longi),5);
geoQuery.addGeoQueryEventListener(new GeoQueryEventListener() {
@Override
public void onKeyEntered(String key, GeoLocation location) {
firebaseDatabase.child(key).addValueEventListener(new ValueEventListener() {
@Override
public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
if(dataSnapshot.exists()){
list=new ArrayList<UserInformation>();
for(DataSnapshot dataSnapshot1:dataSnapshot.getChildren()){
UserInformation uuu=dataSnapshot1.getValue(UserInformation.class);
list.add(uuu);
}
}
adapter=new MyAdapter(NewsFeedActivity.this,list);
recycler.setAdapter(adapter);
}
@Override
public void onCancelled(@NonNull DatabaseError databaseError) {
}
});
}
@Override
public void onKeyExited(String key) {
}
@Override
public void onKeyMoved(String key, GeoLocation location) {
}
@Override
public void onGeoQueryReady() {
}
@Override
public void onGeoQueryError(DatabaseError error) {
}
});
}
});
，我将创建并使用以下ID进行数组
['5'，'0723'，'133'，'1']
public static fromJson(json: any): Ancestry | Ancestry[] {
if (Array.isArray(json)) {
return json.map(Ancestry.fromJson) as Ancestry[];
}
const result = new Ancestry();
const { parents } = json;
parents.forEach(parent => {
parent.parentId = json.id;
});
json.parents = Parent.fromJson(parents);
Object.assign(result, json);
return result;
}
答案 0 :(得分：0)
也许有很多方法可以解决此问题，但是我认为最简单的方法是简单地在数据结构中进行搜索，并以与找到ID时相反的顺序存储ID。这样，输出就是您想要的。
您还可以颠倒其他方法的顺序。
我想指出，json-结构有点奇怪。我希望它只是嵌套了children数组，而不是将它们重命名为parent，children和grandchildren。
let data = [{
"id": "1",
"name": "Deer, spotted",
"parents": [
{
"id": "133",
"name": "Jaime Coldrick",
"children": [
{
"id": "0723",
"name": "Ardys Kurten",
"grandchildren": [
{
"id": "384",
"name": "Madelle Bauman"
},
{
"id": "0576",
"name": "Pincas Maas"
},
{
"id": "5",
"name": "Corrie Beacock"
}
]
}]
}]
}]
const expectedResults = ['5', '0723', '133', '1']
function traverseInverseResults(inputId, childArray) {
if(!childArray){ return }
for (const parent of childArray) {
if(parent.id === inputId){
return [parent.id]
} else {
let res = traverseInverseResults(inputId, parent.parents || parent.children || parent.grandchildren) // This part is a bit hacky, simply to accommodate the strange JSON structure.
if(res) {
res.push(parent.id)
return res
}
}
}
return
}
let result = traverseInverseResults('5', data)
console.log('results', result)
console.log('Got expected results?', expectedResults.length === result.length && expectedResults.every(function(value, index) { return value === result[index]}))