问题
我有一个ObjectId保存在字段-incomingFriendRequests-中，我想将其移动到另一个字段-friends-。
是否有比将其从ReceivedFriendRequests中删除然后再执行findOneAndUpdate并将其添加到朋友更好的方法呢？我觉得这可以通过一个数据库调用而不是两个数据库调用来完成。
代码
猫鼬模型：
import javax.swing.*;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.*;
public class Program4_v2 extends JFrame { //extending Jframe for visual tree
private static void write(LinkedBinaryTreeNode<String> root, String path){
File file=new File(path);
try (PrintWriter writer=new PrintWriter(file)){
root.traversePreorder(node -> {
if(node.getParent()==null){
writer.print(node.getData());
}
else{
writer.print("\n"+node.getData());
}
});
}
catch (FileNotFoundException e){
e.printStackTrace();
}
}
public static void add(BinaryTreeNode<String> oldNode, String data) { //if statements helping to determine if there are more nodes
BinaryTreeNode<String> newNode = new LinkedBinaryTreeNode<>(data); // failing line
if (oldNode.hasLeftChild() && oldNode.getLeft().getData().contains("Q:")) {
add(oldNode.getLeft(), data);
} else if (oldNode.hasRightChild() && oldNode.hasLeftChild() && oldNode.getRight().getData().contains("Q")) {
add(oldNode.getRight(), data);
} else if (!oldNode.hasLeftChild() && oldNode.getData().contains("Q")) {
oldNode.setLeft(newNode);
} else if (!oldNode.hasRightChild() && oldNode.getData().contains("Q")) {
oldNode.setRight(newNode);
} else {
boolean finished = false;
BinaryTreeNode<String> node = oldNode.getParent();
while (oldNode.getParent().getRight() != node) {
if (node.getData().contains("A:")) {
node = node.getParent();
} else if (!node.hasRightChild() && node.getData().contains("Q")) {
finished = true;
node.setRight(newNode);
break;
} else if (node.getRight().getData().contains("A:") || node.getRight() == oldNode) {
node = node.getParent();
} else {
finished = true;
add(node.getRight(), data);
break;
}
}
if (!finished) { //if there are more nodes, find them
if (!node.hasRightChild()) {
node.setRight(newNode);
} else {
add(node.getRight(), data);
}
}
}
}
private static LinkedBinaryTreeNode<String>load(String path){
LinkedBinaryTreeNode<String>root=new LinkedBinaryTreeNode<>(null); // failing line
try {
Scanner file = new Scanner(new File(path));
String string = file.nextLine();
root=new LinkedBinaryTreeNode<>(string); // failing line
while(file.hasNext()){
string=file.nextLine();
add(root,string);
}
}
catch (FileNotFoundException e){
e.printStackTrace();
}
return root;
}
private static String question(String question){
System.out.println(question);
Scanner scanner=new Scanner(System.in);
String object=scanner.nextLine();
if (object.contains("?")){
return "Q:"+object;
}
else{
return "A:"+object;
}
}
private static void question(LinkedBinaryTreeNode<String>node){
System.out.println(node.getData().substring(2));
Scanner scanner=new Scanner(System.in); //scanner in to along for computer to learn
String string=scanner.next();
if (string.contains("no")&& node.isLeaf()){
String object= question("What was the correct answer?"); //printed text for when computer guesses wrong
String question=question("What is the difference between "+node.getData().substring(2)+" and "+object.substring(2)+"?");
String answer=question("What is the correct answer?");
LinkedBinaryTreeNode<String>oldObject=new LinkedBinaryTreeNode<>(node.getData()); // failing line
LinkedBinaryTreeNode<String>newObject=new LinkedBinaryTreeNode<>(object()); // failing line
node.setData(question);
if (answer.contains("yes")){
node.setRight(newObject);
node.setLeft(oldObject);
}
else{
node.setRight(newObject);
node.setLeft(oldObject);
}
}
else if (string.contains("no")){
question((LinkedBinaryTreeNode)node.getLeft());
}
else if (string.contains("yes")&& node.isLeaf()){ //you, or the computer, did it
System.out.println("Excellent");
}
else{
question((LinkedBinaryTreeNode)node.getRight());
}
}
static JFrame frame;
public static void main(String[] args) {
System.out.println("20-Questions Game!"); //launching game
LinkedBinaryTreeNode<String>root=load(args[0]);
boolean play=true;
while(play){
System.out.println("Play?"); //asking to play
Scanner scanner=new Scanner(System.in);
if (scanner.next().equals("no")){ //you dont want to play! abort!
play=false;
}
else{
question(root);
}
}
write(root,"20questions.data"); //writing game to new directory for future playing
frame=new JFrame("20 Questions");
JPanel panel=new TreeDisplayPanel(root);
frame.add(panel);
frame.setSize(1600,900);
frame.show();
}
}
What should be expected is a twenty-questions game that the program helps you make by sort-of playing. a GUI is also supposed to pop up but should work fine and seems unrelated to the problem
答案 0 :(得分：0)
将值从一个字段移动到另一个字段肯定需要进行n +1次调用。可以使用
find({}).forEach(function(doc) {
// your doc update
doc.save();
});
或者您甚至可以执行批量写入。
如果您的目标是删除列incomingFriendRequests，则可以将列重命名为好友，而不是复制所有数据。
update({}, { $rename: {incomingFriendRequests: friends }});
答案 1 :(得分：0)
在单个查询中，您需要首先对传入的FriendRequests字段执行$ pull以删除所选的objectId，然后需要对好友执行另一个更新查询以添加所选的objectId，这是不可能的。另外，您可以使用$ addToSet来防止重复ID。如果在这种情况下使用$ push进行添加，则还会存储重复的objectId。
查询：-
1）删除：-db.getCollection（'testdata'）。update（{id：ObjectId（'5cc3fa66a04275096417f9ca'）}，{“ $ pull”：{“ incomingFriendRequests”：“ 5cc3fa66a04275096417f9da”}}}））; < / p>
2）添加：- db.getCollection（'testdata'）。update（{id：ObjectId（'5cc3fa66a04275096417f9ca'）}，{“ $ addToSet”：{“ friends”：“ 5cc3fa66a04275096417f9da”}}））;; //也可以使用$ push