- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
我有 3 个实体:类别
、文件夹
和卡片
。每个类别都可以包含文件夹,您可以在其中包含卡片。我可以通过使用外键成功关联类别和文件夹实体,它工作得很好。但是当我创建卡片实体并尝试将其与相同的逻辑关联时,工作室会抛出一个错误。还有其他方法可以链接这 3 个吗?实体代码如下:
@Entity(tableName = "categories")
public class Categories {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "category_id")
private int categoryId;
@ColumnInfo(name = "category_name")
private String categoryName;
@Ignore
public Categories() {
}
文件夹
@Entity(tableName = "folders", foreignKeys = @ForeignKey(entity = Categories.class,
parentColumns = "category_id", childColumns = "current_category_id", onDelete = ForeignKey.CASCADE))
public class Folders {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "folder_id")
private int folderId = 0;
@ColumnInfo(name = "folder_name")
private String folderName;
@ColumnInfo(name = "current_category_id")
private int categoryId;
卡片
@Entity(tableName = "cards",foreignKeys = @ForeignKey(entity = Folders.class,
parentColumns = "folder_id", childColumns = "current_folder_id", onDelete = ForeignKey.CASCADE))
public class Cards {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "card_id")
private int cardId = 0;
@ColumnInfo(name = "card_name")
private String cardName;
@ColumnInfo(name = "card_value")
private String cardValue;
@ColumnInfo(name = "current_folder_id")
private int folderId;
@Ignore
public Cards() {
}
最佳答案
要完成您正在尝试的操作,您可以使用:-
@Entity(tableName = "cards",foreignKeys = {
@ForeignKey(entity = Folders.class, parentColumns = "folder_id", childColumns = "current_folder_id", onDelete = CASCADE),
@ForeignKey(entity = Categories.class, parentColumns = "category_id", childColumns = "current_category_id", onDelete = CASCADE)})
public class Cards {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "card_id")
private int cardId = 0;
@ColumnInfo(name = "card_name")
private String cardName;
@ColumnInfo(name = "card_value")
private String cardValue;
@ColumnInfo(name = "current_folder_id")
private int folderId;
@ColumnInfo(name = "current_category_id") /*<<<<<<<<<< ADDED column */
private int category_id;
@Ignore
public Cards() {
}
}
但是,因为文件夹是类别的子级,那么从卡片到类别的链接是隐式的(即文件夹必须有类别行作为家长)以下内容应该足够了(并且没有不必要的列):-
@Entity(tableName = "cards",foreignKeys = {
@ForeignKey(entity = Folders.class, parentColumns = "folder_id", childColumns = "current_folder_id", onDelete = CASCADE)})
public class Cards {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "card_id")
private int cardId = 0;
@ColumnInfo(name = "card_name")
private String cardName;
@ColumnInfo(name = "card_value")
private String cardValue;
@ColumnInfo(name = "current_folder_id")
private int folderId;
@Ignore
public Cards() {
}
}
<小时/>
以下基本应用程序基于您的代码和第二个答案。它构建了一副扑克牌。
反射(reflect)花色红心、黑桃的类别......(请注意,一些白痴无意中(故意)引入了花色名称“哎呀”,所以包中有 65 张牌)
文件夹可以是宫廷牌或普通牌(2-10张),每张牌都有面值。
然后将每张卡牌的信息(名称、值、ID、文件夹和类别(套装))输出到日志中。
Categories.java
@Entity(tableName = "categories")
public class Categories {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "category_id")
private int categoryId;
@ColumnInfo(name = "category_name")
private String categoryName;
public Categories() {
}
@Ignore
public Categories(String categoryName) {
this.categoryName = categoryName;
}
public int getCategoryId() {
return categoryId;
}
public void setCategoryId(int categoryId) {
this.categoryId = categoryId;
}
public String getCategoryName() {
return categoryName;
}
public void setCategoryName(String categoryName) {
this.categoryName = categoryName;
}
}
Folders.java
@Entity(tableName = "folders", foreignKeys = @ForeignKey(entity = Categories.class,
parentColumns = "category_id", childColumns = "current_category_id", onDelete = ForeignKey.CASCADE))
public class Folders {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "folder_id")
private int folderId = 0;
@ColumnInfo(name = "folder_name")
private String folderName;
@ColumnInfo(name = "current_category_id")
private int categoryId;
public Folders() {}
@Ignore
public Folders(int categoryId, String folderName) {
this.categoryId = categoryId;
this.folderName = folderName;
}
public int getFolderId() {
return folderId;
}
public void setFolderId(int folderId) {
this.folderId = folderId;
}
public String getFolderName() {
return folderName;
}
public void setFolderName(String folderName) {
this.folderName = folderName;
}
public int getCategoryId() {
return categoryId;
}
public void setCategoryId(int categoryId) {
this.categoryId = categoryId;
}
public Categories getOwningCategories(GamesDatabase gdb) {
return gdb.getGamesDBDao().getACategoriesRowById(this.categoryId);
}
}
getOwningCategories
方法(GamesDatabase 是 @Database 类,getGaMesDao 是访问 @Dao 的方法(全部都在一个中)文件 GamesDao.java))Cards.java
@Entity(tableName = "cards",foreignKeys = {
@ForeignKey(entity = Folders.class, parentColumns = "folder_id", childColumns = "current_folder_id", onDelete = CASCADE)})
public class Cards {
@PrimaryKey(autoGenerate = true)
@ColumnInfo(name = "card_id")
private int cardId = 0;
@ColumnInfo(name = "card_name")
private String cardName;
@ColumnInfo(name = "card_value")
private String cardValue;
@ColumnInfo(name = "current_folder_id")
private int folderId;
public Cards() {}
@Ignore
public Cards(String cardName, String cardValue, int folderId) {
this.cardName = cardName;
this.cardValue = cardValue;
this.folderId = folderId;
}
public Folders getOwningFolders(GamesDatabase gdb) {
return gdb.getGamesDBDao().getAFoldersRowById(folderId);
}
public Categories getOwningCategories(GamesDatabase gdb) {
return gdb.getGamesDBDao().getACategoriesRowById(getOwningFolders(gdb).getCategoryId());
}
public int getCardId() {
return cardId;
}
public void setCardId(int cardId) {
this.cardId = cardId;
}
public String getCardName() {
return cardName;
}
public void setCardName(String cardName) {
this.cardName = cardName;
}
public String getCardValue() {
return cardValue;
}
public void setCardValue(String cardValue) {
this.cardValue = cardValue;
}
public int getFolderId() {
return folderId;
}
public void setFolderId(int folderId) {
this.folderId = folderId;
}
}
GamesDao.java
@Dao
public interface GamesDao {
@Insert()
long[] insertCategoriesRows(Categories... categories);
@Insert()
long insertCategoriesRow(Categories categories);
@Insert()
long[] insertFoldersRows(Folders... folders);
@Insert()
long insertFoldersRow(Folders folders);
@Insert
long[] insertCardsRows(Cards... cards);
@Insert()
long insertCardsRow(Cards cards);
@Delete()
int deleteManyCategoriesRows(Categories... categories);
@Delete()
int deleteACategoriesRow(Categories categories);
@Delete()
int deleteManyFoldersRows(Folders... folders);
@Delete()
int deleteAFolders(Folders folders);
@Delete()
int deleteManyCardsRows(Cards... cards);
@Delete()
int deleteACardRow(Cards cards);
@Query("SELECT * FROM categories")
Categories[] getAllCategoriesRows();
@Query("SELECT * FROM folders")
Folders[] getAllFoldersRows();
@Query("SELECT * FROM cards")
Cards[] getAllCards();
@Query("SELECT * FROM categories WHERE category_id = :id")
Categories getACategoriesRowById(int id);
@Query("SELECT * FROM folders WHERE folder_id = :id")
Folders getAFoldersRowById(int id);
@Query("SELECT * FROM cards WHERE card_id = :id")
Cards getACardsRowById(int id);
@Query("SELECT * FROM folders WHERE current_category_id = :id")
Folders[] getFoldersPerCategoriesId(int id);
}
GamesDatabase.java
@Database(version = 1,exportSchema = true, entities = {Categories.class,Folders.class,Cards.class})
public abstract class GamesDatabase extends RoomDatabase {
public static final String DBNAME = "games";
public abstract GamesDao getGamesDBDao();
}
MainActivity.java
public class MainActivity extends AppCompatActivity {
GamesDatabase mGamesDB;
GamesDao mGamesDao;
private static final String[] card_suits = new String[]{"Spades","Hearts","Clubs","Diamonds","ooops"};
private static final String[] courtcard_values = new String[]{
"Jack","Queen","King","Ace"
};
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
mGamesDB = Room.databaseBuilder(this,GamesDatabase.class,GamesDatabase.DBNAME)
.allowMainThreadQueries()
.build();
// Suits
mGamesDao = mGamesDB.getGamesDBDao();
for (String s: card_suits) {
mGamesDao.insertCategoriesRow(new Categories(s));
}
for (Categories c: mGamesDao.getAllCategoriesRows()) {
mGamesDao.insertFoldersRow(new Folders(c.getCategoryId(),"CourtCards"));
mGamesDao.insertFoldersRow(new Folders(c.getCategoryId(),"NormalCards"));
for (Folders f: mGamesDao.getFoldersPerCategoriesId(c.getCategoryId())) {
if (f.getFolderName().equals("CourtCards")) {
for (String s: courtcard_values) {
mGamesDao.insertCardsRow(new Cards(s,getCourtCardValueFromName(s),f.getFolderId()));
}
} else {
for (int i=2; i < 11; i++) {
mGamesDao.insertCardsRow(new Cards(String.valueOf(i),String.valueOf(i),f.getFolderId()));
}
}
}
}
for (Cards card: mGamesDao.getAllCards()) {
logCardsInfo(card);
}
mGamesDB.close();
}
private String getCourtCardValueFromName(String courtCardName) {
int basevalue = 11;
for (String s: courtcard_values) {
if (s.equals(courtCardName)) {
return String.valueOf(basevalue);
}
basevalue++;
if (basevalue > 13) basevalue = 1;
}
return "UNKNOWN";
}
private void logCardsInfo(Cards card) {
Log.d("CARDINFO",
"Card Name is " + card.getCardName() +
"\n\tValue is " + card.getCardValue() +
"\n\tCardID is " + card.getCardId() +
"\n\tIn Folder " + card.getOwningFolders(mGamesDB).getFolderName() +
"\n\t\tIn Category " + card.getOwningCategories(mGamesDB).getCategoryName()
);
}
}
- 注意为了方便起见allowMainThreadQueries()
已被使用。
2019-09-16 08:58:23.530 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is Jack
Value is 11
CardID is 1
In Folder CourtCards
In Category Spades
2019-09-16 08:58:23.534 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is Queen
Value is 12
CardID is 2
In Folder CourtCards
In Category Spades
2019-09-16 08:58:23.538 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is King
Value is 13
CardID is 3
In Folder CourtCards
In Category Spades
2019-09-16 08:58:23.546 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is Ace
Value is 1
CardID is 4
In Folder CourtCards
In Category Spades
2019-09-16 08:58:23.553 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is 2
Value is 2
CardID is 5
In Folder NormalCards
In Category Spades
2019-09-16 08:58:23.559 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is 3
Value is 3
CardID is 6
In Folder NormalCards
In Category Spades
......
2019-09-16 08:58:23.822 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is 8
Value is 8
CardID is 63
In Folder NormalCards
In Category ooops
2019-09-16 08:58:23.826 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is 9
Value is 9
CardID is 64
In Folder NormalCards
In Category ooops
2019-09-16 08:58:23.832 17468-17468/aso.so57943963foreignkeys D/CARDINFO: Card Name is 10
Value is 10
CardID is 65
In Folder NormalCards
In Category ooops
关于java - 3层数据库。如何将 3 个实体相互关联?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57943963/
我正在编写一个具有以下签名的 Java 方法。 void Logger(Method method, Object[] args); 如果一个方法(例如 ABC() )调用此方法 Logger,它应该
我是 Java 新手。 我的问题是我的 Java 程序找不到我试图用作的图像文件一个 JButton。 (目前这段代码什么也没做,因为我只是得到了想要的外观第一的)。这是我的主课 代码: packag
好的,今天我在接受采访,我已经编写 Java 代码多年了。采访中说“Java 垃圾收集是一个棘手的问题,我有几个 friend 一直在努力弄清楚。你在这方面做得怎么样?”。她是想骗我吗?还是我的一生都
我的 friend 给了我一个谜语让我解开。它是这样的: There are 100 people. Each one of them, in his turn, does the following
如果我将使用 Java 5 代码的应用程序编译成字节码,生成的 .class 文件是否能够在 Java 1.4 下运行? 如果后者可以工作并且我正在尝试在我的 Java 1.4 应用程序中使用 Jav
有关于why Java doesn't support unsigned types的问题以及一些关于处理无符号类型的问题。我做了一些搜索,似乎 Scala 也不支持无符号数据类型。限制是Java和S
我只是想知道在一个 java 版本中生成的字节码是否可以在其他 java 版本上运行 最佳答案 通常,字节码无需修改即可在 较新 版本的 Java 上运行。它不会在旧版本上运行,除非您使用特殊参数 (
我有一个关于在命令提示符下执行 java 程序的基本问题。 在某些机器上我们需要指定 -cp 。 (类路径)同时执行java程序 (test为java文件名与.class文件存在于同一目录下) jav
我已经阅读 StackOverflow 有一段时间了,现在我才鼓起勇气提出问题。我今年 20 岁,目前在我的家乡(罗马尼亚克卢日-纳波卡)就读 IT 大学。足以介绍:D。 基本上,我有一家提供簿记应用
我有 public JSONObject parseXML(String xml) { JSONObject jsonObject = XML.toJSONObject(xml); r
我已经在 Java 中实现了带有动态类型的简单解释语言。不幸的是我遇到了以下问题。测试时如下代码: def main() { def ks = Map[[1, 2]].keySet()
一直提示输入 1 到 10 的数字 - 结果应将 st、rd、th 和 nd 添加到数字中。编写一个程序,提示用户输入 1 到 10 之间的任意整数,然后以序数形式显示该整数并附加后缀。 public
我有这个 DownloadFile.java 并按预期下载该文件: import java.io.*; import java.net.URL; public class DownloadFile {
我想在 GUI 上添加延迟。我放置了 2 个 for 循环,然后重新绘制了一个标签,但这 2 个 for 循环一个接一个地执行,并且标签被重新绘制到最后一个。 我能做什么? for(int i=0;
我正在对对象 Student 的列表项进行一些测试,但是我更喜欢在 java 类对象中创建硬编码列表,然后从那里提取数据,而不是连接到数据库并在结果集中选择记录。然而,自从我这样做以来已经很长时间了,
我知道对象创建分为三个部分: 声明 实例化 初始化 classA{} classB extends classA{} classA obj = new classB(1,1); 实例化 它必须使用
我有兴趣使用 GPRS 构建车辆跟踪系统。但是,我有一些问题要问以前做过此操作的人: GPRS 是最好的技术吗?人们意识到任何问题吗? 我计划使用 Java/Java EE - 有更好的技术吗? 如果
我可以通过递归方法反转数组,例如:数组={1,2,3,4,5} 数组结果={5,4,3,2,1}但我的结果是相同的数组,我不知道为什么,请帮助我。 public class Recursion { p
有这样的标准方式吗? 包括 Java源代码-测试代码- Ant 或 Maven联合单元持续集成(可能是巡航控制)ClearCase 版本控制工具部署到应用服务器 最后我希望有一个自动构建和集成环境。
我什至不知道这是否可能,我非常怀疑它是否可能,但如果可以,您能告诉我怎么做吗?我只是想知道如何从打印机打印一些文本。 有什么想法吗? 最佳答案 这里有更简单的事情。 import javax.swin
我是一名优秀的程序员,十分优秀!