我正在尝试创建一个数据库(目前)有 4 个表 {USERS,TOPICS,GROUPS,CREATED_TOPICS}
。当我尝试运行它时,我不断收到此错误...
syntax error (code 1): , while compiling: create table created_topics_table (CREATED_TOPIC_ID INTEGER PRIMARY KEY AUTOINCREMENT, TOPIC_ID INTEGER,USER_ID INTEGER, FOREIGN KEY(TOPIC_ID) REFERENCES topics_table(_ID),FOREIGN KEY(USER_ID) REFERENCES users_table(_ID).
这是代码...
public class databaseHelper extends SQLiteOpenHelper
{
//DATABASE NAME
public static final String DATABASE_NAME = "users.db";
//USERS TABLE
public static final String TABLE_NAME_USERS = "users_table";
public static final String COL_USER_ID = "_ID";
public static final String COL_USER_NAME = "USER_NAME";
public static final String COL_USER_EMAIL = "USER_EMAIL";
public static final String COL_USER_PASSWORD = "USER_PASSWORD";
//TOPICS TABLE
public static final String TABLE_NAME_TOPICS = "topics_table";
public static final String COL_TOPIC_ID = "_ID";
public static final String COL_TOPIC_NAME = "TOPIC_NAME";
//GROUPS TABLE
public static final String TABLE_NAME_GROUPS = "groups_table";
public static final String COL_GROUP_ID = "GROUP_ID";
public static final String COL_GROUP_NAME = "GROUP_NAME";
public static final String TABLE_NAME_CREATED_TOPICS =
"created_topics_table";
public static final String COL_CREATED_TOPIC_ID = "CREATED_TOPIC_ID";
public static final String COL_FK_TOPIC_TOPIC_ID = "TOPIC_ID";
public static final String COL_FK_TOPIC_USER_ID = "USER_ID";
public void onCreate(SQLiteDatabase db) {
db.execSQL("create table " + TABLE_NAME_USERS + " (_ID INTEGER PRIMARY KEY
AUTOINCREMENT,USER_NAME TEXT,USER_EMAIL TEXT,USER_PASSWORD TEXT)");
db.execSQL("create table " + TABLE_NAME_TOPICS + " (_ID INTEGER PRIMARY KEY
AUTOINCREMENT,TOPIC_NAME TEXT)");
db.execSQL("create table " + TABLE_NAME_GROUPS + " (GROUPS_ID INTEGER PRIMARY
KEY AUTOINCREMENT,GROUP_NAME TEXT)");
db.execSQL("create table " + TABLE_NAME_CREATED_TOPICS + " (CREATED_TOPIC_ID
INTEGER PRIMARY KEY AUTOINCREMENT, TOPIC_ID INTEGER,USER_ID INTEGER, FOREIGN
KEY(TOPIC_ID) REFERENCES " +TABLE_NAME_TOPICS + "(_ID)," +
"FOREIGN KEY(USER_ID) REFERENCES " +TABLE_NAME_USERS + "(_ID)");
}
最佳答案
你错过了最后的右括号:
create table created_topics_table (CREATED_TOPIC_ID INTEGER PRIMARY KEY AUTOINCREMENT, TOPIC_ID INTEGER,USER_ID INTEGER, FOREIGN KEY(TOPIC_ID) REFERENCES topics_table(_ID),FOREIGN KEY (USER_ID) REFERENCES users_table(_ID))
关于 sqlite 外键的另一个注释。默认情况下它们是关闭的。每次打开数据库连接时都需要使用"PRAGMA foreign_keys=ON"
编辑
改变:
db.execSQL("create table " + TABLE_NAME_CREATED_TOPICS + " (CREATED_TOPIC_ID
INTEGER PRIMARY KEY AUTOINCREMENT, TOPIC_ID INTEGER,USER_ID INTEGER, FOREIGN
KEY (TOPIC_ID) REFERENCES " +TABLE_NAME_TOPICS + " (_ID)," +
"FOREIGN KEY (USER_ID) REFERENCES " +TABLE_NAME_USERS + " (_ID))");
关于android - 使用外键 ANDROID 创建数据库表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36974850/