我测试Android中的外键,我有问题,我不明白:的Android SQLite和外键遇事
创建表(通过硬编码值)
db.execSQL("CREATE TABLE IF NOT EXISTS table_A (" +
"_id long primary key , value1 long );");
db.execSQL("CREATE TABLE IF NOT EXISTS table_B (" +
"_id long primary key , value1fk long , value2 long,"+
"FOREIGN KEY (value1fk) REFERENCES table_A (value1) ON DELETE CASCADE);");
然后我执行:
ContentValues values = new ContentValues();
values.put("_id", 1);
values.put("value1", 200);
long result = mDb.insert("table_A", null, values);
Log.e("","done (" + result + ")");
values = new ContentValues();
values.put("_id", 1);
values.put("value1fk", 200);
values.put("value2", 10);
result= mDb.insert("table_B", null, values);
Log.e("","done (" + result + ")");
输出是
done(1)
done(-1)
给予二次插入的外键不匹配
E/SQLiteDatabase(25476): Error inserting _id=1 value1fk=200 value2=10
E/SQLiteDatabase(25476): android.database.sqlite.SQLiteException: foreign key mismatch: , while compiling: INSERT INTO table_B(_id,value1fk,value2) VALUES (?,?,?)
这是为什么发生?
这是在副本中的错误与后来的代码(编辑) – Addev 2012-04-13 15:53:55