小编Vol*_*man的帖子

更新到Google Play Services 6.5.87后,由于缺少LocationCLient类,我的应用程序无法编译.

的文档链接 是在瞬间损坏(404未找到)

我该如何解决？我想收到位置更新,使用地理围栏等.

房间1.1.0版.

迁移后第一次运行时出现此错误.

如果我关闭应用程序并再次启动它,它运行良好.

ROOM: Cannot run invalidation tracker. Is the db closed?
java.lang.IllegalStateException: Cannot perform this operation because the connection pool has been closed.
    at android.database.sqlite.SQLiteConnectionPool.throwIfClosedLocked(SQLiteConnectionPool.java:1182)
    at android.database.sqlite.SQLiteConnectionPool.waitForConnection(SQLiteConnectionPool.java:792)
    at android.database.sqlite.SQLiteConnectionPool.acquireConnection(SQLiteConnectionPool.java:518)

Caused By : SQL(query) error or missing database.
    (no such table: room_table_modification_log (code 1): , while compiling: DELETE FROM proposals WHERE hotel_id = ?)
#################################################################
    at android.database.sqlite.SQLiteConnection.nativePrepareStatement(Native Method)
    at android.database.sqlite.SQLiteConnection.acquirePreparedStatement(SQLiteConnection.java:1095)
    at android.database.sqlite.SQLiteConnection.prepare(SQLiteConnection.java:660)
    at android.database.sqlite.SQLiteSession.prepare(SQLiteSession.java:588)
    at android.database.sqlite.SQLiteProgram.<init>(SQLiteProgram.java:59)
    at android.database.sqlite.SQLiteStatement.<init>(SQLiteStatement.java:31)
    at android.database.sqlite.SQLiteDatabase.compileStatement(SQLiteDatabase.java:1417)
    at android.arch.persistence.db.framework.FrameworkSQLiteDatabase.compileStatement(FrameworkSQLiteDatabase.java:64)
    at android.arch.persistence.room.RoomDatabase.compileStatement(RoomDatabase.java:244)
    at android.arch.persistence.room.SharedSQLiteStatement.createNewStatement(SharedSQLiteStatement.java:65)
    at android.arch.persistence.room.SharedSQLiteStatement.getStmt(SharedSQLiteStatement.java:77)
    at android.arch.persistence.room.SharedSQLiteStatement.acquire(SharedSQLiteStatement.java:87) …

以下代码仅在 observable2 完成后才从 observable1 发出项目。

observable1.startWith(observable2)
           .subscribe()

我需要实现另一种行为

observable1 ->       0   1   2   3
observable2 -> 1   2   3   4   5   6

observable1.startWithDefault(observable2)
            -> 1   2 0   1   2   3

第二个可观察对象仅在第一个可观察对象为空时发出项目，然后发出第一个可观察对象中的项目。

我无法仅使用基本运算符找到正确的解决方案，自定义运算符 startWithDefault 的正确 RxJava 2 实现应该是什么样子？

聚苯乙烯

observable1.subscribe()
observable2.takeUntil(observable1).subscribe()

不是正确的解决方案，因为在 observable1 立即发出时存在竞争