blocks|key|4357441|text|我也遇到过类似的问题。这是我的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4357442|您可以使用一个额外的实体(ReservationGuest)来保持Guest和Reservation之间的关系。|offset|length|style|CODE|4357443|@Entity+data+class+Guest(
++++@PrimaryKey+val+id:+Long,
++++val+name:+String,
++++val+email:+String
)

@Entity+data+class+Reservation(
++++@PrimaryKey+val+id:+Long,
++++val+table:+String
)

@Entity+data+class+ReservationGuest(
++++@PrimaryKey(autoGenerate+=+true)+val+id:+Long,
++++val+reservationId:+Long,
++++val+guestId:+Long
)|code-block|syntax|javascript|4357444|您可以通过他们的guestId列表(而不是来宾对象)获得预订。|4357445|data+class+ReservationWithGuests(
++++@Embedded+val+reservation:Reservation,
++++@Relation(
++++++++parentColumn+=+"id",
++++++++entityColumn+=+"reservationId",
++++++++entity+=+ReservationGuest::class,
++++++++projection+=+"guestId"
++++)+val+guestIdList:+List<Long>
)|4357446|您还可以使用客户的预订列表(而不是reservationId对象)来获取客户|4357447|data+class+GuestWithReservations(
++++@Embedded+val+guest:Guest,
++++@Relation(
++++++++parentColumn+=+"id",
++++++++entityColumn+=+"guestId",
++++++++entity+=+ReservationGuest::class,
++++++++projection+=+"reservationId"
+++)+val+reservationIdList:+List<Long>
)|4357448|由于可以获得guestId和reservationId，因此可以使用它们查询Reservation和Guest实体。|4357449|如果我找到一种简单的方法来获取预订和访客对象列表，而不是它们的ids，我会更新我的答案。|4357450|Similar+answer|4357451|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/44330452/android-persistence-room-cannot-figure-out-how-to-read-this-field-from-a-curso/44424148#44424148^0|0|D|G|X|5|13|B|0|0|8|7|0|0|H|D|0|0|6|7|E|D|12|B|1E|5|0|0|0|E|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@$D|1A|E|1B|F|G]|$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1G|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1H|8|@$D|1I|E|1J|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|1K|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|1L|8|@$D|1M|E|1N|F|G]]|9|@]|A|$]]|$1|S|3|T|5|J|7|1O|8|@]|9|@]|A|$K|L]]|$1|U|3|V|5|6|7|1P|8|@$D|1Q|E|1R|F|G]|$D|1S|E|1T|F|G]|$D|1U|E|1V|F|G]|$D|1W|E|1X|F|G]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1Z|8|@]|9|@$D|20|E|21|1|22]]|A|$]]|$1|10|3|-4|5|6|7|23|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]]]

I had a similar issue. Here is my solution.

You can use an extra entity (<code>ReservationGuest</code>) which keeps the relation between <code>Guest</code> and <code>Reservation</code>.

<pre><code>@Entity data class Guest(
 @PrimaryKey val id: Long,
 val name: String,
 val email: String
)

@Entity data class Reservation(
 @PrimaryKey val id: Long,
 val table: String
)

@Entity data class ReservationGuest(
 @PrimaryKey(autoGenerate = true) val id: Long,
 val reservationId: Long,
 val guestId: Long
)
</code></pre>

You can get reservations with their list of <code>guestId</code>s. (Not the guest objects)

<pre><code>data class ReservationWithGuests(
 @Embedded val reservation:Reservation,
 @Relation(
 parentColumn = "id",
 entityColumn = "reservationId",
 entity = ReservationGuest::class,
 projection = "guestId"
 ) val guestIdList: List&lt;Long&gt;
)
</code></pre>

You can also get guests with their list of <code>reservationId</code>s. (Not the reservation objects)

<pre><code>data class GuestWithReservations(
 @Embedded val guest:Guest,
 @Relation(
 parentColumn = "id",
 entityColumn = "guestId",
 entity = ReservationGuest::class,
 projection = "reservationId"
 ) val reservationIdList: List&lt;Long&gt;
)
</code></pre>

Since you can get the <code>guestId</code>s and <code>reservationId</code>s, you can query <code>Reservation</code> and <code>Guest</code> entities with those.

I'll update my answer if I find an easy way to fetch Reservation and Guest object list instead of their ids.

<a href="https://stackoverflow.com/questions/44330452/android-persistence-room-cannot-figure-out-how-to-read-this-field-from-a-curso/44424148#44424148">Similar answer</a>

blocks|key|4357499|text|基于上面的答案：https://stackoverflow.com/a/44428451/4992598仅通过在实体之间保留单独的字段名，您可以返回模型(而不仅仅是ids)。你所需要做的就是：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4357500|@Entity+data+class+ReservationGuest(
++++@PrimaryKey(autoGenerate+=+true)+val+id:+Long,
++++val+reservationId:+Long,
++++@Embedded
++++val+guest:+Guest
)|code-block|syntax|javascript|4357501|是的，只要你不保留重复的字段，实体就可以相互嵌入。因此，ReservationWithGuests类可能如下所示。|4357502|data+class+ReservationWithGuests(
++++@Embedded+val+reservation:Reservation,
++++@Relation(
++++++++parentColumn+=+"id",
++++++++entityColumn+=+"reservationId",
++++++++entity+=+ReservationGuest::class,
++++++++projection+=+"guestId"
++++)+val+guestList:+List<Guest>
)|4357503|因此，此时您可以使用val+guestIdList:+List，因为您的ReservationGuest实体实际上将ids映射到实体模型。|4357504|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/44428451/4992598^0|8|18|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|11|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

Based on the answer above: <a href="https://stackoverflow.com/a/44428451/4992598">https://stackoverflow.com/a/44428451/4992598</a> only by keeping separate field names between entities
you can have models returned (not just ids). All you need to do is:

<pre><code>@Entity data class ReservationGuest(
 @PrimaryKey(autoGenerate = true) val id: Long,
 val reservationId: Long,
 @Embedded
 val guest: Guest
)
</code></pre>

And yes entities can be embedded in one another as long as you don't keep duplicate fields. So in consequence the ReservationWithGuests class can look like this.

<pre><code>data class ReservationWithGuests(
 @Embedded val reservation:Reservation,
 @Relation(
 parentColumn = "id",
 entityColumn = "reservationId",
 entity = ReservationGuest::class,
 projection = "guestId"
 ) val guestList: List&lt;Guest&gt;
)
</code></pre>

So at this point you can use val guestIdList: List because your ReservationGuest entity actually maps ids with entity models.

blocks|key|4357146|text|实际上还有一种获取Guest列表的可能性，而不仅仅是@Devrim答案中的id。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4357147|首先定义表示Guest和Reservation之间连接的类。|4357148|@Entity(primaryKeys+=+["reservationId",+"guestId"],
++++++++foreignKeys+=+[
++++++++++++ForeignKey(entity+=+Reservation::class,
++++++++++++++++++++parentColumns+=+["id"],
++++++++++++++++++++childColumns+=+["reservationId"]),
++++++++++++ForeignKey(entity+=+Guest::class,
++++++++++++++++++++parentColumns+=+["id"],
++++++++++++++++++++childColumns+=+["guestId"])
++++++++])
data+class+ReservationGuestJoin(
++++val+reservationId:+Long,
++++val+guestId:+Long
)|code-block|syntax|javascript|4357149|每次要插入新的Reservation时，都必须插入ReservationGuestJoin对象以满足外键约束。现在，如果你想得到Guest列表，你可以使用SQL+query的强大功能：|4357150|@Dao
interface+ReservationGuestJoinDao+{

++++@SuppressWarnings(RoomWarnings.CURSOR_MISMATCH)
++++@Query("""
++++++++SELECT+*+FROM+guest+INNER+JOIN+reservationGuestJoin+ON
++++++++guest.id+=+reservationGuestJoin.guestId+WHERE
++++++++reservationGuestJoin.reservationId+=+:reservationId
++++++++""")
++++fun+getGuestsWithReservationId(reservationId:+Long):+List<Guest>
}|4357151|要查看更多详细信息，请访问this+blog。|4357152|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/44428451/4386668|1|https://android.jlelse.eu/android-architecture-components-room-relationships-bf473510c14a^0|9|5|Q|7|0|0|6|5|C|B|0|0|7|B|P|K|1S|5|0|0|D|9|1|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1L|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1M|8|@]|D|@$9|1N|A|1O|1|1P]]|E|$]]|$1|S|3|-4|5|6|7|1Q|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]|10|$5|V|W|X|E|$Y|11]]]]

Actually there is one more possibility to get <code>Guest</code> list, not only id's like in <a href="https://stackoverflow.com/a/44428451/4386668">@Devrim</a> answer.

First define class which will represent the connection between <code>Guest</code> and <code>Reservation</code>.

<pre><code>@Entity(primaryKeys = ["reservationId", "guestId"],
 foreignKeys = [
 ForeignKey(entity = Reservation::class,
 parentColumns = ["id"],
 childColumns = ["reservationId"]),
 ForeignKey(entity = Guest::class,
 parentColumns = ["id"],
 childColumns = ["guestId"])
 ])
data class ReservationGuestJoin(
 val reservationId: Long,
 val guestId: Long
)
</code></pre>

Each time you will be inserting new <code>Reservation</code>, you will have to insert <code>ReservationGuestJoin</code> object in order to fulfill foreign key constraint. 
And now if you want to get <code>Guest</code> list you can use power of SQL query:

<pre><code>@Dao
interface ReservationGuestJoinDao {

 @SuppressWarnings(RoomWarnings.CURSOR_MISMATCH)
 @Query("""
 SELECT * FROM guest INNER JOIN reservationGuestJoin ON
 guest.id = reservationGuestJoin.guestId WHERE
 reservationGuestJoin.reservationId = :reservationId
 """)
 fun getGuestsWithReservationId(reservationId: Long): List&lt;Guest&gt;
}
</code></pre>

To see more details visit <a href="https://android.jlelse.eu/android-architecture-components-room-relationships-bf473510c14a" rel="noreferrer">this blog</a>.

blocks|key|4357264|text|对于连接表实体，我建议使用复合ID索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4357265|@Entity(
++++primaryKeys+=+["reservationId",+"guestId"],
++++indices+=+[Index(value+=["reservationId",+"guestId"],+unique+=+true)]
)
data+class+ReservationGuestJoin(
++++@PrimaryKey(autoGenerate+=+true)+var+id:+Long,
++++var+reservationId:+Long+=+0,
++++var+guestId:+Long+=+0
)|code-block|syntax|javascript|4357266|GuestDao.kt：|4357267|@Dao
@TypeConverters(GuestDao.Converters::class)
interface+GuestDao+{

++++@Query(QUERY_STRING)
++++fun+listWithReservations():+LiveData<List<GuestWithReservations>>

++++data+class+GuestWithReservations(
++++++++var+id:+Long?+=+null,
++++++++var+name:+String?+=+null,
++++++++var+email:+String?+=+null,
++++++++var+reservations:+List<Reservation>+=+emptyList()
++++)

++++class+Converters{
++++++++@TypeConverter
++++++++fun+listReservationFromConcatString(value:+String?):+List<Reservation>?+=+value?.let+{+value+->
++++++++++++++++.split("%5E%5E")
++++++++++++++++.map+{+it.split("%5E_")+}
++++++++++++++++.map+{+Reservation(id+=+it.getOrNull(0)?.toLongOrNull(),+name+=+it.getOrNull(1))+}
++++++++}+?:+emptyList()
++++}
}|4357268|QUERY_STRING。我们做一个内部连接来产生一个包含来自两个实体的数据的大表，我们将来自Reservation的数据连接成一个列字符串，最后我们根据客户ID+group_concat行，用不同的分隔符连接保留字符串，我们的转换器将负责将其重新构建为一个实体：|offset|length|style|CODE|4357269|SELECT+
++++t.id,+t.name,+t.email,+GROUP_CONCAT(t.reservation,+'%5E%5E')+as+reservations+
FROM+(
++++SELECT+
++++++++guestId+as+id,+name,+email,+(reservationId+%7C%7C+'%5E_'+%7C%7C+reservationTable)+as+reservation+
++++FROM++
++++++++GuestReservationJoin
++++++++INNER+JOIN+Guest+ON+Guest.id+=+GuestReservationJoin.guestId+
++++++++INNER+JOIN+Reservation+ON+Reservation.id+=+GuestReservationJoin.reservationId
++++)+as+t+
GROUP+BY+t.id|4357270|请注意，我更改了您的列table名称，因为我认为SQLite不允许您使用保留名称。|4357271|与使用更扁平的实体(没有连接的另一种选择)相比，我没有测试所有这些的性能。如果我这样做了，我会更新我的答案。|4357272|entityMap^0|0|0|0|0|0|C|1B|B|0|0|B|5|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@$M|13|N|14|O|P]|$M|15|N|16|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|18|8|@$M|19|N|1A|O|P]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|X|$]]

For the join table entity, I suggest to use a composite ID indexed:

<pre><code>@Entity(
 primaryKeys = ["reservationId", "guestId"],
 indices = [Index(value =["reservationId", "guestId"], unique = true)]
)
data class ReservationGuestJoin(
 @PrimaryKey(autoGenerate = true) var id: Long,
 var reservationId: Long = 0,
 var guestId: Long = 0
)
</code></pre>

The GuestDao.kt:

<pre><code>@Dao
@TypeConverters(GuestDao.Converters::class)
interface GuestDao {

 @Query(QUERY_STRING)
 fun listWithReservations(): LiveData&lt;List&lt;GuestWithReservations&gt;&gt;

 data class GuestWithReservations(
 var id: Long? = null,
 var name: String? = null,
 var email: String? = null,
 var reservations: List&lt;Reservation&gt; = emptyList()
 )

 class Converters{
 @TypeConverter
 fun listReservationFromConcatString(value: String?): List&lt;Reservation&gt;? = value?.let { value -&gt;
 .split("^^")
 .map { it.split("^_") }
 .map { Reservation(id = it.getOrNull(0)?.toLongOrNull(), name = it.getOrNull(1)) }
 } ?: emptyList()
 }
}
</code></pre>

The <code>QUERY_STRING</code>. We make a inner joins to produce a big table with data from both entities, them we concatenate the data from <code>Reservation</code> as a column string and lastly we group_concat the rows by the guest ID, concatenating the reservation strings with different separators, our converter will take care of rebuild it as a entity:

<pre><code>SELECT 
 t.id, t.name, t.email, GROUP_CONCAT(t.reservation, '^^') as reservations 
FROM (
 SELECT 
 guestId as id, name, email, (reservationId || '^_' || reservationTable) as reservation 
 FROM 
 GuestReservationJoin
 INNER JOIN Guest ON Guest.id = GuestReservationJoin.guestId 
 INNER JOIN Reservation ON Reservation.id = GuestReservationJoin.reservationId
 ) as t 
GROUP BY t.id
</code></pre>

Note that I changed your column <code>table</code> name because I think Room do not allow you to use SQLite reserved names.

I didn't test the performance of all this compared with have more flat entity (another option without the concatenations). If I do, I'll update my answer.

blocks|key|4357345|text|通过在房间中介绍Junction，您可以轻松处理多对多关系。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4357346|正如@Devrim所说，您可以使用一个额外的实体(ReservationGuest)来保持客户和预订之间的关系(也称为关联表、连接表或连接表)。|4357347|@Entity
data+class+Guest(
++@PrimaryKey
++val+gId:+Long,
++val+name:+String,
++val+email:+String
)

@Entity
data+class+Reservation(
++@PrimaryKey
++val+rId:+Long,
++val+table:+String
)

@Entity(
++primaryKeys+=+["reservationId",+"guestId"]
)
data+class+ReservationGuest(+++++
++val+reservationId:+Long,
++val+guestId:+Long
)|code-block|syntax|javascript|4357348|现在，您可以使用此模型与客人进行预订：|4357349|data+class+ReservationWithGuests+(
++++@Embedded
++++val+reservation:+Reservation,
++++@Relation(
++++++++++++parentColumn+=+"rId",
++++++++++++entity+=+Guest::class,
++++++++++++entityColumn+=+"gId",
++++++++++++associateBy+=+Junction(
++++++++++++++++++++value+=+ReservationGuest::class,
++++++++++++++++++++parentColumn+=+"reservationId",
++++++++++++++++++++entityColumn+=+"guestId"
++++++++++++)
++++)
++++val+guests:+List<Guest>
)|4357350|您也可以通过客人的预订列表获得他们的预订。|4357351|data+class+GuestWithReservations+(
++@Embedded
++val+guest:+Guest,
++@Relation(
++++++++parentColumn+=+"gId",
++++++++entity+=+Reservation::class,
++++++++entityColumn+=+"rId",
++++++++associateBy+=+Junction(
++++++++++++++++value+=+ReservationGuest::class,
++++++++++++++++parentColumn+=+"guestId",
++++++++++++++++entityColumn+=+"reservationId"
++++++++)
++)
++val+reservations:+List<Reservation>
)|4357352|现在您可以在数据库中查询结果，如下所示：|4357353|@Dao
interface+GuestReservationDao+{
++@Query("SELECT+*+FROM+Reservation")
++fun+getReservationWithGuests():+LiveData<List<ReservationWithGuests>>

++@Query("SELECT+*+FROM+Guest")
++fun+getGuestWithReservations():+LiveData<List<GuestWithReservations>>

}|4357354|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.android.com/reference/androidx/room/Junction^0|8|8|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|6|7|18|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|19|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|1B|8|@]|9|@]|C|$I|J]]|$1|O|3|P|5|6|7|1C|8|@]|9|@]|C|$]]|$1|Q|3|R|5|H|7|1D|8|@]|9|@]|C|$I|J]]|$1|S|3|T|5|6|7|1E|8|@]|9|@]|C|$]]|$1|U|3|V|5|H|7|1F|8|@]|9|@]|C|$I|J]]|$1|W|3|-4|5|6|7|1G|8|@]|9|@]|C|$]]]|X|$Y|$5|Z|10|11|C|$12|13]]]]

With the introduction to <a href="https://developer.android.com/reference/androidx/room/Junction" rel="noreferrer">Junction</a> in room you can handle many-to-many relationship with ease.

As @Devrim stated you can use an extra entity (ReservationGuest) which keeps the relation between Guest and Reservation(also know as associative table or junction table or join table).

<pre><code>@Entity
data class Guest(
 @PrimaryKey
 val gId: Long,
 val name: String,
 val email: String
)

@Entity
data class Reservation(
 @PrimaryKey
 val rId: Long,
 val table: String
)

@Entity(
 primaryKeys = ["reservationId", "guestId"]
)
data class ReservationGuest( 
 val reservationId: Long,
 val guestId: Long
)
</code></pre>

Now you can get reservation with guests using this model: 

<pre><code>data class ReservationWithGuests (
 @Embedded
 val reservation: Reservation,
 @Relation(
 parentColumn = "rId",
 entity = Guest::class,
 entityColumn = "gId",
 associateBy = Junction(
 value = ReservationGuest::class,
 parentColumn = "reservationId",
 entityColumn = "guestId"
 )
 )
 val guests: List&lt;Guest&gt;
)
</code></pre>

You can also get guest with their list of reservations as.

<pre><code>data class GuestWithReservations (
 @Embedded
 val guest: Guest,
 @Relation(
 parentColumn = "gId",
 entity = Reservation::class,
 entityColumn = "rId",
 associateBy = Junction(
 value = ReservationGuest::class,
 parentColumn = "guestId",
 entityColumn = "reservationId"
 )
 )
 val reservations: List&lt;Reservation&gt;
)
</code></pre>

Now you can query database for the result as:

<pre><code>@Dao
interface GuestReservationDao {
 @Query("SELECT * FROM Reservation")
 fun getReservationWithGuests(): LiveData&lt;List&lt;ReservationWithGuests&gt;&gt;

 @Query("SELECT * FROM Guest")
 fun getGuestWithReservations(): LiveData&lt;List&lt;GuestWithReservations&gt;&gt;

}
</code></pre>

How can I represent a &quot;many to many&quot; relation with <code>Room</code>? My column names are also the same.
e.g. I have <code>Guest</code> and <code>Reservation</code>. <code>Reservation</code> can have many <code>Guest</code>'s and a <code>Guest</code> can be part of many Reservations.
Here are my entity definitions:
<pre><code>@Entity data class Reservation(
 @PrimaryKey val id: Long,
 val table: String,
 val guests: List&lt;Guest&gt;
)

@Entity data class Guest(
 @PrimaryKey val id: Long,
 val name: String,
 val email: String
)
</code></pre>
While looking into docs I came across <a href="https://developer.android.com/reference/android/arch/persistence/room/Relation.html" rel="nofollow noreferrer"><code>@Relation</code></a>. I found it really confusing though.
According to this I would want to create a POJO and add the relationships there. So, with my example I did the following:
<pre><code>data class ReservationForGuest(
 @Embedded val reservation: Reservation,
 @Relation(
 parentColumn = &quot;reservation.id&quot;, 
 entityColumn = &quot;id&quot;, 
 entity = Guest::class
 ) val guestList: List&lt;Guest&gt;
)
</code></pre>
With above I get the compiler error:
<pre><code>&gt; Cannot figure out how to read this field from a cursor.
</code></pre>
I wasn't able to find a working sample of <code>@Relation</code>.

How can I represent a "many to many" relation with Android Room when column names are same?

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如何用Room表示“多对多”关系？我的列名也是一样的。例如，我有Guest和Reservation。Reservation可以有多个Guest，一个Guest可以是许多预留的一部分。以下是我的实体定义：@Entity data class Reservation(    @PrimaryKey val id: Long...

问在列名相同的情况下，如何表示Android Room的多对多关系？
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问在列名相同的情况下，如何表示Android Room的多对多关系？
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