Question

这是我在casssandra中的用户定义类型：

  create type app
  (   app_id uuid,
      app_name varchar,
      maker varchar
   );

我的表格中包含此类User DefinedType的地图

 create table device 
 (   device_id timeuuid, 
     apps map<uuid,frozen<app>>,

     primary key (device_id) 
);

Java中的映射

@Table("device")
public class Device {

   @PrimaryKey
   @Column("device_id")
   private UUID device_id;


   @Column("apps")
   private Map<UUID, App> apps;

}

和

@UserDefinedType
public class App {

   @Column("app_id")
   private UUID app_id;

   @Column("app_name")
   private String app_name;

   @Column("maker")
   private String maker;

}

现在使用标准的Sprind-data-cassandra CrudRepository：

public interface DeviceRepository extends CrudRepository<Device, UUID>{

当我试图保存到它时

    Device st1 = new Device();
    st1.setDevice_id(MyUtils.getRandomTimestampUUID());

    /** Apps **/

    App c1 = new App();
    c1.setApp_id(MyUtils.getRandomTimestampUUID());
    c1.setMaker("Maker of game 1");
    c1.setApp_name("game 1");

    App c2 = new App();
    c2.setApp_id(MyUtils.getRandomTimestampUUID());
    c2.setMaker("Maker of game 2");
    c2.setApp_name("game 2");

    Map<UUID, App> apps = new LinkedHashMap<UUID, App>();
    apps.put(MyUtils.getRandomTimestampUUID(), c1);
    apps.put(MyUtils.getRandomTimestampUUID(), c2);

    st1.setApps(apps);

    _deviceRepository.save(st1);

我得到了这个例外：

 java.lang.IllegalArgumentException: Target bean of type java.util.LinkedHashMap is not of type of the persistent entity (com.pfellwock.cassandra.type.App)!
    at org.springframework.util.Assert.isTrue(Assert.java:68) ~[spring-core-4.3.2.RELEASE.jar:4.3.2.RELEASE]
    at org.springframework.data.mapping.model.BasicPersistentEntity.getPropertyAccessor(BasicPersistentEntity.java:397) ~[spring-data-commons-1.12.2.RELEASE.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.getConvertingAccessor(MappingCassandraConverter.java:608) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.write(MappingCassandraConverter.java:322) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.getWriteValue(MappingCassandraConverter.java:713) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.getWriteValue(MappingCassandraConverter.java:665) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.access$000(MappingCassandraConverter.java:86) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter$2.doWithPersistentProperty(MappingCassandraConverter.java:340) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter$2.doWithPersistentProperty(MappingCassandraConverter.java:335) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.mapping.model.BasicPersistentEntity.doWithProperties(BasicPersistentEntity.java:312) ~[spring-data-commons-1.12.2.RELEASE.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.writeInsertFromWrapper(MappingCassandraConverter.java:335) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.writeInsertFromObject(MappingCassandraConverter.java:329) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.write(MappingCassandraConverter.java:314) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.convert.MappingCassandraConverter.write(MappingCassandraConverter.java:298) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.core.CassandraTemplate.createInsertQuery(CassandraTemplate.java:948) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.core.CassandraTemplate.createInsertQuery(CassandraTemplate.java:717) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.core.CassandraTemplate.doInsert(CassandraTemplate.java:708) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.core.CassandraTemplate.insert(CassandraTemplate.java:290) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]
    at org.springframework.data.cassandra.core.CassandraTemplate.insert(CassandraTemplate.java:285) ~[spring-data-cassandra-1.5.0.BUILD-SNAPSHOT.jar:na]

以下是gradle的设置：

compile("com.datastax.cassandra:cassandra-driver-core:3.1.1")
compile('org.springframework.boot:spring-boot-starter-data-cassandra')
compile("org.springframework.data:spring-data-cassandra:1.5.0.BUILD-SNAPSHOT")
compile("org.springframework.data:spring-cql:1.5.0.BUILD-SNAPSHOT")

Answer 1

Currentlly spring数据不支持自定义数据类型作为地图“app”中的值。即使你纠正了这个错误，你也会偶然发现其他类似的实体[Device]中类型[interface java.util.Map]的属性[app]的Collections中允许只有原始类型。

就此而言，错误将通过“@CassandraType（type = Name.MAP，userTypeName =”address“，typeArguments = {Name.TEXT，Name.CUSTOM}）”

而消失。

在类设备中的app属性上。注意* map中的typeArguments仅支持原始类型，假设CUSTOM是假设的。

具有用户定义类型的映射的Sprint-data-Cassandra映射失败，其中＆＃34;不是持久实体的类型＆＃34;

1 个答案: