具有用户定义类型的映射的Sprint-data-Cassandra映射失败，并显示“不是持久实体的类型”。 [英] Sprint-data-Cassandra Mapping of Map with User Defined Type fails with "not of type of the persistent entity"
问题描述
这是我在casssandra中的用户定义类型：
Here is my User Defined Type in casssandra:
create type app
( app_id uuid,
app_name varchar,
maker varchar
);
我的表格以及此类用户定义类型的地图
And my Table with a Map of such User DefinedType
create table device
( device_id timeuuid,
apps map<uuid,frozen<app>>,
primary key (device_id)
);
其Java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Sprind-data-cassandra CrudRepository：
Now using a standard Sprind-data-cassandra CrudRepository:
public interface DeviceRepository extends CrudRepository<Device, UUID>{
当我尝试将其保存为这样时
When I try to save to it as such
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);
我收到此异常：
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]
这是我从gradle进行的设置：
Here is my setup from gradle:
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")
推荐答案
当前spring数据不支持自定义数据类型，因为地图中 app中的值你的情况。
即使纠正了此错误，您也可能会碰到其他错误，例如在实体[Device]中类型[interface java.util.Map]的属性[app]的Collections中仅允许原始类型。
Currentlly spring data does not support custom data type as values in map "app" in your case. Even if you rectify this error you stumble upon other like Only primitive types are allowed inside Collections for property [app] of type [interface java.util.Map] in entity [Device].
就此而言，通过使用 @CassandraType（type = Name.MAP，userTypeName = address，typeArguments = {Name.TEXT，Name.CUSTOM}），错误将消失。
As far as this is concern the error will be gone by having "@CassandraType(type = Name.MAP, userTypeName = "address",typeArguments={Name.TEXT,Name.CUSTOM})"
在类设备上的应用程序属性中。
注意* type如果地图仅支持基本类型，则假定添加了CUSTOM。
on app property in class device. NOTE* typeArguments in case of map supports only primitive type do CUSTOM is hypothetically added.
这篇关于具有用户定义类型的映射的Sprint-data-Cassandra映射失败，并显示“不是持久实体的类型”。的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
