使用 spring 数据从 mongodb 中随机选取条目
Picking random entries from mongodb using spring data
我正在尝试使用 spring 从 mongodb 中获取 x 个随机条目。
我的存储库如下所示
public interface StoryRepository extends MongoRepository<Story, Long> {
@Query("{$sample: {size: ?0} }")
List<Story> findRandom(int quantity);
}
我得到的错误看起来像这样
com.mongodb.BasicDBObject cannot be cast to org.springframework.data.domain.Example
我也尝试了以下给出完全相同的错误
public List<Story> findRandom(final int quantity) {
CustomAggregationOperation customAggregationOperation = new CustomAggregationOperation(new BasicDBObject("$sample", new BasicDBObject("size", quantity)));
TypedAggregation<Story> aggregation = new TypedAggregation<>(Story.class, customAggregationOperation);
AggregationResults<Story> aggregationResults = mongoTemplate.aggregate(aggregation, Story.class);
return aggregationResults.getMappedResults();
}
我的故事class如下所示
public class Story {
@Id
private long id;
private String by;
private int descendants;
private List<Long> kids;
private int score;
private long time;
private String title;
private String type;
private String url;
private By author;
public long getId() {
return id;
}
...
}
而我的pom文件如下
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>dk.tons.hackernews.backend</groupId>
<artifactId>tons-hackernews-backend</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>jar</packaging>
<name>Backend</name>
<description>Tons Hacker News Backend</description>
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>1.4.0.RELEASE</version>
<relativePath/> <!-- lookup parent from repository -->
</parent>
<properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<project.reporting.outputEncoding>UTF-8</project.reporting.outputEncoding>
<java.version>1.8</java.version>
</properties>
<dependencies>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-mongodb</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-rest</artifactId>
</dependency>
<dependency>
<groupId>com.h2database</groupId>
<artifactId>h2</artifactId>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-test</artifactId>
<scope>test</scope>
</dependency>
</dependencies>
<build>
<plugins>
<plugin>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-maven-plugin</artifactId>
</plugin>
</plugins>
</build>
有什么线索吗?
为什么会失败
您使用自定义查询 @Query("{$sample: {size: ?0} }")
and/or 定义了您的 CustomAggregationOperation
(使用 context.getMappedObject
):
public class CustomAggregationOperation implements AggregationOperation {
private DBObject operation;
public CustomAggregationOperation (DBObject operation) {
this.operation = operation;
}
@Override
public DBObject toDBObject(final AggregationOperationContext context) {
return context.getMappedObject(operation);
}
}
两者都经过 QueryMapper.getMappedKeyword
,这是引发错误的 spring 方法。如果你打开 spring 的 QueryMapper.getMappedKeyword
,你会看到:
protected DBObject getMappedKeyword(Keyword keyword, MongoPersistentEntity<?> entity) {
...
if (keyword.isSample()) {
return exampleMapper.getMappedExample(keyword.<Example<?>> getValue(), entity);
}
...
}
public boolean isSample() {
return "$sample".equalsIgnoreCase(key);
}
它解析查询并在找到单词 $sample
时尝试使用 Example
。 这解释了您的错误。
现在的问题是:如何在没有 $sample
或绕过这段逻辑的情况下实现您想要的?此外,对 Spring 的 JIRA 提出了改进请求,这将确认 $sample 不受开箱即用的支持:https://jira.spring.io/browse/DATAMONGO-1415
(1) 在不使用 AggregationOperationContext 的情况下实现 CustomSampleOperation
Return 不使用上下文的 $sample 查询:
CustomSampleOperation customSampleOperation = new CustomSampleOperation(1);
TypedAggregation<Story> typedAggr = Aggregation.newAggregation(Story.class,
customSampleperation);
AggregationResults<Story> aggregationResults = mongoTemplate.aggregate(typedAggr, Story.class);
aggregationResults.getMappedResults().get(0);
...
public class CustomSampleOperation implements AggregationOperation {
private int size;
public CustomSampleOperation(int size){
this.size = size;
}
@Override
public DBObject toDBObject(final AggregationOperationContext context){
return new BasicDBObject("$sample", new BasicDBObject("size", size));
}
}
如果你看看其他操作是如何写的,我们就在 (LimitOperation):
public class LimitOperation implements AggregationOperation {
private final long maxElements;
public LimitOperation(long maxElements) {
this.maxElements = maxElements;
}
public DBObject toDBObject(AggregationOperationContext context) {
return new BasicDBObject("$limit", maxElements);
}
}
(2) 如果需要,请将其设为通用
为了让您的 CustomOperation 保持通用,您可以这样定义它:
CustomGenericOperation customGenericOperation =
new CustomGenericOperation(new BasicDBObject("$sample", new BasicDBObject("size", 1)));
...
public class CustomGenericOperation implements AggregationOperation {
private DBObject dbObject;
public CustomGenericOperation(DBObject dbObject){
this.dbObject = dbObject;
}
@Override
public DBObject toDBObject(final AggregationOperationContext context) {
return dbObject;
}
}
(3) 备选方案
您可以:
而不是定义自定义 AggregationOperation
- 在Java中得到一个随机数(假设你先取回集合中的文档数)
- 在聚合查询中
- 限制(随机数)
- 升序排序
- 限制(1)
简而言之,用一个随机数限制,得到最后一个文档:
$ db.story.aggregate([{$limit: RANDOM_NUMBER},{$sort: {_id: 1}}, {$limit: 1}])
我正在尝试使用 spring 从 mongodb 中获取 x 个随机条目。
我的存储库如下所示
public interface StoryRepository extends MongoRepository<Story, Long> {
@Query("{$sample: {size: ?0} }")
List<Story> findRandom(int quantity);
}
我得到的错误看起来像这样
com.mongodb.BasicDBObject cannot be cast to org.springframework.data.domain.Example
我也尝试了以下给出完全相同的错误
public List<Story> findRandom(final int quantity) {
CustomAggregationOperation customAggregationOperation = new CustomAggregationOperation(new BasicDBObject("$sample", new BasicDBObject("size", quantity)));
TypedAggregation<Story> aggregation = new TypedAggregation<>(Story.class, customAggregationOperation);
AggregationResults<Story> aggregationResults = mongoTemplate.aggregate(aggregation, Story.class);
return aggregationResults.getMappedResults();
}
我的故事class如下所示
public class Story {
@Id
private long id;
private String by;
private int descendants;
private List<Long> kids;
private int score;
private long time;
private String title;
private String type;
private String url;
private By author;
public long getId() {
return id;
}
...
}
而我的pom文件如下
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>dk.tons.hackernews.backend</groupId>
<artifactId>tons-hackernews-backend</artifactId>
<version>0.0.1-SNAPSHOT</version>
<packaging>jar</packaging>
<name>Backend</name>
<description>Tons Hacker News Backend</description>
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>1.4.0.RELEASE</version>
<relativePath/> <!-- lookup parent from repository -->
</parent>
<properties>
<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
<project.reporting.outputEncoding>UTF-8</project.reporting.outputEncoding>
<java.version>1.8</java.version>
</properties>
<dependencies>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-mongodb</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-rest</artifactId>
</dependency>
<dependency>
<groupId>com.h2database</groupId>
<artifactId>h2</artifactId>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-test</artifactId>
<scope>test</scope>
</dependency>
</dependencies>
<build>
<plugins>
<plugin>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-maven-plugin</artifactId>
</plugin>
</plugins>
</build>
有什么线索吗?
为什么会失败
您使用自定义查询 @Query("{$sample: {size: ?0} }")
and/or 定义了您的 CustomAggregationOperation
(使用 context.getMappedObject
):
public class CustomAggregationOperation implements AggregationOperation {
private DBObject operation;
public CustomAggregationOperation (DBObject operation) {
this.operation = operation;
}
@Override
public DBObject toDBObject(final AggregationOperationContext context) {
return context.getMappedObject(operation);
}
}
两者都经过 QueryMapper.getMappedKeyword
,这是引发错误的 spring 方法。如果你打开 spring 的 QueryMapper.getMappedKeyword
,你会看到:
protected DBObject getMappedKeyword(Keyword keyword, MongoPersistentEntity<?> entity) {
...
if (keyword.isSample()) {
return exampleMapper.getMappedExample(keyword.<Example<?>> getValue(), entity);
}
...
}
public boolean isSample() {
return "$sample".equalsIgnoreCase(key);
}
它解析查询并在找到单词 $sample
时尝试使用 Example
。 这解释了您的错误。
现在的问题是:如何在没有 $sample
或绕过这段逻辑的情况下实现您想要的?此外,对 Spring 的 JIRA 提出了改进请求,这将确认 $sample 不受开箱即用的支持:https://jira.spring.io/browse/DATAMONGO-1415
(1) 在不使用 AggregationOperationContext 的情况下实现 CustomSampleOperation
Return 不使用上下文的 $sample 查询:
CustomSampleOperation customSampleOperation = new CustomSampleOperation(1);
TypedAggregation<Story> typedAggr = Aggregation.newAggregation(Story.class,
customSampleperation);
AggregationResults<Story> aggregationResults = mongoTemplate.aggregate(typedAggr, Story.class);
aggregationResults.getMappedResults().get(0);
...
public class CustomSampleOperation implements AggregationOperation {
private int size;
public CustomSampleOperation(int size){
this.size = size;
}
@Override
public DBObject toDBObject(final AggregationOperationContext context){
return new BasicDBObject("$sample", new BasicDBObject("size", size));
}
}
如果你看看其他操作是如何写的,我们就在 (LimitOperation):
public class LimitOperation implements AggregationOperation {
private final long maxElements;
public LimitOperation(long maxElements) {
this.maxElements = maxElements;
}
public DBObject toDBObject(AggregationOperationContext context) {
return new BasicDBObject("$limit", maxElements);
}
}
(2) 如果需要,请将其设为通用
为了让您的 CustomOperation 保持通用,您可以这样定义它:
CustomGenericOperation customGenericOperation =
new CustomGenericOperation(new BasicDBObject("$sample", new BasicDBObject("size", 1)));
...
public class CustomGenericOperation implements AggregationOperation {
private DBObject dbObject;
public CustomGenericOperation(DBObject dbObject){
this.dbObject = dbObject;
}
@Override
public DBObject toDBObject(final AggregationOperationContext context) {
return dbObject;
}
}
(3) 备选方案
您可以:
而不是定义自定义 AggregationOperation- 在Java中得到一个随机数(假设你先取回集合中的文档数)
- 在聚合查询中
- 限制(随机数)
- 升序排序
- 限制(1)
简而言之,用一个随机数限制,得到最后一个文档:
$ db.story.aggregate([{$limit: RANDOM_NUMBER},{$sort: {_id: 1}}, {$limit: 1}])