我正在使用spring-data-mongodb 1.8.0; MongoDB 3.0.6; mongo-java-driver 3.1.0; spring-framework.version 4.0.3 .
我想要的是查询某些电话号码的 user
列表 . 用户示例: { "_id" : ObjectId("5625e5c32e1ca013a03f0d1b"), "phone" : "12345535"}
在Mongo Shell db.user.find({phone: { $in: [ "12345535", "123535"]}})
工作正常 . 但在 Spring 天我失败了 . Java类用户(省略了getter / setter):
@Document(collection = "user")
public class User {
@Id
String id;
String phone;
}
我尝试的是:
Query q = new Query(Criteria.where("phone").in("12345535","123535"));
mongoTemplate.find(q, User.class);
它出现了错误:
Exception in thread "main" java.lang.IllegalAccessError: tried to access class org.springframework.beans.PropertyMatches from class org.springframework.data.mapping.PropertyReferenceException
at org.springframework.data.mapping.PropertyReferenceException.detectPotentialMatches(PropertyReferenceException.java:134)
at org.springframework.data.mapping.PropertyReferenceException.<init>(PropertyReferenceException.java:59)
at org.springframework.data.mapping.PropertyPath.<init>(PropertyPath.java:75)
at org.springframework.data.mapping.PropertyPath.create(PropertyPath.java:327)
at org.springframework.data.mapping.PropertyPath.create(PropertyPath.java:307)
at org.springframework.data.mapping.PropertyPath.from(PropertyPath.java:270)
at org.springframework.data.mongodb.core.convert.QueryMapper$MetadataBackedField.getPath(QueryMapper.java:837)
at org.springframework.data.mongodb.core.convert.QueryMapper$MetadataBackedField.<init>(QueryMapper.java:729)
at org.springframework.data.mongodb.core.convert.QueryMapper$MetadataBackedField.with(QueryMapper.java:740)
at org.springframework.data.mongodb.core.convert.QueryMapper$MetadataBackedField.with(QueryMapper.java:686)
at org.springframework.data.mongodb.core.convert.QueryMapper.getMappedKeyword(QueryMapper.java:258)
at org.springframework.data.mongodb.core.convert.QueryMapper.getMappedObjectForField(QueryMapper.java:200)
at org.springframework.data.mongodb.core.convert.QueryMapper.getMappedObject(QueryMapper.java:123)
at org.springframework.data.mongodb.core.MongoTemplate.doFind(MongoTemplate.java:1700)
at org.springframework.data.mongodb.core.MongoTemplate.doFind(MongoTemplate.java:1690)
at org.springframework.data.mongodb.core.MongoTemplate.find(MongoTemplate.java:602)
at org.springframework.data.mongodb.core.MongoTemplate.find(MongoTemplate.java:593)
at com.example.TestMongo.main(TestMongo.java:30)
但是通过将现场电话更改为id,相同的代码工作正常 .
Query q = new Query(Criteria.where("id").in("5625e5c32e1ca013a03f0d1b","f0d1e"));
mongoTemplate.find(q, User.class);
通过调试,我发现它甚至没有进入请求阶段,在查询构建阶段发生了错误 . 似乎 $in
无法由 PropertyPath.create
处理,而在 id
情况下,它可以 .
我怎样才能解决这个问题?我是一个新手,搜索了很多但没有运气 . 可以帮助我 . 每个答案都表示赞赏 . 多谢你们 .
2 回答
如announcement blog和release train wiki所示,Spring Data MongoDB 1.8需要Spring 4.1,理想情况下4.1.8包含一个重要的安全修复程序 .
使用Spring 4.2.3.RELEASE和Spring MongoDB 1.6.1时出现了这个问题
切换到Spring mongoDB 1.8.1解决了这个问题 .
(作为对@ OliverGierke的答案的评论,但由于声誉水平低,无法做到这一点 . )