java – 如何序列化类?
发布时间:2020-05-24 14:36:26 所属栏目:Java 来源:互联网
导读:当我将列表插入 mongodb时,有一个问题: Exception in thread main java.lang.IllegalArgumentException: cant serialize class mongodb.Person at org.bson.BasicBSONEncoder._putObjectField(BasicBSONEncoder.java:234)
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当我将列表插入 mongodb时,有一个问题: Exception in thread "main" java.lang.IllegalArgumentException: can't serialize class mongodb.Person
at org.bson.BasicBSONEncoder._putObjectField(BasicBSONEncoder.java:234)
at org.bson.BasicBSONEncoder.putIterable(BasicBSONEncoder.java:259)
at org.bson.BasicBSONEncoder._putObjectField(BasicBSONEncoder.java:198)
at org.bson.BasicBSONEncoder.putObject(BasicBSONEncoder.java:140)
at org.bson.BasicBSONEncoder.putObject(BasicBSONEncoder.java:86)
at com.mongodb.DefaultDBEncoder.writeObject(DefaultDBEncoder.java:27)
at com.mongodb.OutMessage.putObject(OutMessage.java:142)
at com.mongodb.DBApiLayer$MyCollection.insert(DBApiLayer.java:252)
at com.mongodb.DBApiLayer$MyCollection.insert(DBApiLayer.java:211)
at com.mongodb.DBCollection.insert(DBCollection.java:57)
at com.mongodb.DBCollection.insert(DBCollection.java:87)
at com.mongodb.DBCollection.save(DBCollection.java:716)
at com.mongodb.DBCollection.save(DBCollection.java:691)
at mongodb.MongoDB.main(MongoDB.java:45)
类Person定义如下: class Person{
private String name;
public Person(String name){
this.name = name;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
程序是: DBCollection coll = db.getCollection("test");
DBObject record = new BasicDBObject();
List<Person> persons= new ArrayList<Person>();
persons.add(new Person("Jack"));
record.put("person",persons);
coll.save(record);
我找不到谷歌的答案,所以请帮助我. 解决方法只需在Person类中实现 Serializable界面.在你的类中定义一个serialVersionUID也是很好的. AFAIK,在Java中创建POJO类时,类应该是可序列化的,如果要通过某些流传输,具有默认构造函数,并允许使用getter和setter方法访问属性/字段. 你可能有兴趣阅读:Discover the secrets of the Java Serialization API (编辑:安卓应用网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
