在保存包含多个json的json对象时，该json对象被保存为单个数组而不是多个行。
样本json：
[
{
"id" : 1, -- this is not a primary key and not unique but cannot be null
"name" : "John Doe",
"phone" : [
{ "type" : "home", "ref" : "111-111-1234"},
{ "type" : "work", "ref" : "222-222-2222"}
]
},
{
"id" : 2, -- this is not a primary key and not unique but cannot be null
"name" : "Jane Doe",
"phone" : [
{ "type" : "home", "ref" : "111-111-1234"},
{ "type" : "work", "ref" : "222-222-2222"}
]
}
]
这是我保存到数据库后需要的
id name phone
1 John Doe { "type" : "home", "ref" : "111-111-1234"}
1 John Doe { "type" : "work", "ref" : "222-222-2222"}
2 Jane Doe { "type" : "home", "ref" : "111-111-1234"}
2 Jane Doe { "type" : "work", "ref" : "222-222-2222"}
这就是我要得到的
id name phone
1 John Doe [{ "type" : "home", "ref" : "111-111-1234"},{ "type" : "work", "ref" : "222-222-2222"}]
2 Jane Doe [{ "type" : "home", "ref" : "111-111-1234"},{ "type" : "work", "ref" : "222-222-2222"}]
这是我将json对象解析为pojo并保存到db
的方式@Entity
@Table(name="person")
public class person{
private Integer id;
private String name;
private String phone;
@Transient
JsonNode phoneJson;
private static OhjectMapper mapper = new ObjectMapper();
getter/setter
@Transient
public JsonNode getPhoneJson(){
return phoneJson;
}
public void setPhoneJson(JsonNode phoneJson){
this.phoneJson = phoneJson;
}
@JsonIgnore
@Column(name="phone")
public String getPhone() throws Exception{
return mapper.writeValueAsString(phoneJson);
}
public void setPhone(String phone) throws Exception{
this.phone = mapper.readTree(phone);
}
}
dao-保存
personRepository.save(person)
任何帮助将不胜感激。
更新
多个jSON列
[
{
"id" : 1, -- this primary key and not unique but cannot be null
"name" : { --this element can be empty/null
"first" : "John",
"last" : "Doe"
},
"phone" : [
{ "type" : "home", "ref" : 1111111234},
{ "type" : "work", "ref" : 2222222222}
]
},
{
"id" : 2, -- this primary key and not unique but cannot be null
"name" : {
"first" : "Jane",
"last" : "Doe"
},
"phone" : [
{ "type" : "home", "ref" : 1111111234},
{ "type" : "work", "ref" : 2222222222}
]
}
]
我如何获得如下结果
id name phone
1 [{John},{Doe}] { "type" : "home", "ref" : "111-111-1234"}
1 [{John},{Doe}] { "type" : "work", "ref" : "222-222-2222"}
2 [{Jane},{Doe}] { "type" : "home", "ref" : "111-111-1234"}
2 [{Jane},{Doe}] { "type" : "work", "ref" : "222-222-2222"}
答案 0 :(得分：2)
您需要Person复制n个对象n次，其中phone是JSON数组的大小。为了明确起见，我建议创建两个单独的模型，我们可以分别使用这些模型来解析DB并将其保存在JSON中。您可以在下面找到一个简单的示例，
List<JsonPerson>解析为List<JsonPerson> List<Person>转换为List<Person> DB（您可以将其保存到import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.node.ArrayNode;
import com.fasterxml.jackson.databind.type.CollectionType;
import java.io.File;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;
public class JsonApp {
public static void main(String[] args) throws Exception {
File jsonFile = new File("./resource/test.json").getAbsoluteFile();
ObjectMapper mapper = new ObjectMapper();
CollectionType personsType = mapper.getTypeFactory().constructCollectionType(List.class, JsonPerson.class);
// parse
List<JsonPerson> jsonPersons = mapper.readValue(jsonFile, personsType);
// convert
List<Person> persons = jsonPersons.stream()
.map(p -> p.mapTo(mapper))
.flatMap(List::stream)
.collect(Collectors.toList());
persons.forEach(System.out::println);
// save persons to DB
// ...
}
}
class JsonPerson {
private Integer id;
private String name;
private ArrayNode phone;
public List<Person> mapTo(ObjectMapper mapper) {
List<Person> persons = new ArrayList<>();
phone.elements().forEachRemaining(phone -> {
persons.add(map(mapper, phone));
});
return persons;
}
private Person map(ObjectMapper mapper, JsonNode p) {
Person person = new Person();
person.setId(id);
person.setName(name);
try {
person.setPhone(mapper.writeValueAsString(p));
} catch (JsonProcessingException e) {
throw new IllegalStateException(e);
}
return person;
}
// getters, setters, toString
}
class Person {
private Integer id;
private String name;
private String phone;
// getters, setters, toString
}
）示例：
Person{id=1, name='John Doe', phone='{"type":"home","ref":"111-111-1234"}'}
Person{id=1, name='John Doe', phone='{"type":"work","ref":"222-222-2222"}'}
Person{id=2, name='Jane Doe', phone='{"type":"home","ref":"111-111-1234"}'}
Person{id=2, name='Jane Doe', phone='{"type":"work","ref":"222-222-2222"}'}
上面的代码显示：
JSON以上代码将解析ObjectMapper与其他部分分开。另外，请勿在每个POJO中创建POJO。 ObjectMapper对Jackson和name一无所知。
由于JSON Object是POJO，因此您可以使用Name和first属性创建新的last-phone或将其与{{ 1}}并反序列化为JsonNode：
class JsonPerson {
private Integer id;
private JsonNode name;
private ArrayNode phone;
public List<Person> mapTo(ObjectMapper mapper) {
List<Person> persons = new ArrayList<>();
phone.elements().forEachRemaining(phone -> {
persons.add(map(mapper, phone));
});
return persons;
}
private Person map(ObjectMapper mapper, JsonNode p) {
Person person = new Person();
person.setId(id);
person.setName(getNameAsString());
try {
person.setPhone(mapper.writeValueAsString(p));
} catch (JsonProcessingException e) {
throw new IllegalStateException(e);
}
return person;
}
private String getNameAsString() {
if (name == null) {
return null;
}
StringBuilder builder = new StringBuilder();
if (name.isObject()) {
ObjectNode nameObject = (ObjectNode) name;
builder.append("[");
builder.append("{").append(nameObject.get("first")).append("}");
builder.append(",");
builder.append("{").append(nameObject.get("last")).append("}");
builder.append("]");
}
return builder.toString();
}
// getters, setters, toString
}
更改以上代码后，应打印：
Person{id=1, name='[{"John"},{"Doe"}]', phone='{"type":"home","ref":1111111234}'}
Person{id=1, name='[{"John"},{"Doe"}]', phone='{"type":"work","ref":2222222222}'}
Person{id=2, name='[{"Jane"},{"Doe"}]', phone='{"type":"home","ref":1111111234}'}
Person{id=2, name='[{"Jane"},{"Doe"}]', phone='{"type":"work","ref":2222222222}'}
getNameAsString方法已简化，您需要处理所有极端情况并为String，null和empty节点创建更好的semi-empty表示形式。