英文:
no String-argument constructor/factory method to deserialize from String value - Exception while deserializing json object from restTemplate
问题
在调用以获取 JSON 响应并解析它时遇到了问题。
Person.java 模型:
@Data
@NoArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
public class Person {
private String name;
private String age;
private Address address;
}
Address.java 模型:
@Data
@NoArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
public class Address {
private String state;
private String country;
}
读取数据的代码:
ResponseEntity<List<Person>> response = restTemplate.exchange(builder.toUriString(), HttpMethod.GET, requestEntity, new ParameterizedTypeReference<List<Person>>() {});
然而,我得到以下异常:
在调用 ABS 服务时出现 RestClientException,在提取类型
[java.util.List<com.bp.model.Person>]
和内容类型 [application/json;charset=UTF-8] 的响应时出现 ServiceError,嵌套异常是 org.springframework.http.converter.HttpMessageNotReadableException: JSON 解析错误: 无法构造com.bp.model.Address
的实例(尽管至少存在一个构造函数): 没有从 String 值反序列化的 String-argument 构造函数/工厂方法('{"state":"LA","country":"US"}'); 嵌套异常是 com.fasterxml.jackson.databind.exc.MismatchedInputException: 无法构造com.bp.model.Address
的实例(尽管至少存在一个构造函数): 没有从 String 值反序列化的 String-argument 构造函数/工厂方法('{"state":"IN","brand":"anthem"}')位于 [Source: (PushbackInputStream); line: 1, column: 325](通过引用链: java.util.ArrayList[0]->com.bp.model.Person["address"])。
注意:由于您要求只翻译代码部分,因此我只翻译了代码和异常信息。
英文:
Facing issue while making call to retrieve a json response and parse it.
[
{
"name": "john doe",
"age": "24",
"address": "{\"state\":\"LA\",\"country\":\"US\"}"
}
]
Models:
Person.java
@Data
@NoArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
public class Person {
private String name;
private String age;
private Address address;
}
Address .java
@Data
@NoArgsConstructor
@JsonIgnoreProperties(ignoreUnknown = true)
public class Address {
private String state;
private String country;
}
Code to read this data,
ResponseEntity<List<Person>> response = restTemplate.exchange(builder.toUriString(), HttpMethod.GET,requestEntity,new ParameterizedTypeReference<List<Person>>() {});
However i get below exception,
> RestClientException while invoking ABS ServiceError while extracting response for type [java.util.List<com.bp.model.Person>]
and content type [application/json;charset=UTF-8]; nested exception is org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Cannot construct instance of com.bp.model.Address
(although at least one Creator exists): no String-argument constructor/factory method to deserialize from String value ('{"state":"LA","country":"US"}'); nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of com.bp.model.Address
(although at least one Creator exists): no String-argument constructor/factory method to deserialize from String value ('{"state":"IN","brand":"anthem"}')
at [Source: (PushbackInputStream); line: 1, column: 325] (through reference chain: java.util.ArrayList[0]->com.bp.model.Person["address"])
答案1
得分: 15
代码是正确的,但 JSON 存在问题。地址是一个字符串,而不是 JSON 对象。为了使其正常工作,需要像下面这样:
"address": {"state": "LA", "country": "US"}
去掉外层的引号和转义字符。
英文:
The code is correct but there's a problem with the JSON. The address is a string and not a JSON object. For it to work, it would need to be something like:
"address": {"state": "LA", "country": "US"}
Without the outer quotes and the escape characters.
答案2
得分: 1
**JSON字符串转换为Map:**
String json = "{\"name\":\"mkyong\", \"age\":\"37\"}";
输出 {name=mkyong, age=37}
public static void main(String[] args) {
ObjectMapper mapper = new ObjectMapper();
String json = "{\"name\":\"mkyong\", \"age\":\"37\"}";
try {
// 将JSON字符串转换为Map
Map<String, String> map = mapper.readValue(json, Map.class);
// 它能正常工作
// Map<String, String> map = mapper.readValue(json, new TypeReference<Map<String, String>>() {});
System.out.println(map);
} catch (IOException e) {
e.printStackTrace();
}
}
**Map转换为JSON字符串:**
{"name":"mkyong","age":"37"}
{
"name" : "mkyong",
"age" : "37"
}
public static void main(String[] args) {
ObjectMapper mapper = new ObjectMapper();
Map<String, String> map = new HashMap<>();
map.put("name", "mkyong");
map.put("age", "37");
try {
// 将Map转换为JSON字符串
String json = mapper.writeValueAsString(map);
System.out.println(json); // 紧凑打印
json = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(map);
System.out.println(json); // 格式化打印
} catch (JsonProcessingException e) {
e.printStackTrace();
}
}
在Jackson中,我们可以使用mapper.readValue(json, Map.class)将JSON字符串转换为Map。依赖项
> <dependency>
> <groupId>com.fasterxml.jackson.core</groupId>
> <artifactId>jackson-databind</artifactId>
> <version>2.9.8</version>
> </dependency>
英文:
JSON string to Map:
String json = "{"name":"mkyong", "age":"37"}";
Output {name=mkyong, age=37}
public static void main(String[] args) {
ObjectMapper mapper = new ObjectMapper();
String json = "{\"name\":\"mkyong\", \"age\":\"37\"}";
try {
// convert JSON string to Map
Map<String, String> map = mapper.readValue(json, Map.class);
// it works
//Map<String, String> map = mapper.readValue(json, new TypeReference<Map<String, String>>() {});
System.out.println(map);
} catch (IOException e) {
e.printStackTrace();
}
}
Map to JSON string:
{"name":"mkyong","age":"37"}
{
"name" : "mkyong",
"age" : "37"
}
public static void main(String[] args) {
ObjectMapper mapper = new ObjectMapper();
Map<String, String> map = new HashMap<>();
map.put("name", "mkyong");
map.put("age", "37");
try {
// convert map to JSON string
String json = mapper.writeValueAsString(map);
System.out.println(json); // compact-print
json = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(map);
System.out.println(json); // pretty-print
} catch (JsonProcessingException e) {
e.printStackTrace();
}
}
In Jackson, we can use mapper.readValue(json, Map.class) to convert a JSON string to a Map. Dependency
> <dependency>
> <groupId>com.fasterxml.jackson.core</groupId>
> <artifactId>jackson-databind</artifactId>
> <version>2.9.8</version>
> </dependency>
答案3
得分: 0
这里我们可以尝试在运行时设置数值。
@JsonProperty("address")
public void setCustomAddress(String addressFromJson){
this.address = new Gson().fromJson(addressFromJson, Address.class);
}
英文:
Here we can try to set the values during runtime.
@JsonProperty("address")
public void setCustomAddress(String addressFromJson){
this.address = new Gson().fromJson(addressFromJson, Address.class);
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库,让每个人都能够通过互相帮助和分享经验来进步。
评论