Java 无法读取 JSON：无法从 START_OBJECT 令牌中反序列化 hello.Country[] 的实例
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Could not read JSON: Can not deserialize instance of hello.Country[] out of START_OBJECT token
提问by Volodymyr Levytskyi
I have rest url that gives me all countries - http://api.geonames.org/countryInfoJSON?username=volodiaL.
我有给我所有国家/地区的休息 url - http://api.geonames.org/countryInfoJSON?username=volodiaL。
I use RestTemplate from spring 3 to parse returned json into java objects:
我使用 Spring 3 中的 RestTemplate 将返回的 json 解析为 java 对象：
RestTemplate restTemplate = new RestTemplate();
Country[] countries = restTemplate.getForObject("http://api.geonames.org/countryInfoJSON?username=volodiaL",Country[].class);
When I run this code I get an exception:
当我运行此代码时，出现异常：
Caused by: com.fasterxml.Hymanson.databind.JsonMappingException: Can not deserialize instance of hello.Country[] out of START_OBJECT token
at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@1846149; line: 1, column: 1]
at com.fasterxml.Hymanson.databind.JsonMappingException.from(JsonMappingException.java:164)
at com.fasterxml.Hymanson.databind.DeserializationContext.mappingException(DeserializationContext.java:691)
at com.fasterxml.Hymanson.databind.DeserializationContext.mappingException(DeserializationContext.java:685)
at com.fasterxml.Hymanson.databind.deser.std.ObjectArrayDeserializer.handleNonArray(ObjectArrayDeserializer.java:222)
at com.fasterxml.Hymanson.databind.deser.std.ObjectArrayDeserializer.deserialize(ObjectArrayDeserializer.java:133)
at com.fasterxml.Hymanson.databind.deser.std.ObjectArrayDeserializer.deserialize(ObjectArrayDeserializer.java:18)
at com.fasterxml.Hymanson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:2993)
at com.fasterxml.Hymanson.databind.ObjectMapper.readValue(ObjectMapper.java:2158)
at org.springframework.http.converter.json.MappingHymanson2HttpMessageConverter.readJavaType(MappingHymanson2HttpMessageConverter.java:225)
... 7 more
Finally my Country class:
最后我的乡村课：
import com.fasterxml.Hymanson.annotation.JsonIgnoreProperties;
@JsonIgnoreProperties(ignoreUnknown = true)
public class Country {
private String countryName;
private long geonameId;
public String getCountryName() {
return countryName;
}
public long getGeonameId() {
return geonameId;
}
@Override
public String toString() {
return countryName;
}
}
The problem is that returned json contains root element "geonames" which contains array of country elements like so:
问题是返回的 json 包含根元素“geonames”，其中包含像这样的国家元素数组：
{
"geonames": [
{
"continent": "EU",
"capital": "Andorra la Vella",
"languages": "ca",
"geonameId": 3041565,
"south": 42.42849259876837,
"isoAlpha3": "AND",
"north": 42.65604389629997,
"fipsCode": "AN",
"population": "84000",
"east": 1.7865427778319827,
"isoNumeric": "020",
"areaInSqKm": "468.0",
"countryCode": "AD",
"west": 1.4071867141112762,
"countryName": "Andorra",
"continentName": "Europe",
"currencyCode": "EUR"
}
]
}
How to tell RestTemplateto convert each element of array into Countryobject?
如何告诉RestTemplate将数组的每个元素转换为Country对象？
采纳答案by geoand
You need to do the following:
您需要执行以下操作：
public class CountryInfoResponse {
@JsonProperty("geonames")
private List<Country> countries;
//getter - setter
}
RestTemplate restTemplate = new RestTemplate();
List<Country> countries = restTemplate.getForObject("http://api.geonames.org/countryInfoJSON?username=volodiaL",CountryInfoResponse.class).getCountries();
It would be great if you could use some kind of annotation to allow you to skip levels, but it's not yet possible (see thisand this)
回答by vonox7
Another solution:
另一种解决方案：
public class CountryInfoResponse {
private List<Object> geonames;
}
Usage of a generic Object-List solved my problem, as there were other Datatypes like Boolean too.
通用对象列表的使用解决了我的问题，因为还有其他数据类型，如布尔值。
回答by tyler
For Spring-boot 1.3.3 the method exchange() for List is working as in the related answer
对于 Spring-boot 1.3.3，List 的方法 exchange() 与相关答案一样工作
回答by Filip Savic
In my case, I was getting value of <input type="text">with JQuery and I did it like this:
就我而言，我<input type="text">使用 JQuery获得了价值，我是这样做的：
var newUserInfo = { "lastName": inputLastName[0].value, "userName": inputUsername[0].value,
"firstName": inputFirstName[0] , "email": inputEmail[0].value}
And I was constantly getting this exception
我不断收到这个例外
com.fasterxml.Hymanson.databind.JsonMappingException: Can not deserialize instance of java.lang.String out of START_OBJECT token at [Source: java.io.PushbackInputStream@39cb6c98; line: 1, column: 54] (through reference chain: com.springboot.domain.User["firstName"]).
com.fasterxml.Hymanson.databind.JsonMappingException：无法从 [来源：java.io.PushbackInputStream@39cb6c98; 的 START_OBJECT 令牌中反序列化 java.lang.String 的实例；行：1，列：54]（通过参考链：com.springboot.domain.User["firstName"]）。
And I banged my head for like an hour until I realised that I forgot to write .valueafter this"firstName": inputFirstName[0].
我敲了一个小时的头，直到我意识到我.value在这之后忘记写了"firstName": inputFirstName[0]。
So, the correct solution was:
所以，正确的解决方案是：
var newUserInfo = { "lastName": inputLastName[0].value, "userName": inputUsername[0].value,
"firstName": inputFirstName[0].value , "email": inputEmail[0].value}
I came here because I had this problem and I hope I save someone else hours of misery.
我来这里是因为我遇到了这个问题，我希望我能拯救别人几个小时的痛苦。
Cheers :)
干杯:)
回答by clD
If you want to avoid using an extra Classand List<Object> genomesyou could simply use a Map.
如果你想避免使用额外的Class，List<Object> genomes你可以简单地使用Map.
The data structure translates into Map<String, List<Country>>
数据结构转化为 Map<String, List<Country>>
String resourceEndpoint = "http://api.geonames.org/countryInfoJSON?username=volodiaL";
Map<String, List<Country>> geonames = restTemplate.getForObject(resourceEndpoint, Map.class);
List<Country> countries = geonames.get("geonames");
