use*_*048 5 java json gson deserialization json-deserialization
我从API长的json接收,例如:
{
"field1": "val1",
"field2": "val2",
...
"SOME_FIELD": " ABC ",
...
"fieldx": "valx"
}
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我想和Gson反序列化。一切正常,但字段的“ SOME_FIELD”值始终带有烦人的空格。我想trim()此字段值(无法更改API)。我知道我可以使用JsonDeserializer,但随后我必须手动读取所有字段。反序列化时是否只能编辑一个内部字段,而其余部分使用自动反序列化?
这里我正在编写一个演示类,通过忽略嵌入在JSON.
在这里,我使用ObjectMapperClass 将其JSONObject反序列化为Object.
//configuration to enables us to ignore non-used Unknow Properties.
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
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测试类(将 Json 转换为类对象的代码)。
import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.ObjectMapper;
public class Test {
/**
* @param args
* @throws IOException
* @throws JsonProcessingException
* @throws JsonMappingException
* @throws JsonParseException
*/
public static void main(String[] args) throws JsonParseException, JsonMappingException, JsonProcessingException, IOException {
Test o = new Test();
o.GetJsonAsObject(o.putJson());
}
//function to generate a json for demo Program.
private String putJson() throws JsonProcessingException{
HashMap<String, String> v_Obj = new HashMap<>();
v_Obj.put("field1", "Vikrant");
v_Obj.put("field2", "Kashyap");
return new ObjectMapper().writeValueAsString(v_Obj); // change the HashMap as JSONString
}
//function to Convert a json Object in Class Object for demo Program.
private void GetJsonAsObject(String value) throws JsonParseException, JsonMappingException, IOException{
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
Test1 obj = mapper.readValue(value, Test1.class);
System.out.println(obj);
}
}
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Test1.java(转换POJO类)
class Test1{
private String field1;
public String getField1() {
return field1;
}
public void setField1(String field1) {
this.field1 = field1;
}
public String toString(){
return this.field1.toString();
}
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}
正确阅读评论……希望您明白了。
谢谢
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