如何,如果我不需要它,我可以忽略根名称?我只需要法案。
如果我从JSON做工精细删除“版本” ..
我在控制台日志错误
2019-07-27 19:20:14.874 WARN 12516 --- [p-nio-80-exec-2] .w.s.m.s.DefaultHandlerExceptionResolver : Resolved [org.springframework.http.converter.HttpMessageNotReadableException: JSON parse error: Unexpected token (FIELD_NAME), expected END_OBJECT: Current token not END_OBJECT (to match wrapper object with root name 'bill'), but FIELD_NAME; nested exception is com.fasterxml.jackson.databind.exc.MismatchedInputException: Unexpected token (FIELD_NAME), expected END_OBJECT: Current token not END_OBJECT (to match wrapper object with root name 'bill'), but FIELD_NAME
at [Source: (PushbackInputStream); line: 8, column: 2]]
我的JSON这个样子的
{
"bill":
{
"siteId":"gkfhuj-00",
"billId":"d6334954-d1c2-4b51-bb10-11953d9511ea"
},
"version":"1"
}
我对JSON类我尝试使用JsonIgnoreProperties但不要帮忙还我写的“版本”
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonRootName(value = "bill")
public class Bill {
private String siteId;
private String billId;
//getters and setters
我的帖子方法丽森对象比尔
@PostMapping("/bill")
@ResponseBody
public ResponseEntity<String> getBill(@RequestBody Bill bill)
当你正在依靠Spring boot
通过注释和Jackson
定制解串器将在这里工作完美无瑕。你必须创建解串器类,如下所示
public class BillDeserializer extends StdDeserializer<Bill> {
public BillDeserializer() {
this(null);
}
public BillDeserializer(Class<?> vc) {
super(vc);
}
@Override
public Bill deserialize(JsonParser jp, DeserializationContext ctxt)
throws IOException, JsonProcessingException {
JsonNode billNode = jp.getCodec().readTree(jp);
Bill bill = new Bill();
bill.setSiteId(billNode.get("bill").get("siteId").textValue());
bill.setBillId(billNode.get("bill").get("billId").textValue());
return bill;
}
}
现在,你需要指导Jackson
使用该解串器,而不是默认的为Bill
类。这是由注册desearilizer完成。它可以通过一个简单的注解在做Bill
类似的类@JsonDeserialize(using = BillDeserializer.class)
你的Bill
类通常看起来如下规定
@JsonDeserialize(using = BillDeserializer.class)
public class Bill {
private String siteId;
private String billId;
//getters and setters
}
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我来说两句