我正在尝试通过jquery POST将一些json数据发送到本地计算机上的泽西REST服务。
在我的服务器端,我使用Jersey方法来使用POST的这个JSON。
@Path("/question")
public class QuestionAPI {
private final static Logger LOGGER = Logger.getLogger(HelloWorldApi.class .getName());
@POST
@Path("/askquestion")
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_JSON)
public TQARequest askquestion(TQARequest tqaRequest, @Context HttpServletRequest request) {
LOGGER.info("Inside-->askquestion-->TQARequest"+tqaRequest.getQuestion());
return tqaRequest;
}
}
我正在请求中包装json数据。因此,在服务器中,我可以在该包装类中获取请求中发送的所有数据。我的请求包装类是
public class TQARequest {
private Question question;
public Question getQuestion() {
return question;
}
public void setQuestion(Question question) {
this.question = question;
}
@Override
public String toString() {
return "TQARequest [question=" + question + "]";
}
}
问题pojo类
public class Question {
@Id
private Long questionID;
private String questionText;
private long createdOn;
private String questionURL;
private String questionTrackingURL;
@Override
public String toString() {
return "Question [questionID=" + questionID + ", questionText="
+ questionText + ", createdOn=" + createdOn + ", questionURL="
+ questionURL + ", questionTrackingURL=" + questionTrackingURL
+ "]";
}
public Question(String questionText, long createdOn, String questionURL,
String questionTrackingURL) {
super();
this.questionText = questionText;
this.createdOn = createdOn;
this.questionURL = questionURL;
this.questionTrackingURL = questionTrackingURL;
}
public Long getQuestionID() {
return questionID;
}
public void setQuestionID(Long questionID) {
this.questionID = questionID;
}
public String getQuestionText() {
return questionText;
}
public void setQuestionText(String questionText) {
this.questionText = questionText;
}
public long getCreatedOn() {
return createdOn;
}
public void setCreatedOn(long createdOn) {
this.createdOn = createdOn;
}
public String getQuestionURL() {
return questionURL;
}
public void setQuestionURL(String questionURL) {
this.questionURL = questionURL;
}
public String getQuestionTrackingURL() {
return questionTrackingURL;
}
public void setQuestionTrackingURL(String questionTrackingURL) {
this.questionTrackingURL = questionTrackingURL;
}
public Question(){
}
}
每当我从jquery发出请求时,如下所示,
function askQuestion(){
$.ajax({
type: "POST",
contentType: "application/json; charset=utf-8",
url: "/api/question/askquestion",
data:
JSON.stringify({
"tqaRequest" : {
"question" : {
"createdOn" : "sfddsf",
"questionText" : "fsdfsd",
"questionTrackingURL" : "http://www.google.com",
"questionURL" : "ssdf"
}
}
}
),
dataType: "json",
success: function(response){
console.log(response);
}
});
}
我在控制台中收到此错误:
WARNING: /api/question/askquestion: org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "tqaRequest" (Class com.netsquid.tqa.entity.TQARequest), not marked as ignorable
at [Source: org.mortbay.jetty.HttpParser$Input@899e3e; line: 1, column: 16] (through reference chain: com.netsquid.tqa.entity.TQARequest["tqaRequest"])
我可以通过从jquery发送Question json并接受方法中的question参数来解决这个问题。 但我需要在TQARequest中包装所有jquery请求并接受所有请求作为TQARequest,然后从中提取问题对象。我该怎么做?
我在web.xml中的POJO映射是:
<init-param>
<param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
<param-value>true</param-value>
</init-param>
答案 0 :(得分:7)
我相信您可以按如下方式简化JSON文档:
{
"question" : {
"createdOn" : "sfddsf",
"questionText" : "fsdfsd",
"questionTrackingURL" : "http://www.google.com",
"questionURL" : "ssdf"
}
}
这种形式仍然是“tqaRequest”对象。
如果你想支持一系列问题,你的JSON可能看起来像这样(JSON数组放在方括号内):
{
"questions" : [
{
"createdOn" : "date 1",
"questionText" : "question 1",
"questionTrackingURL" : "http://www.google.com",
"questionURL" : "question 1 url"
},
{
"createdOn" : "date 2",
"questionText" : "question 2",
"questionTrackingURL" : "http://www.google.com",
"questionURL" : "question 2 url"
}]
}
}
你会调整你的TQARequest类来引用:
private List<Question> questions;
而不是
private Question question;
答案 1 :(得分:4)
希望这能解决问题。
@POST
@Path("/askquestion")
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_JSON)
public TQARequest askquestion(String jsonRequest){
TQARequest tqaRequest = MapperUtil
.readAsObjectOf(TQARequest.class, jsonRequest);
}
<强> MapperUtil.java 强>
com.fasterxml.jackson.databind.ObjectMapper MAPPER = new ObjectMapper();
public static <T> T readAsObjectOf(Class<T> clazz, String value)
throws MYPException {
try {
return MAPPER.readValue(value, clazz);
} catch (Exception e) {
LOGGER.error("{}, {}", e.getMessage(), e.fillInStackTrace());
}
}