我的jsp和jquery代码为
var article = new Object();
article.title = "abc";
article.url = "abc";
article.categories = [1,2,3];
article.tags = [1,2,3];
console.log('hi');
$.ajax({
type: 'POST',
url: URL,
contentType:"application/json",
data: JSON.stringify(article),
dataType: 'json',
success: function(result) {
console.log(result);
},
error: function(e){
alert('Error in Processing');
}
});
和我的java代码为
BufferedReader br = new BufferedReader(new InputStreamReader(request.getInputStream()));
String json = "";
if(br != null){
json = br.readLine();
}
// 2. initiate jackson mapper
ObjectMapper mapper = new ObjectMapper();
//mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
// 3. Convert received JSON to Article
Article article = mapper.readValue(json, Article.class);
现在我的Arcticle课是
public class Article {
private String title;
private String url;
private List<String> categories;
private List<String> tags;
//getters and setters
}
现在我正在将异常作为
在线
String json = "";
if(br != null){
json = br.readLine();
}
我得到的JSON如下
{"title":"abc","url":"abc","categories":"[1, 2, 3]","tags":"[1, 2, 3]"}
实际上应该是
{"title":"abc","url":"abc","categories":[1, 2, 3],"tags":[1, 2, 3]}
我不明白发生了什么
因此我得到了例外
com.fasterxml.jackson.databind.JsonMappingException:无法从VALUE_STRING令牌中反序列化java.util.ArrayList实例
在[来源:java.io.StringReader@f94ca;行:1,列:27](通过参考链:com.ihexa.common.admin.cabsharing.action.Article [“ categories”])
我按照以下方式解决了答案
文章用户= mapper.readValue(point,Article.class);
System.out.println(user.getRouteFirst());
Gson gson =新的Gson();
TypeToken>令牌=新的TypeToken>(){};
列出personList = gson.fromJson(user.getRouteFirst(),token.getType());
文章类别为
公共类文章{
private String routeFirst;
private String routeSecond;
//设置者和获取者
}
jsp的jQuery代码为
var article = new Object();
article.routeFirst = newRoute1;
article.routeSecond = newRoute2;
$.ajax({
type: 'POST',
url: '../..//admin/cabsharing/findIntersesctionPoint.do',
data : "point="+JSON.stringify(article),
dataType: 'json',
success: function(result) {
console.log("success");
},
error: function(e){
console.log("error");
}
});
最佳答案
我的jsp
将是,
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>Calculator</title>
</head>
<script
src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js">
</script>
<script type="text/javascript">
$( document ).ready(function() {
//alert("DOM is ready");
});
function sendJsonData() {
var article = new Object();
article.title = "abc";
article.url = "abc";
article.categories = [1,2,3];
article.tags = [1,2,3];
//alert("JSON string :"+ JSON.stringify(article));
$.ajax({
type: 'POST',
url: "JsonServlet",
//contentType:"application/json",
//data: {point:point},
data : "point="+encodeURIComponent(JSON.stringify(article)),
dataType: 'json',
success: function(result) {
},
error: function(e){
//alert('Error in Processing');
}
});
}
</script>
<body>
<button id="jsonButton" onclick="sendJsonData()">send jdon Data</button>
</body>
</html>
我的
servlet
将是,public class JsonServlet extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String point = request.getParameter("point");
System.out.println("Point : " + point );
if(point != null){
ObjectMapper mapper = new ObjectMapper();
try {
// read from string, convert it to Article class object
Article user = mapper.readValue(point, Article.class);
// Conver the Article class object in to the JSON string
System.out.println("Output Json String is :::::::::::> "+mapper.writeValueAsString(user));
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
这是我进入控制台的
output
Point : {"title":"abc","url":"abc","categories":[1,2,3],"tags":[1,2,3]}
Json String is :::::::::::> {"title":"abc","url":"abc","categories":["1","2","3"],"tags":["1","2","3"]}
希望这可以帮助。