如何将JSON对象映射到Spring对象...我有AJAX,将JSON对象发布到我的Spring Controller但是如何让Spring将JSON转换为java中的Object
Java代码:
@RequestMapping(method=RequestMethod.POST, value="/employee")
public ModelAndView addEmployee(@RequestBody String body) {
System.out.println("in post: " + body);
Source source = new StreamSource(new StringReader(body));
System.out.println("source: " + source.toString());
//
// how do I turn the JSON String into a Java Object?
//
return new ModelAndView(XML_VIEW_NAME, "object", body);
}
JavaSript / html代码:
<%@ page language="java" contentType="text/html; charset=UTF-8"
pageEncoding="UTF-8"%>
<!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=UTF-8">
<title>This is a project to show how to use RESTful</title>
</head>
<body>
<script type="text/javascript">var contexPath = "<%=request.getContextPath()%>";</script>
<script src="<%=request.getContextPath()%>/js/jquery.js"></script>
<script type="text/javascript">
function doAjaxPost() {
var queryString = $('#htmlform').serialize();
alert("doAjaxPost Called :" + queryString +":");
$.ajax({
contentType : "application/json",
dataType : 'json',
type : "POST",
url : contexPath + "/service/employee",
data : queryString, //json serialization (like array.serializeArray() etc)
success : function(data) {
alert("Thanks for submitting. \n\n" + response.result);
// response
},
error : function(request, status, error) {
alert('Error: ' + e);
}
});
}
</script>
<H1>Add Employee</H1>
<p>
<form name="htmlform" id="htmlform">
<table border=1>
<thead><tr>
<th>ID</th>
<th>Name</th>
<th>Email</th>
</tr></thead>
<tr>
<td><input type="text" name="ID" maxlength="5" size="3"></td>
<td><input type="text" name="Name" maxlength="10" size="10"></td>
<td><input type="text" name="Email" maxlength="10" size="10"></td>
</tr>
</table>
<input type="button" value="Save Employee" onclick="doAjaxPost();" />
<p>
<p>
</form>
[<a href="http://localhost:8080/RESTful/service/employees">List all Employees</a> | <a href="add.jsp">Employee Form Test</a>]
</body>
</html>
答案 0 :(得分:1)
请务必:
然后,您可以在控制器的方法签名处添加@RequestBody Employee employee
。
答案 1 :(得分:0)
我想你使用的是Spring 3.0+。看一下JSON simplifications上的Spring Source博客文章。看来你已经走了一半了。
答案 2 :(得分:0)
您的代码大致如下:
Employee e;
try {
e = objectMapper.readValue(body, Employee .class);
} catch (IOException e) {
throw new IllegalArgumentException("Couldn't parse json into a employee", e);
}
如果您使用的是Spring 3或更高版本,则可以采用更简单的方法: http://blog.springsource.org/2010/01/25/ajax-simplifications-in-spring-3-0/