java 如何将 JSON 对象映射到 Spring 对象
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How to map JSON object to Spring Object
提问by SJS
How to map JSON object to Spring Object... I have AJAX, posting a JSON object to my Spring Controller but how do I make Spring turn that JSON into a Object in java
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 控制器，但如何让 Spring 将该 JSON 转换为 Java 中的对象
Java Code:
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 code:
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>
回答by Ravi Khakhkhar
Make sure to:
确保：
- send JSON object, not JSON String
- included Hymanson on your classpath
- 发送 JSON 对象，而不是 JSON 字符串
- 将 Hymanson 包含在您的类路径中
Then you can add @RequestBody Employee employeeat the controller's method signature.
然后您可以添加 @RequestBody Employee employee控制器的方法签名。
回答by nobeh
I suppose you're using Spring 3.0+. Take a look at Spring Source blog post on JSON simplifications. It seems that you're half way there already.
我想你正在使用 Spring 3.0+。查看有关JSON 简化的Spring Source 博客文章。看来你已经成功了一半。
回答by Eugene Ryzhikov
Use JSON mapping librarysuch as Hymansonor Gson
Your code would look approximately as :
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);
}
If you using Spring 3 or higher, there is even simpler way to do it: http://blog.springsource.org/2010/01/25/ajax-simplifications-in-spring-3-0/
如果您使用 Spring 3 或更高版本，还有更简单的方法：http: //blog.springsource.org/2010/01/25/ajax-simplifications-in-spring-3-0/
