Java Spring MVC Rest中处理JSon时如何处理POJO嵌套对象
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How to handle POJO nested objects when dealing with JSon in Spring MVC Rest
提问by Alexio Cassani
I'm trying to figure out how to better deal with JSon serialization/deserialization of nested Java objects in Spring MVC.
我试图弄清楚如何更好地处理 Spring MVC 中嵌套 Java 对象的 JSon 序列化/反序列化。
My domain model is the following:
我的域模型如下：
public class Cart {
private String id;
private Customer customerID;
private Checkout checkoutID;
private List<CartProduct> itemCatalogList;
*** ... getters & setters ... ***
}
public class ProductCart {
private String sku;
private String color;
private String sizeBase
private int qty;
*** ... getters & setters ... ***
}
public class Checkout {
private String id;
private String billingAddress;
private String shippingAddress;
private Cart cartID;
*** ... getters & setters ... ***
}
The JSon I was thinking is something like this:
我在想的 JSon 是这样的：
checkout:
查看：
{
"cart": {
"$oid": "51f631cb84812abb04000006"
},
"shippingAddress" : "5h avenue - new york",
"billingAddress" : "5h avenue - new york"
}
cart:
大车：
{
"customer": {
"$oid": "5174da574940368a9126e8dc"
},
"items_catalog": [
{
"sku": "00075161",
"color": "ff99cc",
"size_base": "IT_25",
"qty": 3,
},
{
"sku": "00075161",
"color": "ff99cc",
"size_base": "IT_27",
"qty": 2,
},
{
"sku": "00075161",
"color": "ff99cc",
"size_base": "IT_29",
"qty": 1,
}
}
Assuming this is a viable domain model & json document, how in Spring I could create a checkout starting from a JSon?
假设这是一个可行的域模型和 json 文档，我如何在 Spring 中创建从 JSon 开始的结帐？
My problem is that I don't know how to "explode" the $oid in the checkout & cart json in order to create checkout & cart Java Beans:
我的问题是我不知道如何“分解”结帐和购物车 json 中的 $oid 以创建结帐和购物车 Java Bean：
is there a way to do it automatically with Hymanson?
or should I create a sort of Interceptor to handle a, for example, checkout json in order to retrieve the cart and then perform the mapping to the POJO?
有没有办法用Hyman逊自动做到这一点？
或者我应该创建一种拦截器来处理例如结帐 json 以检索购物车然后执行到 POJO 的映射？
(- or there is a 3rd way?)
（ - 或者有第三种方式？）
Thanks a lot for any advice.
非常感谢您的任何建议。
采纳答案by Ernestas Kardzys
If I understood you correctly, you could do something like this (I'm using Spring 3.2.3.RELEASE & Hymanson 1.9.12).
如果我理解正确，你可以做这样的事情（我使用的是 Spring 3.2.3.RELEASE & Hymanson 1.9.12）。
In your applicationContext.xml you have:
在你的 applicationContext.xml 你有：
<bean id="HymansonMessageConverter"
class="org.springframework.http.converter.json.MappingHymansonHttpMessageConverter"/>
<bean class="org.springframework.web.servlet.mvc.annotation.AnnotationMethodHandlerAdapter">
<property name="messageConverters">
<list>
<ref bean="HymansonMessageConverter"/>
</list>
</property>
</bean>
You have Spring controller which looks like this:
您有如下所示的 Spring 控制器：
package test;
import org.apache.log4j.Logger;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestBody;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import org.springframework.web.bind.annotation.ResponseBody;
@Controller
@RequestMapping("/json")
public class JsonParsingController {
private final static Logger log = Logger.getLogger(JsonParsingController.class);
@RequestMapping(value = "/cart.do", method = RequestMethod.POST, produces = "application/json; charset=utf-8")
@ResponseBody public CartResponse handleCart(@RequestBody Cart cart) {
if (cart != null) {
log.debug(cart);
}
return new CartResponse("OK!");
}
}
and three POJOs:
和三个 POJO：
package test;
public class Cart {
private String id;
private Checkout checkoutID;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public Checkout getCheckoutID() {
return checkoutID;
}
public void setCheckoutID(Checkout checkoutID) {
this.checkoutID = checkoutID;
}
@Override
public String toString() {
return "Cart{" +
"id='" + id + '\'' +
", checkoutID=" + checkoutID +
'}';
}
}
package test;
public class Checkout {
private String id;
private String billingAddress;
private String shippingAddress;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public String getBillingAddress() {
return billingAddress;
}
public void setBillingAddress(String billingAddress) {
this.billingAddress = billingAddress;
}
public String getShippingAddress() {
return shippingAddress;
}
public void setShippingAddress(String shippingAddress) {
this.shippingAddress = shippingAddress;
}
@Override
public String toString() {
return "Checkout{" +
"id='" + id + '\'' +
", billingAddress='" + billingAddress + '\'' +
", shippingAddress='" + shippingAddress + '\'' +
'}';
}
}
package test;
public class CartResponse {
private String result;
public CartResponse(String result) {
this.result = result;
}
public String getResult() {
return result;
}
public void setResult(String result) {
this.result = result;
}
}
Then in your HTML page you can do something like this:
然后在您的 HTML 页面中，您可以执行以下操作：
<script language="JavaScript" type="text/javascript">
$(document).ready(function () {
// Your data
var arr = {
id: '51f631cb84812abb04000006',
checkoutID: {
id: '123456789',
"shippingAddress" : "5h avenue - new york",
"billingAddress" : "5h avenue - new york"
}
};
$.ajax({
url: '/json/cart.do',
type: 'POST',
data: JSON.stringify(arr),
contentType: 'application/json; charset=utf-8',
dataType: 'json',
async: false,
success: function (msg) {
alert(msg.result);
}
});
});
</script>
At least as for me - it works :)
至少对我来说 - 它有效:)
回答by a_r_a_s
It's enough to return data structure (model) in controller file. JSON will be generate automatically based on structure in your model. Take a look here for example: http://www.mkyong.com/spring-mvc/spring-3-mvc-and-json-example/(section 3 - controller, and then, 5 - results)
在控制器文件中返回数据结构（模型）就足够了。JSON 将根据模型中的结构自动生成。看看这里的例子：http: //www.mkyong.com/spring-mvc/spring-3-mvc-and-json-example/（第3节 - 控制器，然后，5 - 结果）
