跪求懂json的大神！

[xulianzhen]

如何将复杂的json字符串转java对象？
json字符串：
{"createTime":"2016-07-05 08:08:08","cityName":"杭州","orderItems":"[{\"skuId\":\"1008604\",\"count\":2,\"unitPrice\":8.88},{\"skuId\":\"1008605\",\"count\":1,\"unitPrice\":6.66}]"}]

javaBean：
public class Order{
private String cityName;
private String createTime;
private List<OrderItem> orderItemList;

// 省略getter和setter方法。。。
}

工具：fastjson
使用Order Order = (Order) JSON.parseObject(data, Order.class);该方法转换发现orderItemList为空，如果处理才能正确解析让orderItemList不为空呢？

[missMeyo]

package com.imooc.tag;

import com.alibaba.fastjson.JSON;

public class TagMain {
	public static void main(String[] args) {
		Order order = JSON.parseObject("{\"createTime\":\"2016-07-05 08:08:08\",\"cityName\":\"杭州\",\"orderItems\":[{\"skuId\":\"1008604\",\"count\":2,\"unitPrice\":8.88},{\"skuId\":\"1008605\",\"count\":1,\"unitPrice\":6.66}]}",
				   Order.class);
	    
		for (OrderItem orderItem : order.getOrderItems()) {
			System.out.println(orderItem);
		}
	}
}

package com.imooc.tag;

import java.util.List;

public class Order {
	private String cityName;
	private String createTime;
	private List<OrderItem> orderItems;
	
	public String getCityName() {
		return cityName;
	}
	public void setCityName(String cityName) {
		this.cityName = cityName;
	}
	public String getCreateTime() {
		return createTime;
	}
	public void setCreateTime(String createTime) {
		this.createTime = createTime;
	}
	public List<OrderItem> getOrderItems() {
		return orderItems;
	}
	public void setOrderItems(List<OrderItem> orderItems) {
		this.orderItems = orderItems;
	}
}

package com.imooc.tag;

public class OrderItem {
	private String skuId;
	private Integer count;
	private Integer unitPrice;
	
	public String getSkuId() {
		return skuId;
	}
	public void setSkuId(String skuId) {
		this.skuId = skuId;
	}
	public Integer getCount() {
		return count;
	}
	public void setCount(Integer count) {
		this.count = count;
	}
	public Integer getUnitPrice() {
		return unitPrice;
	}
	public void setUnitPrice(Integer unitPrice) {
		this.unitPrice = unitPrice;
	}
	
	@Override
	public String toString() {
		return "OrderItem [skuId=" + skuId + ", count=" + count + ", unitPrice=" + unitPrice + "]";
	}
}

[bree06]

楼上是正解, 参数名字不一样. 对象里是orderItemList 但是json串里是orderItems,

[于大圣]

在对JSON格式的字符串<->Java对象互相转化时，一定要注意Java类属性的名字和JSON字符串里面的属性的名字一定要一致，切记！！然后底层解析框架运用反射将解析出来的属性值构造完一个对象，你这边解析为空，说明没有做到一致，确认下！

[qq_35517879]

sad水电费水电费

[qq_35517879]

我觉得这种情况用jackson比较好一点，因为jackson是逐层解析的。兼容性比较高，如果用fastjson的话碰到你这样的情况还是要再次转换的。