这是我的URL中的JSON https://api.myjson.com/bins/142jr
[
{
"serviceNo":"SR0000000001",
"serDate":"17",
"serMonth":"DEC",
"serYear":"2015",
"serTime":"02.30 AM",
"serApartmentName":"Galaxy Apartments"
},
{
"serviceNo":"SR0000000002",
"serDate":"19",
"serMonth":"JUN",
"serYear":"2016",
"serTime":"03.30 AM",
"serApartmentName":"The Great Apartments"
}
]
我有一个ListView我希望从在线JSON填充详细信息,上面我给出了一个链接和示例json任何人在java中给出样本jackson代码
谢谢你提前, Rajesh Rajendiran
答案 0 :(得分:2)
要使用jackson,您需要创建一个模型类:
[
{
"serviceNo":"SR0000000001",
"serDate":"17",
"serMonth":"DEC",
"serYear":"2015",
"serTime":"02.30 AM",
"serApartmentName":"Galaxy Apartments"
},
{
"serviceNo":"SR0000000002",
"serDate":"19",
"serMonth":"JUN",
"serYear":"2016",
"serTime":"03.30 AM",
"serApartmentName":"The Great Apartments"
}
]
对于上面的json,模型类将是:
public class SomeClass {
private String serviceNo;
private String serDate;
private String serMonth;
private String serYear;
private String serTime;
private String serApartmentName;
@JsonProperty("serviceNo") //to bind it to serviceNo attribute of the json string
public String getServiceNo() {
return serviceNo;
}
public void setServiceNo(String sNo) { //@JsonProperty need not be specified again
serviceNo = sNo;
}
//create getter setters like above for all the properties.
//if you want to avoid a key-value from getting parsed use @JsonIgnore annotation
}
现在每当你将上面的json作为存储在变量中的字符串说jsonString时,使用以下代码来解析它:
ObjectMapper mapper = new ObjectMapper(); // create once, reuse
ArrayList<SomeClass> results = mapper.readValue(jsonString,
new TypeReference<ArrayList<ResultValue>>() { } );
结果现在应该包含两个SomeClass对象,将上面的json解析为相应的对象。
PS:自从我使用Jackson进行解析以来已经很长时间了,所以这段代码可能需要一些改进。
答案 1 :(得分:0)
如果您将此作为http响应,那么我建议使用spring rest template for android。 它支持消息转换器。这就是编组和解组的责任。
[更新] 这是一个相同的博客:http://www.journaldev.com/2552/spring-restful-web-service-example-with-json-jackson-and-client-program
请参阅文档以获取更多详细信息:
http://docs.spring.io/spring-android/docs/current/reference/html/rest-template.html