Java 在Android中解析JSON数据
我有一个JSON数据,格式如下Java 在Android中解析JSON数据,java,android,json,Java,Android,Json,我有一个JSON数据,格式如下 [ { "name": "France", "date_time": "2015-05-17 19:59:00", "dewpoint": "17", "air_temp": "10.8" }, { "name": "England", "date_time": "2015-05-17 19:58:48", "dewpoint":
[
{
"name": "France",
"date_time": "2015-05-17 19:59:00",
"dewpoint": "17",
"air_temp": "10.8"
},
{
"name": "England",
"date_time": "2015-05-17 19:58:48",
"dewpoint": "13",
"air_temp": "10.6"
},
{
"name": "Ireland",
"date_time": "2015-05-17 19:58:50",
"dewpoint": "15",
"air_temp": "11.1"
}
]
我已经为Android应用程序设置了谷歌地图,所以我有一个在两个活动之间传递名称值(GoogleMaps.java和WeatherInfo.java),现在当我单击谷歌地图中的一个点时,它会将名称传递到WeatherInfo.java,我想获取该名称的天气数据
例如:我单击地图上的法国点,WeatherInfo.class将得到名称为“France”,并打印出该点的“日期、时间、露点、气温”
我的问题是,如何才能只解析我在地图中单击的点的Json数据?有人可以查看我的WeatherInfo.java类中的for循环吗
WeatherInfo.java
protected Void doInBackground(Void... arg0) {
// Creating service handler class instance
ServiceHandler sh = new ServiceHandler();
// Making a request to url and getting response
String jsonStr = sh.makeServiceCall(url, ServiceHandler.GET);
Log.d("Response: ", "> " + jsonStr);
if (jsonStr != null) {
try {
JSONObject jsonObj = new JSONObject(jsonStr);
// Getting JSON Array node
contacts = jsonObj.getJSONArray("");
// looping through All Contacts
for (int i = 0; i < contacts.length(); i++) {
JSONObject c = contacts.getJSONObject(i);
String name = c.getString(TAG_NAME);
String date_time = c.getString(TAG_DATE);
String temp = c.getString(TAG_TEMP);
String dewpoint = c.getString(TAG_DEWPOINT);
// tmp hashmap for single contact
HashMap<String, String> contact = new HashMap<String, String>();
// adding each child node to HashMap key => value
contact.put(TAG_NAME, name);
contact.put(TAG_DATE, date_time);
contact.put(TAG_TEMP, temp);
contact.put(TAG_DEWPOINT, dewpoint);
// adding contact to contact list
contactList.add(contact);
}
} catch (JSONException e) {
e.printStackTrace();
}
} else {
Log.e("ServiceHandler", "Couldn't get any data from the url");
}
return null;
}
受保护的Void doInBackground(Void…arg0){
//创建服务处理程序类实例
ServiceHandler sh=新的ServiceHandler();
//向url发出请求并获得响应
字符串jsonStr=sh.makeServiceCall(url,ServiceHandler.GET);
Log.d(“响应:”、“>”+jsonStr);
if(jsonStr!=null){
试一试{
JSONObject jsonObj=新的JSONObject(jsonStr);
//获取JSON数组节点
contacts=jsonObj.getJSONArray(“”);
//通过所有触点循环
对于(int i=0;ivalue
联系人:put(标签名称、姓名);
联系人。放置(标记日期、日期和时间);
触点放置(标签温度、温度);
接触。放置(标记露点、露点);
//将联系人添加到联系人列表
联系人列表。添加(联系人);
}
}捕获(JSONException e){
e、 printStackTrace();
}
}否则{
Log.e(“ServiceHandler”,“无法从url获取任何数据”);
}
返回null;
}
JSONArray数组=(JSONArray)新的JSONTokener(jsonStr).nextValue();
对于(int i=0;i您可以使用Jackson json解析器,如下所示:-
点数据需要一个值对象
import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonProperty;
public class Point {
private final String name;
private final String dateTime;
private final int dewpoint;
private final double airTemp;
@JsonCreator
public Point(@JsonProperty("name") final String name, @JsonProperty("date_time") final String dateTime, @JsonProperty("dewpoint") final int dewpoint, @JsonProperty("air_temp") final double airTemp) {
this.name = name;
this.dateTime = dateTime;
this.dewpoint = dewpoint;
this.airTemp = airTemp;
}
public String getName() {
return name;
}
public String getDateTime() {
return dateTime;
}
public int getDewpoint() {
return dewpoint;
}
public double getAirTemp() {
return airTemp;
}
}
然后这个Jackson对象映射器
// 2. Convert JSON to Java object
ObjectMapper mapper = new ObjectMapper();
Point[] points = mapper.readValue(new File("points.json"), Point[].class);
for (Point point : points) {
System.out.println("" + point.getName());
System.out.println("" + point.getDateTime());
System.out.println("" + point.getDewpoint());
System.out.println("" + point.getAirTemp());
}
我如何才能得到仅针对我在地图中单击的点解析的Json数据?只需使用库解析所有Json,并在需要时从解析结果中检索相关数据。联系人列表是否包含所有位置以及日期,即法国、英国等我使用的库,只是认为我的for循环不正确,可以吗看我在类中的for循环吗?是的,网页上显示的contec列表是我在上面放的,只有一个数组,包含这三个点名称,日期时间,露点,空气温度。你的json以JsonArray[]开始,而不是JSONObject{}你是对的,我是按照下面的链接来做这门课的,我想实现一些东西,但仅仅因为我的json数据格式与教程不同,我就迷路了。嗨@vrbsm,我怎么说如果我从GoogleMaps获得的数据是法国,而不是我想打印name=“France”的天气信息(日期、时间、露点、气温)只是,我怎样才能做到呢?
import com.fasterxml.jackson.annotation.JsonCreator;
import com.fasterxml.jackson.annotation.JsonProperty;
public class Point {
private final String name;
private final String dateTime;
private final int dewpoint;
private final double airTemp;
@JsonCreator
public Point(@JsonProperty("name") final String name, @JsonProperty("date_time") final String dateTime, @JsonProperty("dewpoint") final int dewpoint, @JsonProperty("air_temp") final double airTemp) {
this.name = name;
this.dateTime = dateTime;
this.dewpoint = dewpoint;
this.airTemp = airTemp;
}
public String getName() {
return name;
}
public String getDateTime() {
return dateTime;
}
public int getDewpoint() {
return dewpoint;
}
public double getAirTemp() {
return airTemp;
}
}
// 2. Convert JSON to Java object
ObjectMapper mapper = new ObjectMapper();
Point[] points = mapper.readValue(new File("points.json"), Point[].class);
for (Point point : points) {
System.out.println("" + point.getName());
System.out.println("" + point.getDateTime());
System.out.println("" + point.getDewpoint());
System.out.println("" + point.getAirTemp());
}