我正在创建天气应用程序,我可以选择用户搜索任何城市的天气。当用户输入有效的城市名称我的应用程序工作正常,但当用户输入无效字符串ex: asndfs,bdfbsj 时,我的应用程序终止。
如何处理?以便我的应用程序不会终止而是向用户说明请输入有效的城市名称
这是我的代码,当用户输入字符串并且我调用openWeatherApi时。
private void getWeatherDataByCity(String city) {
params = new HashMap<>();
params.put(Constants.CITY, city);
params.put(Constants.UNIT, Constants.UNITS);
params.put(Constants.APP_ID, Constants.API_KEY);
RetrofitClient.getData(params).enqueue(new Callback<Result>() {
@Override
public void onResponse(Call<Result> call, Response<Result> response) {
cityName.setText(response.body().getCityName() + "," + response.body().getSys().getCounty());
String temperature = String.format("%.2f", response.body().getMain().getTemp()) + "°C";
temp.setText(temperature);
}
@Override
public void onFailure(Call<Result> call, Throwable t) {
}
});
}
用户输入有效城市时Json响应:
{
"coord": {
"lon": 77.22,
"lat": 28.65
},
"weather": [
{
"id": 721,
"main": "Haze",
"description": "haze",
"icon": "50d"
}
],
"base": "stations",
"main": {
"temp": 305.15,
"pressure": 1008,
"humidity": 43,
"temp_min": 305.15,
"temp_max": 305.15
},
"visibility": 3000,
"wind": {
"speed": 5.7,
"deg": 110
},
"clouds": {
"all": 20
},
"dt": 1525413600,
"sys": {
"type": 1,
"id": 7809,
"message": 0.0996,
"country": "IN",
"sunrise": 1525392476,
"sunset": 1525440495
},
"id": 1273294,
"name": "Delhi",
"cod": 200
}
当用户输入无效城市时,Json会回复:
{
"cod": "404",
"message": "city not found"
}
Result.java
public class Result {
@SerializedName("main")
private Temprature main;
@SerializedName("sys")
private Sys sys;
@SerializedName("cod")
private int cod;
@SerializedName("message")
private String message;
@SerializedName("name")
private String cityName;
@SerializedName("weather")
private List<Weather> weather = new ArrayList<>();
public List<Weather> getWeather() {
return weather;
}
public Temprature getMain() {
return main;
}
public Sys getSys() {
return sys;
}
public String getCityName() {
return cityName;
}
public int getCod() {
return cod;
}
public String getMessage() {
return message;
}
}
答案 0 :(得分:2)
您的应用终止,因为它正在寻找因名称错误而不存在的json字段。试试这个:
public void onResponse(Call<Result> call, Response<Result> response) {
if (response.code() == 404) {
Toast.makeText(MainActivity.this, response.message(), Toast.LENGTH_SHORT).show();
cityName.setText(response.message());
temp.setText("0" + "°C");
} else {
cityName.setText(response.body().getCityName() + "," +
response.body().getSys().getCounty());
String temperature = String.format("%.2f", response.body().getMain().getTemp()) + "°C";
temp.setText(temperature);
}
}
编辑:我不知道如何从类型结果的响应中取出json,现在我不能继续寻找,但是你需要做的就是检查你的回答的“鳕鱼”是否与404不同解析json
答案 1 :(得分:1)
请使用此
public void onResponse(Call<Result> call, Response<Result> response){
try {
cityName.setText(response.body().getCityName() + "," +
response.body().getSys().getCounty());
String temperature = String.format("%.2f",
response.body().getMain().getTemp()) + "°C";
temp.setText(temperature);
} catch(Exception e {
e.printStackTrace();
}