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我想第一次使用翻新并丢失简单的逻辑..请帮助我解决这个问题。带有查询参数的翻新网址
这是我的用户类
public class User {
private String name, email, password;
public User(){
}
public User(String name, String email, String password){
this.name = name;
this.email = email;
this.password = password;
}
public String getName() {
return name;
}
public String getEmail() {
return email;
}
public String getPassword() {
return password;
}
public void setName(String name) {
this.name = name;
}
public void setEmail(String email) {
this.email = email;
}
public void setPassword(String password) {
this.password = password;
}
}
API接口
public interface MyApiEndpointInterface {
// Request method and URL specified in the annotation
// Callback for the parsed response is the last parameter
@GET("users?email={email}")
Call<User> getUser(@Query("email") String email);
}
这是我得到的细节:
public void getUserDetails()
{
String email = inputEmail.getText().toString()
Call<User> call = apiService.getUser(email);
call.enqueue(new Callback<User>() {
@Override
public void onResponse(Call<User>call, Response<User> response) {
if(response.body()!=null)
{
Log.d("TAG", "Name: " + response.body().getName());
Log.d("TAG", "Password: " + response.body().getPassword());
}
else
{
Toast.makeText(getApplicationContext(), "User does not exist", Toast.LENGTH_SHORT).show();
Log.d("TAG", "User details does not exist");
}
}
@Override
public void onFailure(Call<User>call, Throwable t) {
Log.e("TAG", t.toString());
}
});
}
现在我的问题是我有网页API,是托管在服务器上,它看起来像:
,我们将根据电子邮件用户提供细节我试图用这个:
http://www.somesite.com/api/user?email= {EMAIL}
现在我该怎样设置的网址,API接口作为其返回null?
啊傻......刚才我也看到它,因为它失踪了...谢谢哥们:) – coder