问题：
Android改装无响应
聂溪叠
HI我只是新来的改造服务，并遵循本教程https://www.simplifiedcoding.net/retrofit-android-tutorial-to-get-json-from-server/它工作得很好，想创建我自己的，所以我使用了一个新的json网络http://api.androidhive.info/contacts/，其中包含
{
"contacts": [
{
"id": "c200",
"name": "Ravi Tamada",
"email": "ravi@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c201",
"name": "Johnny Depp",
"email": "johnny_depp@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c202",
"name": "Leonardo Dicaprio",
"email": "leonardo_dicaprio@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c203",
"name": "John Wayne",
"email": "john_wayne@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c204",
"name": "Angelina Jolie",
"email": "angelina_jolie@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "female",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c205",
"name": "Dido",
"email": "dido@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "female",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c206",
"name": "Adele",
"email": "adele@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "female",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c207",
"name": "Hugh Jackman",
"email": "hugh_jackman@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c208",
"name": "Will Smith",
"email": "will_smith@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c209",
"name": "Clint Eastwood",
"email": "clint_eastwood@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c2010",
"name": "Barack Obama",
"email": "barack_obama@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c2011",
"name": "Kate Winslet",
"email": "kate_winslet@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "female",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
},
{
"id": "c2012",
"name": "Eminem",
"email": "eminem@gmail.com",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
}
]
我实现了自己的接口ContactAPI。Java语言
public interface ContactsAPI {
@GET("/contacts/")
public void getContacts(Callback<List<Contact>> response);}
并实现了像这样的模型类Contact.java
public class Contact {
@SerializedName("id")
@Expose
private String id;
@SerializedName("name")
@Expose
private String name;
@SerializedName("email")
@Expose
private String email;
@SerializedName("address")
@Expose
private String address;
@SerializedName("gender")
@Expose
private String gender;
public String getId() {return id;}
public void setId(String id) {this.id = id;}
public String getName() {return name;}
public void setName(String name) {this.name = name;}
public String getEmail() {return email;}
public void setEmail(String email) {this.email = email;}
public String getAddress() {return address;}
public void setAddress(String address) {this.address = address;}
public String getGender() {return gender;}
public void setGender(String gender) {this.gender = gender;}}
然后最终在MainActivity中实现了我的Restadapter。班
public static final String ROOT_URL = "http://api.androidhive.info";
private ListView listView;
private List<Contact> contacts;
RestAdapter adapter = new RestAdapter.Builder().setEndpoint(ROOT_URL).build();
ContactsAPI api = adapter.create(ContactsAPI.class);
api.getContacts(new Callback<List<Contact>>() {
@Override
public void success(List<Contact> list, Response response) {
Toast.makeText(MainActivity.this,list.toString(),Toast.LENGTH_SHORT).show();
showList();
}
@Override
public void failure(RetrofitError error) {
//you can handle the errors here
Toast.makeText(MainActivity.this,"Error Occured:"+error.toString(),Toast.LENGTH_SHORT).show();
}
});
该应用程序运行平稳，但在4秒延迟后，它会提示一个错误，这是公共无效失败（RetrofitError错误）我不知道我错过了什么，我检查了我的代码，没有发现任何错误，请提前帮助我谢谢。
共有1个答案
武嘉祥
如果json是this，您需要一个类：
public class Contacts {
@SerializedName("contacts")
@Expose
private List<Contact> contacts = new ArrayList<Contact>();
/**
*
* @return
* The contacts
*/
public List<Contact> getContacts() {
return contacts;
}
/**
*
* @param contacts
* The contacts
*/
public void setContacts(List<Contact> contacts) {
this.contacts = contacts;
}
}
也在您的接口ContactAPI中。Java语言
public interface ContactsAPI {
@GET("/contacts/")
public void getContacts(Callback<Contacts> response);
}
同时检查此链接，因为“”有问题，此错误
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