嗨,我想学习Android,现在实施改造1.9我的休息POST和GET请求可以有人帮助我如何模型给定的JSON对象和字符串?即时通讯很困惑的一些教程,我已经学会了如何做一个POJO此JSON对象Android Retrofit POJO模型
{
"contacts": [
{
"id": "c200",
"name": "Ravi Tamada",
"email": "[email protected]",
"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": "[email protected]",
"address": "xx-xx-xxxx,x - street, x - country",
"gender" : "male",
"phone": {
"mobile": "+91 0000000000",
"home": "00 000000",
"office": "00 000000"
}
}
}}]}
采用这种模式
Contacts.class
public class Contacts {
@SerializedName("contacts")
@Expose
private List<Contact> contacts = new ArrayList<Contact>();
public List<Contact> getContacts() {
return contacts;
}
public void setContacts(List<Contact> contacts) {
this.contacts = contacts;
}
和Contact.class为对象
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;}}
并在我的MainActivity.class上使用此方法调用列表
private void getContacts() {
final ProgressDialog loading = ProgressDialog.show(this, "Fetching Data", "Please wait...", false, false);
RestAdapter adapter = new RestAdapter.Builder().setEndpoint(ROOT_URL).build();
ContactsAPI api = adapter.create(ContactsAPI.class);
api.getContacts(new Callback<Contacts>() {
@Override
public void success(Contacts contacts, Response response) {
loading.dismiss();
List<Contact> contactList = contacts.getContacts();
String[] items = new String[contactList.size()];
for (int i = 0; i < contactList.size(); i++) {
items[i] = contactList.get(i).getName();
}
ArrayAdapter adapter = new ArrayAdapter<String>(MainActivity.this, R.layout.simple_list,R.id.textview, items);
//Setting adapter to listviesw
listView.setAdapter(adapter);
}
@Override
public void failure(RetrofitError error) {
}
});
}
我的问题是我应该如何使这个数组对象的模型?
{
"-KNea90tV5nZlkeqxc3Q": {
"accountName": "Mark Papyrus",
"accountNumber": "12435656443",
"accountType": "Peso Savings"
},
"-KNeaPmBoTXV4mQC6cia": {
"accountName": "Mark Dremeur",
"accountNumber": "12435656444",
"accountType": "Peso Checking"
}
我发现它混淆了如何使模型和给定的json数组的区别pls指导我的感谢。
这里有一个普罗蒂普 - 它使用http://www.jsonschema2pojo.org/例如产生。 – Shark
@ Shark-实际上做到了,但我的问题是JSON像这样的键“-KNea90tV5nZlkeqxc3Q”是随机生成的,所以我不能让它保持不变 – user3262438
这将使JSON成为一个糟糕的选择,除非随机生成的salt部分成为成员。我说这很糟糕,因为模型类最终将包含所有随机值并将值分配给其中的一个,而您不知道要检查哪一个值......它需要与后端团队进行交谈,以向您提供进一步的指示。 – Shark