我对我的某个网络服务有以下回应，并且我使用了Retrofit和GSON。
{
"error": false,
"Timeline": {
"Date": "2040-06-15",
"bandList": {
"breakfast": {
"dosageList": {
"01": {
"packed": "true",
"medicineList": [
{
"medicine": {
"id": "01",
"name": "glipizide 5 mg tablet, 100 ",
"category": "regular",
"image": null,
"indication": "NIDDM",
"packed": true,
"med_id": "352",
"dosage": 1
}
},
{
"medicine": {
"id": "04",
"name": "Frusemide (Terry White Chemists) 20 mg uncoated tablet, 100 ",
"category": "regular",
"image": null,
"indication": "Fluid",
"packed": true,
"med_id": "4",
"dosage": 2
}
}
]
},
"02": {
"packed": "false",
"medicineList": [
{
"medicine": {
"id": "05",
"name": "Refresh Tears Plus 0.5% eye drops solution, 15 mL ",
"category": "regular",
"image": null,
"indication": "Dry Eyes",
"packed": false,
"med_id": "372",
"dosage": 1
}
}
]
}
}
}
}
}
}
Q1。 有没有办法使用模型类（POJO）解析上面的响应或没有它们？我坚持为上述结构生成模型类。如何为JSON生成POJO？
Q2。我能够说服发送以下回复，JSON的正确结构/格式是什么？是否有任何JSON标准我可以向Web开发人员展示获取此JSON格式？ （注意：我可以解析这个结构）
{
"error": false,
"Timeline": {
"Date": "2040-06-15",
"band": [
{
"name": "breakfast",
"dosage": [
{
"id": "01",
"packed": "true",
"medicine": [
{
"id": "01",
"name": "glipizide 5 mg tablet, 100 ",
"category": "regular",
"image": null,
"indication": "NIDDM",
"packed": true,
"med_id": "52",
"dosage": 1
},
{
"id": "04",
"name": "Frusemide (Terry White Chemists) 20 mg uncoated tablet, 100 ",
"category": "regular",
"image": null,
"indication": "Fluid",
"packed": true,
"med_id": "54",
"dosage": 2
}
]
},
{
"id": "02",
"packed": "false",
"medicine": [
{
"id": "05",
"name": "Refresh Tears Plus 0.5% eye drops solution, 15 mL ",
"category": "regular",
"image": null,
"indication": "Dry Eyes",
"packed": false,
"med_id": "372",
"dosage": 1
}
]
}
]
}
]
}
}
提前谢谢。
修改
我使用这些网站自动生成POJO，但是它为某些类提供了以下响应。如何将其转换为适当的类？
package ;
public class DosageList
{
private 01 01;
private 02 02;
public void set01(01 01){
this.01 = 01;
}
public 01 get01(){
return this.01;
}
public void set02(02 02){
this.02 = 02;
}
public 02 get02(){
return this.02;
}
}
编辑2
我几乎完成了解析第一个JSON，但仍然坚持到这里。
for (String bandName: event.getTimeline().getBand().keySet()) {
Log.d("<<<--Band-->>>", "Value " + event.getTimeline().getBand().get(bandName));
Band band = event.getTimeline().getBand().get(bandName);
for (String dosageName:band.getDosage().keySet()) {
Dosage dosage = band.getDosage().get(dosageName);
Log.d("<<<--Dosage-->>>", "Value " + dosage.getMedicine());
for (Medicine medicine: dosage.getMedicine()) {
Log.d("<<<--Medicine-->>>", "Value " + dosage.getMedicine().get(0));
}
}
}
如何检索药物值？
答案 0 :(得分：1)
public class Medicine
{
private String id;
private String name;
private String category;
private String image;
private String indication;
private boolean packed;
private String med_id;
private int dosage;
public void setId(String id){
this.id = id;
}
public String getId(){
return this.id;
}
public void setName(String name){
this.name = name;
}
public String getName(){
return this.name;
}
public void setCategory(String category){
this.category = category;
}
public String getCategory(){
return this.category;
}
public void setImage(String image){
this.image = image;
}
public String getImage(){
return this.image;
}
public void setIndication(String indication){
this.indication = indication;
}
public String getIndication(){
return this.indication;
}
public void setPacked(boolean packed){
this.packed = packed;
}
public boolean getPacked(){
return this.packed;
}
public void setMed_id(String med_id){
this.med_id = med_id;
}
public String getMed_id(){
return this.med_id;
}
public void setDosage(int dosage){
this.dosage = dosage;
}
public int getDosage(){
return this.dosage;
}
}
import java.util.ArrayList;
import java.util.List;
public class Dosage
{
private String id;
private String packed;
private List<Medicine> medicine;
public void setId(String id){
this.id = id;
}
public String getId(){
return this.id;
}
public void setPacked(String packed){
this.packed = packed;
}
public String getPacked(){
return this.packed;
}
public void setMedicine(List<Medicine> medicine){
this.medicine = medicine;
}
public List<Medicine> getMedicine(){
return this.medicine;
}
}
import java.util.ArrayList;
import java.util.List;
public class Band
{
private String name;
private List<Dosage> dosage;
public void setName(String name){
this.name = name;
}
public String getName(){
return this.name;
}
public void setDosage(List<Dosage> dosage){
this.dosage = dosage;
}
public List<Dosage> getDosage(){
return this.dosage;
}
}
import java.util.ArrayList;
import java.util.List;
public class Timeline
{
private DateTime Date;
private List<Band> band;
public void setDate(DateTime Date){
this.Date = Date;
}
public DateTime getDate(){
return this.Date;
}
public void setBand(List<Band> band){
this.band = band;
}
public List<Band> getBand(){
return this.band;
}
}
public class Root
{
private boolean error;
private Timeline Timeline;
public void setError(boolean error){
this.error = error;
}
public boolean getError(){
return this.error;
}
public void setTimeline(Timeline Timeline){
this.Timeline = Timeline;
}
public Timeline getTimeline(){
return this.Timeline;
}
}
...享受...
答案 1 :(得分：0)
对于第一个问题：
您可以在没有POJO类的情况下解析JSON，但建议使用它 他们和关于你一直在坚持从JSON生成它们 我想你可以使用API 这是最好的。
第二个：
是否有JSON标准，您可以在jsonschema2pojo网站找到它们。
答案 2 :(得分：0)
步骤1：首先从最内层的json对象开始，我可以看到＆＃34; medicine＆＃34;
创建一个像这样的POJO类
uninstall第2步：为＆＃34; medicineList＆＃34;创建一个类。这有点像这样
public class Medicine implements android.os.Parcelable {
@SerializedName("id")
private String id;
// Getter setter method for id
// Do it for all the JSON tags
}
以类似的方式在JSON响应中向外移动到基本标记。这让我很容易理解。我没有发布完整的解决方案，因为那是你在EOD的作业。
答案 3 :(得分：0)
Q1的解决方案 -
是的，您可以通过在Android Studio中安装DTO插件来解析使用模型类的上述响应。该插件将自动为响应创建POJO类。