Android中的java解析动态json对象
我正在尝试解析下面的json
,并使用BaseAdapter
Android在ListView
中显示它。这里的对象4和1是动态的。如果这些值不是动态的,我知道如何创建模型类。我尝试使用以下方法获取json
,但它不返回任何值在Model.java
中有什么错误吗?还是应该更改json的格式强>
{
"effect_list":[
{
"4":[
{
"effects_id":"18",
"effects_name":"Band 1"
},
{
"effects_id":"19",
"effects_name":"Band 2"
}
],
"1":[
{
"effects_id":"1",
"effects_name":"Background Blur"
},
{
"effects_id":"4",
"effects_name":"Blemish Removal"
}
]
}
]
}
模型。java
public class Model{
@SerializedName("effect_list")
@Expose
List<Map<String,List<EffectList>>> effectlist;
public List<Map<String, List<EffectList>>> getEffectlist() {
return effectlist;
}
public void setEffectlist(List<Map<String, List<EffectList>>> effectlist) {
this.effectlist = effectlist;
}
}
效应列表。java
public class EffectList {
@SerializedName("effects_id")
@Expose
private String effectsId;
@SerializedName("effects_name")
@Expose
private String effectsName;
//GETTERS AND SETTERS
}
mycontactadapter 2。java
public class MyContactAdapter2 extends BaseAdapter {
ArrayList<Map<String, List<EffectList>>> contactList;
Context context;
private LayoutInflater mInflater;
public MyContactAdapter2(Context context, ArrayList<Map<String, List<EffectList>>> data) {
this.context = context;
this.mInflater = LayoutInflater.from(context);
contactList = data;
}
@Override
public int getCount() {
return 0;
}
@Override
public List<HashMap<String, List<EffectList>>> getItem(int position) {
return (List<HashMap<String, List<EffectList>>>) contactList.get(position);
}
@Override
public long getItemId(int i) {
return 0;
}
@Override
public View getView(int position, View convertView, ViewGroup parent) {
final ViewHolder1 vh1;
if (convertView == null) {
View view = mInflater.inflate(R.layout.get_layout_row_view, parent, false);
vh1 = ViewHolder1.create((RelativeLayout) view);
view.setTag(vh1);
} else {
vh1 = (ViewHolder1) convertView.getTag();
}
EffectList item = (EffectList) getItem(position);
// vh.textViewName.setText(item.getEffectsId());
vh1.textViewName.setText(item.getEffectsName());
vh1.textViewEmail.setText(item.getEffectsId());
// Picasso.with(context).load(item.getProfilePic()).placeholder(R.mipmap.ic_launcher).error(R.mipmap.ic_launcher).into(vh.imageView);
return vh1.rootView;
}
private static class ViewHolder1 {
public final RelativeLayout rootView;
public final ImageView imageView;
public final TextView textViewName;
public final TextView textViewEmail;
private ViewHolder1(RelativeLayout rootView, ImageView imageView, TextView textViewName, TextView textViewEmail) {
this.rootView = rootView;
this.imageView = imageView;
this.textViewName = textViewName;
this.textViewEmail = textViewEmail;
}
public static MyContactAdapter2.ViewHolder1 create(RelativeLayout rootView) {
ImageView imageView = (ImageView) rootView.findViewById(R.id.imageView);
TextView textViewName = (TextView) rootView.findViewById(R.id.textViewName);
TextView textViewEmail = (TextView) rootView.findViewById(R.id.textViewEmail);
return new MyContactAdapter2.ViewHolder1(rootView, imageView, textViewName, textViewEmail);
}
}
}
MyContactAdapter2有什么问题吗强>
# 1 楼答案
Have a look at this
我在使用GSON解析时也遇到了同样的问题,但通过使用StringRequest实现<&燃气轮机;而不是GSONRequest<&燃气轮机; 请核对我的答案
# 2 楼答案
您可以从本网站的
JSON
创建POJO Model class
http://www.jsonschema2pojo.org/
这是我用你的
json
做的!虽然你的json
应该更简单-com。实例效应列表。爪哇-
# 3 楼答案
我认为你应该改变这种方法
希望这有帮助