结果变量包含已更正的已解析JSON。
但是,在反序列化 List
包含正确数量的项目后,所有都为空。
如何解决?
Gson gson = new Gson();
List<UnitView> unitViews = new ArrayList<UnitView>();
// https://stackoverflow.com/questions/5554217/google-gson-deserialize-listclass-object-generic-type
Type typeToken = new TypeToken<List<UnitView>>() { }.getType();
unitViews = gson.fromJson(result,typeToken);
即使我喜欢
UnitView[] unitViews = gson.fromJson(result, UnitView[].class);
项目的字段也是空的。
UnitView
public class UnitView implements Serializable {
public String id ;
public String name ;
public String description ;
public String deviceTypeName ;
public String categoryID ;
public String lastOnline ;
public String latitude ;
public String longitude ;
public String atTime ;
public String getId() {
return id;
}
public String getName() {
return name;
}
public String getDescription() {
return description;
}
public String getDeviceTypeName() {
return deviceTypeName;
}
public String getCategoryID() {
return categoryID;
}
public String getLastOnline() {
return lastOnline;
}
public String getLatitude() {
return latitude;
}
public String getLongitude() {
return longitude;
}
public String getAtTime() {
return atTime;
}
public void setId(String id) {
this.id = id;
}
public void setName(String name) {
this.name = name;
}
public void setDescription(String description) {
this.description = description;
}
public void setDeviceTypeName(String deviceTypeName) {
this.deviceTypeName = deviceTypeName;
}
public void setCategoryID(String categoryID) {
this.categoryID = categoryID;
}
public void setLastOnline(String lastOnline) {
this.lastOnline = lastOnline;
}
public void setLatitude(String latitude) {
this.latitude = latitude;
}
public void setLongitude(String longitude) {
this.longitude = longitude;
}
public void setAtTime(String atTime) {
this.atTime = atTime;
}
}
JSON DATA
[{"ID":"294","Name":"Foton Tunland № F110","Description":null,"DeviceTypeName":"Техника ТО","CategoryID":"18","LastOnline":"19.12.2017 20:38:04","Latitude":"11,40119","Longitude":"11,42403","AtTime":"19.12.2017 20:38:04"},{"ID":"295","Name":"DML LP1200 № 9793","Description":null,"DeviceTypeName":"Буровой станок дизельный","CategoryID":"15","LastOnline":null,"Latitude":null,"Longitude":null,"AtTime":null}]
答案 0 :(得分:2)
好的,问题是解析器区分大小写,您可以更改属性的名称以匹配您可以使用SerializedName
注释的json值的名称,如下所示:
@SerializedName("ID")
public String id ;
@SerializedName("Name")
public String name ;
@SerializedName("Description")
public String description;
...
或
public String ID ;
public String Name ;
public String Description ;
...
答案 1 :(得分:-2)
我认为你因为json中的空值而遇到了这个问题。 核实。 Source