我从json
获取价值(API
),并且像下面这样创建了model
:
public class ClassesModel implements Parcelable {
public static final Creator<ClassesModel> CREATOR = new Creator<ClassesModel>() {
@Override
public ClassesModel createFromParcel(Parcel in) {
return new ClassesModel(in);
}
@Override
public ClassesModel[] newArray(int size) {
return new ClassesModel[size];
}
};
@SerializedName("status")
@Expose
private String status;
@SerializedName("0")
@Expose
private _0 _0 = null;
public ClassesModel() {
}
protected ClassesModel(Parcel in) {
readFromParcel(in);
}
public String getStatus() {
return status;
}
public _0 get0() {
return _0;
}
@Override
public int describeContents() {
return 0;
}
@Override
public void writeToParcel(Parcel parcel, int flags) {
parcel.writeString(status);
parcel.writeParcelable(_0, flags);
}
private void readFromParcel(Parcel in) {
this.status = in.readString();
this._0 = in.readParcelable(_0.class.getClassLoader());
}
public static class _0 implements Parcelable{
//Some code
}
public static class Userinfo implements Parcelable{
//Some code
}
public static class Teacher implements Parcelable{
//Some code
}
public static class Quiz implements Parcelable{
//Some code
}
public static class Enroll implements Parcelable{
//Some code
}
public static class Detail__ implements Parcelable{
//Some code
}
public static class Detail_ implements Parcelable{
//Some code
}
public static class Detail implements Parcelable{
//Some code
}
public static class Adobeconnect implements Parcelable{
//Some code
}
public static class CourseInfo implements Parcelable{
//Some code
}
public static class Assign implements Parcelable{
//Some code
}
}
但是我正在将源代码转换为architect component
。
如何将上述模型转换为实体以保存在sqlite
中?
当我将所有值保存在Sqlite
中时,我需要选择所有这些值。
在上述模型中,我有多个类,例如Assign
和CourseInfo
,并且...它们具有多个值。我认为我应该为每个类创建单独的table
,并且所有tables
都应具有查询关系!
我该怎么办?