持久列表的最有效方法是什么？

I am creating an Android application that allows the user to create different lists of friends. For each list, the items have a name and 2 parameters...

我正在创建一个Android应用程序，允许用户创建不同的朋友列表。对于每个列表，项目都有一个名称和2个参数......

EXAMPLE

例

List 1: (Steve, param 1, param 2), (Lisa, param 1, param 2)... etc

清单1：（Steve，param 1，param 2），（Lisa，param 1，param 2）......等

List 2: (John, param 1, param 2), (Steve, param 1, param 2)... etc ...

清单2 :( John，param 1，param 2），（Steve，param 1，param 2）......等等......

What is the best way of dynamically persisting all these lists created by the user?

动态保存用户创建的所有列表的最佳方法是什么？

I was thinking about using SQLite since it is quite easy to implement.

我正在考虑使用SQLite，因为它很容易实现。

I also though about working with plain JSON files, but this approach seemed too resource consuming, because the application deals with adding, reading, modifying stuff...

我也使用普通的JSON文件，但这种方法似乎太耗费资源，因为应用程序处理添加，读取，修改内容......

So, what is the best way of persisting these objects?

那么，持久化这些对象的最佳方法是什么？

If I use a database, I thought it should have the following scheme:

如果我使用数据库，我认为它应该有以下方案：

PERSONS - columns: Person_ID, Person_Name

PERSONS - 列：Person_ID，Person_Name

LISTS - columns: list

列表 - 列：列表

For LISTS, each row is just formed by some sort of set of tuples, representing a single list.

对于LISTS，每一行都是由某种元组集组成，代表一个列表。

EXAMPLE

例

Row 1: [(Person_ID, param1, param2), (Person_ID, param1, param2), ...]

第1行：[（Person_ID，param1，param2），（Person_ID，param1，param2），...]

Where Row 1 is a single list with all the people contained in it.

第1行是单个列表，其中包含所有人员。

Unfortunately this doesn't seem very efficient....

不幸的是，这似乎不是很有效....

I also thought about creating an extra table

我还想过创建一个额外的表

LIST_ROW - columns: row_id, Person_ID, param1, param2

LIST_ROW - 列：row_id，Person_ID，param1，param2

and then each row in the LISTS table would just be a tuple of IDs for all the rows that belong to that list...

然后LISTS表中的每一行都只是属于该列表的所有行的ID元组...

EXAMPLE

例

Row 1 would just become: (row_id, row_id....) representing the elements in the list 1.

第1行将成为：（row_id，row_id ....），表示列表1中的元素。

Unfortunately, I am not sure on how efficient this is.

不幸的是，我不确定这是多么有效。

Can someone tell me what is the best way of achieving this?

有人能告诉我实现这个目标的最佳方式是什么？

2 个解决方案

#1

You can use shared preferences for this. Convert your list to a String(comma sepereated objects and values) then add this is shared prefrences. In same way you can get this and convert to your list.

您可以使用共享首选项。将列表转换为String（逗号分隔的对象和值）然后添加这是共享的prefrences。以同样的方式，您可以获得此并转换为您的列表。

#2

if the data is no so large,you can save them in sharedpreference, for example: first define the data bean, second add the data bean to arraylist and then save to sp:

如果数据不是那么大，可以将它们保存在sharedpreference中，例如：首先定义数据bean，然后将数据bean添加到arraylist然后保存到sp：

saveObject("list_one", list);



/**
 * Save object to SharedPreferences
 *
 * @param key
 * @param obj
 */
protected void saveObject(String key, Object obj) {
    try {
        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        ObjectOutputStream oos = new ObjectOutputStream(baos);
        oos.writeObject(obj);
        String str = Base64.encodeToString(baos.toByteArray(), Base64.DEFAULT);
        SharedPreferences.Editor editor = sp.edit();
        editor.putString(key, str);
        editor.commit();
    } catch (IOException e) {
        e.toString();
    }
}

/**
 * Read object from SharedPreferences
 *
 * @param key
 * @return object
 */
protected Object getObject(String key) {
    try {
        String str = sp.getString(key, "");
        ByteArrayInputStream baim = new ByteArrayInputStream(Base64.decode(str, Base64.DEFAULT));
        ObjectInputStream ois = new ObjectInputStream(baim);
        return ois.readObject();
    } catch (Exception e) {
    }
    return null;
}