Android - 带有可序列化对象的SharedPreferences

Question

我知道SharedPreferences有putString(),putFloat(),putLong(),putInt()和putBoolean().但我需要存储一个对象,它是类型的Serializable在SharedPreferences.我怎样才能做到这一点？

Answer 1

总之,你不能尝试将对象序列化为私有文件,这相当于同样的事情.样本类如下:

import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;

import android.app.Activity;
import android.content.Context;

/**
 *
 * Writes/reads an object to/from a private local file
 * 
 *
 */
public class LocalPersistence {


    /**
     * 
     * @param context
     * @param object
     * @param filename
     */
    public static void witeObjectToFile(Context context, Object object, String filename) {

        ObjectOutputStream objectOut = null;
        try {

            FileOutputStream fileOut = context.openFileOutput(filename, Activity.MODE_PRIVATE);
            objectOut = new ObjectOutputStream(fileOut);
            objectOut.writeObject(object);
            fileOut.getFD().sync();

        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            if (objectOut != null) {
                try {
                    objectOut.close();
                } catch (IOException e) {
                    // do nowt
                }
            }
        }
    }


    /**
     * 
     * @param context
     * @param filename
     * @return
     */
    public static Object readObjectFromFile(Context context, String filename) {

        ObjectInputStream objectIn = null;
        Object object = null;
        try {

            FileInputStream fileIn = context.getApplicationContext().openFileInput(filename);
            objectIn = new ObjectInputStream(fileIn);
            object = objectIn.readObject();

        } catch (FileNotFoundException e) {
            // Do nothing
        } catch (IOException e) {
            e.printStackTrace();
        } catch (ClassNotFoundException e) {
            e.printStackTrace();
        } finally {
            if (objectIn != null) {
                try {
                    objectIn.close();
                } catch (IOException e) {
                    // do nowt
                }
            }
        }

        return object;
    }

}

Answer 2

接受的答案具有误导性,我们可以使用GSON将可序列化对象存储到SharedPreferences中.在google-gson上阅读更多相关信息.

您可以在Gradle文件中添加GSON依赖项:

compile 'com.google.code.gson:gson:2.7'

这里的片段:

首先,创建您通常的sharedPreferences:

//Creating a shared preference
SharedPreferences  mPrefs = getPreferences(MODE_PRIVATE);

从可序列化对象保存到首选项:

 Editor prefsEditor = mPrefs.edit();
 Gson gson = new Gson();
 String json = gson.toJson(YourSerializableObject);
 prefsEditor.putString("SerializableObject", json);
 prefsEditor.commit();

从首选项获取可序列化对象:

Gson gson = new Gson();
String json = mPrefs.getString("SerializableObject", "");
yourSerializableObject = gson.fromJson(json, YourSerializableObject.class);

Answer 3

如果对象是简单的POJO,则可以将对象转换为JSON字符串,并使用putString()将其保存在共享首选项中.

Answer 4

没有文件就可以做到这一点.

我正在将信息序列化到base64,像这样我可以将它保存为首选项中的字符串.

以下代码是将序列化对象序列化为base64字符串,反之亦然:import android.util.Base64;

import java.io.ByteArrayInputStream;
import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.io.Serializable;


public class ObjectSerializerHelper {
    static public String objectToString(Serializable object) {
        String encoded = null;
        try {
            ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
            ObjectOutputStream objectOutputStream = new ObjectOutputStream(byteArrayOutputStream);
            objectOutputStream.writeObject(object);
            objectOutputStream.close();
            encoded = new String(Base64.encodeToString(byteArrayOutputStream.toByteArray(),0));
        } catch (IOException e) {
            e.printStackTrace();
        }
        return encoded;
    }

    @SuppressWarnings("unchecked")
    static public Serializable stringToObject(String string){
        byte[] bytes = Base64.decode(string,0);
        Serializable object = null;
        try {
            ObjectInputStream objectInputStream = new ObjectInputStream( new ByteArrayInputStream(bytes) );
            object = (Serializable)objectInputStream.readObject();
        } catch (IOException e) {
            e.printStackTrace();
        } catch (ClassNotFoundException e) {
            e.printStackTrace();
        } catch (ClassCastException e) {
            e.printStackTrace();
        }
        return object;
    }

}

Answer 5

我们可以使用 Kotlin 创建易于使用的语法。

@Throws(JsonIOException::class)
fun Serializable.toJson(): String {
   return Gson().toJson(this)
}

@Throws(JsonSyntaxException::class)
 fun <T> String.to(type: Class<T>): T where T : Serializable {
 return Gson().fromJson(this, type)
}

@Throws(JsonIOException::class)
fun SharedPreferences.Editor.putSerializable(key: String, o: Serializable?) = apply {
   putString(key, o?.toJson())
}

@Throws(JsonSyntaxException::class)
   fun <T> SharedPreferences.getSerializable(key: String, type: Class<T>): T? where T : Serializable {
    return getString(key, null)?.to(type)
}

然后SharedPreferences使用类似的方法保存任何可序列化的get/put()

此处完整要点将可序列化文件保存在与 Kotlin 和 GSON 共享的首选项中

正如其他答案中提到的，当数据类结构发生变化时，您可能必须考虑迁移。或者至少必须更改您用来存储的密钥。