Gson().tojson() 返回 {} 大括号

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英文:

Gson().tojson() returning {} braces

问题

请有人帮助我找出问题所在...我想将 BluetoothDevice 对象存储到 SharedPrefrences 中,这样下次打开应用程序时就不需要再次选择配对设备。为此,我必须将这个 BluetoothDevice 对象转换为 String 变量,然后将其存储到 SharedPrefrence 中,我使用了以下代码:

Gson gson = new Gson();
String json = gson.toJson(selectedDevice, BluetoothDevice.class);

其中 selectedDevice 是 BluetoothDevice 类的一个对象。但是在运行代码后,我在 json 变量中得到了 {} 大括号。有人知道为什么会出现这种情况吗?

英文:

Please somebody help me to figure out the problem..I want to store BluetoothDevice object in to SharedPrefrences so that next time when I open the app not need to select the paring device again.For this I have to convert this BluetoothDevice object in to String variable and then store it in to SharedPrefrence for that I have used

Gson gson = new Gson();
String json = gson.toJson(selectedDevice,BluetoothDevice.class) ;

Where selectedDevice is an object of BluetoothDevice class. But after running the code I got the {} braces in json variable. Can anyone have any idea why this is happening.

答案1

得分: 1

几乎 android.bluetooth.BluetoothDevice 中的所有属性都是 static final 的。因此,这些属性不依赖于实例。

有一个 address 属性可用,取决于实例变量。您可以使用 getAddress() 将此 address 属性保存到共享首选项中。

如果蓝牙已启用,还可以使用 getName() 获取设备名称。

因此,我建议您将类似以下内容保存在共享首选项中:

class BluetoothDeviceInformation {

    String address;
    String name;

    @RequiresPermission(Manifest.permission.BLUETOOTH)
    public BluetoothDeviceInformation (
        BluetoothDevice device
    ) {
        this.address = device.getAddress();
        this.name = device.getName();
    }
}

您可以使用以下方式创建此实例:

BluetoothDeviceInformation deviceInformation = 
    new BluetoothDeviceInformation(selectedBluetoothDevice);

还有其他方法,如 getType()getAlias() 等。如果您需要它们的信息,只需将它们添加到您的 BluetoothDeviceInformation 类中。

英文:

Almost all of the properties in android.bluetooth.BluetoothDevice are static final. So, these properties do not depend on an instance.

One address property is available and depends on the instance variable. You can just save this address property to your shared-preferences using getAddress().

If bluetooth is enabled, you can, also, get device name with getName().

So, I suggest you save something like this in your shared-preferences;

class BluetoothDeviceInformation {

    String address;
    String name;

    @RequiresPermission(Manifest.permission.BLUETOOTH)
    public BluetoothDeviceInformation (
        BluetoothDevice device
    ) {
        this.address = device.getAddress();
        this.name = device.getName();
    }
}

You can create this instance using:

BluetoothDeviceInformation deviceInformation = 
    new BluetoothDeviceInformation(selectedBluetoothDevice);

There are other methods like getType(), getAlias() etc. If you need their information, just add them to your BluetoothDeviceInformation class.

答案2

得分: 0

假设您指的是 https://android.googlesource.com/platform/frameworks/base/+/master/core/java/android/bluetooth/BluetoothDevice.java,其中包含静态字段,但是,

Java序列化仅对对象的非静态和非瞬态字段进行序列化

文档

但是如果您真的需要,您可以尝试相同的方法:

BluetoothDevice selectedDevice = new BluetoothDevice();
GsonBuilder gsonBuilder = new GsonBuilder();
gsonBuilder.excludeFieldsWithModifiers(java.lang.reflect.Modifier.TRANSIENT);
Gson gson = gsonBuilder.create();
String json = gson.toJson(selectedDevice, BluetoothDevice.class);
英文:

Assuming you are referring https://android.googlesource.com/platform/frameworks/base/+/master/core/java/android/bluetooth/BluetoothDevice.java, it contains static fields but,

Java Serialization only serialize object's non-static and non-transient fields

Doc

But if you really need it same way you can try:

BluetoothDevice selectedDevice = new BluetoothDevice();
GsonBuilder gsonBuilder = new GsonBuilder();
gsonBuilder.excludeFieldsWithModifiers(java.lang.reflect.Modifier.TRANSIENT);
Gson gson = gsonBuilder.create();
String json = gson.toJson(selectedDevice, BluetoothDevice.class);

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  • 本文由 发表于 2020年9月25日 15:12:42
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