android – 如何在Service和Activity之间传递复杂数据？

>首先序列化要传递的对象.
>将序列化对象放在intent extras中.
>在接收端,只需获取序列化对象,然后对其进行反序列化.

说,

employee employee = new employee();

然后,

intent.putExtra("employee",serializeObject(employeE));

接收时,

byte[] semployee = extras.getByteArray("employee");

employee =(employeE)deserializeObject(semployeE);

仅供参考,

public static byte[] serializeObject(Object o) throws Exception,IOException {
    ByteArrayOutputStream bos = new ByteArrayOutputStream();
    ObjectOutput out = new ObjectOutputStream(bos);
    try {
        out.writeObject(o);
        // Get the bytes of the serialized object
        byte[] buf = bos.toByteArray();

        return buf;
    } catch (IOException E) {
        Log.e(LOG_TAG,"serializeObject",E);
        throw new Exception(E);
    } finally {
        if (out != null) {
            out.close();
        }
    }
}

public static Object deserializeObject(byte[] b)
        throws StreamCorruptedException,IOException,ClassnotFoundException,Exception {
    ObjecTinputStream in = new ObjecTinputStream(
            new ByteArrayInputStream(b));
    try {
        Object object = in.readObject();
        return object;
    } catch (Exception E) {
        Log.e(LOG_TAG,"deserializeObject",E);
        throw new Exception(E);
    } finally {
        if (in != null) {
            in.close();
        }
    }
}