Question

这是我的主要活动：

from string import *

function = False
words_split = []
lineNumber=0
final_value = []

def indexer(t):
    global words_split
    words = t.split();
    for word in words:
        words_split.append(word)

def dict_print():
    for keys in Dict:
            output = keys + " " + str(Dict[keys][0])
            i= 1
            while i < len(Dict[keys]):
                output = output + ", " + str(Dict[keys][i])
                i = i + 1
            print(output)

print("Please type a line and hit 'Enter' or type a single fullstop followed by 'Enter' to exit \n")
text = ""
while function == False :
    if(text == "."):
        print("The index is:")
        function = True
        dict_print()

    else:
        Dict = {}
        text = input()
        lineNumber += 1
        for word in words_split:
            if word in Dict:
                if lineNumber not in Dict[word]:
                    Dict[word] = Dict[word] + [lineNumber]
            else:
                Dict[word] = [lineNumber]
        indexer(text)

这是我的另一项活动：

public class MainMenu extends AppCompatActivity implements BluetoothSerialListener, BluetoothDeviceListDialog.OnDeviceSelectedListener {

          public static BluetoothSerial bluetoothSerial;

          public onCreate(Bundle savedInstanceState) {
                 super.onCreate(savedInstanceState);
                 [...]
                 bluetoothSerial = new BluetoothSerial(this, this);
                 bluetoothSerial.setup();
          }

问题是当我想在bluetoothTest（第二个活动）上做一些东西时，对象为空，我需要在两个活动之间保持蓝牙配置......

如何让这个对象在它们之间保持持久？

此致

Answer 1

使用Singleton模式使BluetoothSerial成为单身。确保BluetoothSerial没有保留上下文，视图或其他系统资源。如果需要，您还可以使BluetoothSerial线程安全。

其他活动中的对象为null

1 个答案: