0

我对 Java/Android 开发有点陌生，我需要比较两次。其中一个时间值存储在一个字符串中，另一个是当前系统时间（长） - 我怎样才能找出两个时间之间的差异（以毫秒为单位）并将其存储为一个数字？

编辑：

尝试＃1（响应懒惰的忍者）

public class Rules extends Activity {

private String password;

private PendingIntent mPendingIntent;
String TIMELIMIT = "10";

TextView textSsid, textSpeed, textRssi, Time;
private static final int NOTIFY_ME_ID = 1337;
private int count = 0;
private NotificationManager notifyMgr = null;
public Handler mHandler = new Handler();
public long mStartRX = 0;
public long mStartTX = 0;
public long txBytes;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.rules);

    String NDEF_PREF = "prefs";
    SharedPreferences prefs = getSharedPreferences(NDEF_PREF,
            Context.MODE_PRIVATE);
    String name = prefs.getString("name", "");
    String code = prefs.getString("corename", "");
    String time = prefs.getString("time", "");
    String ssid = prefs.getString("restricted", "");
    Time = (TextView) findViewById(R.id.Time);
    Time.setText(time);

     String dtStart = "09:27:37";
    SimpleDateFormat format = new SimpleDateFormat("HH:mm:ss");
    try {
        Date date = format.parse(time);
    } catch (ParseException e) {
        e.printStackTrace();
    }

    Long convertedLong = date.getTimeInMillis();

    long currentTimeLong = System.currentTimeMillis(); 
    long difference = currentTimeLong - convertedLong;
     String strDiff = String.valueOf(difference);

问题：

日期无法解析 Rules.java Java 问题

4

1 回答 1

1

好的，将您的字符串转换为日期时间：

String dtStart = "09:27:37";  
SimpleDateFormat  format = new SimpleDateFormat("HH:mm:ss");  
try {  
    Date date = format.parse(dtStart);  
 } catch (ParseException e) {  
    // TODO Auto-generated catch block  
    e.printStackTrace();  
}

然后将其转换为毫秒

long convertedLong = date.getTimeInMillis();

然后

long currentTimeLong = System.currentTimeMillis(); // will get you current time in milli
long difference = currentTimeLong - convertedLong;
 String strDiff = String.valueOf(difference);

从那里数学 currentTimeLong - convertLong。

于 2013-04-23T02:48:06.290 回答