int a() {
Date start_date = null;
Date end_date = null;
try { DateFormat df = new SimpleDateFormat("yyyy-MM-dd hh:mm");
start_date = df.parse("2012-01-11 11:10");
end_date = df.parse("2012-01-11 12:08");
Log.v("a", "start" + start_date.getTime());
Log.v("a", "end " + end_date.getTime());
}
catch(Exception e) {
return -1;
}
return (int) ((end_date.getTime() - start_date.getTime()) / (1000 * 60));如果时间在中午11点和12点之间就会有问题 他返回的时差不对 其他的时间段暂时未发现问题
当前代码返回值是-662
求解
Date start_date = null;
Date end_date = null;
try { DateFormat df = new SimpleDateFormat("yyyy-MM-dd hh:mm");
start_date = df.parse("2012-01-11 11:10");
end_date = df.parse("2012-01-11 12:08");
Log.v("a", "start" + start_date.getTime());
Log.v("a", "end " + end_date.getTime());
}
catch(Exception e) {
return -1;
}
return (int) ((end_date.getTime() - start_date.getTime()) / (1000 * 60));如果时间在中午11点和12点之间就会有问题 他返回的时差不对 其他的时间段暂时未发现问题
当前代码返回值是-662
求解
我还发现另一种方法
int a() {
Date start_date = null;
Date end_date = null;
try {
start_date = Timestamp.valueOf(year.split(" ")[0] + " " + hour+ ":00");
end_date = Timestamp.valueOf(yearr.split(" ")[0] + " " + hourr+ ":00");
}
catch(Exception e) {
return -1;
}
return (int) ((end_date.getTime() - start_date.getTime()) / (1000 * 60));
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为什么问个问题都没几个人回答