Question

从两个时间选择器获得总小时数的最佳方法是什么？我有两个不同的时间选择器，每次选择器有时间和超时。如果两个时间选择器已经填满，则下面的编码工作正常，但如果只填充一个时间选择器则不起作用。任何建议都会非常感激。

 SimpleDateFormat format = new SimpleDateFormat("HH:mm");
            Date dateb = null; //time in
            Date datec = null; //time out
            Date dateb1 = null; //time in1
            Date datec1 = null; // time out2

 try {
                dateb = format.parse(b);
                datec = format.parse(c);
                long difference = datec.getTime() - dateb.getTime();
                int minutes = (int) ((difference / (1000 * 60)) % 60);
                int hours = (int) ((difference / (1000 * 60 * 60)) % 24) - 1;
                editTextH1.setText((hours + ":" + minutes));
            } catch (Exception e) {

                System.err.println("ouch!");
            }
            try {

                dateb1 = format.parse(d);
                datec1 = format.parse(e1);
                long difference1 = datec1.getTime() - dateb1.getTime();
                int minutes1 = (int) ((difference1 / (1000 * 60)) % 60);
                int hours1 = (int) ((difference1 / (1000 * 60 * 60)) % 24) - 1;
                editTextH2.setText((hours1 + ":" + minutes1));
            } catch (Exception e) {

                System.err.println("ouch!");
            }

 try {

                long dateb_sum = dateb.getTime() + dateb1.getTime();
                long datec_sum = datec.getTime() + datec1.getTime();
                long difference4 = datec_sum - dateb_sum;
                int minutes4 = (int) ((difference4 / (1000 * 60)) % 60);
                int hours4 = (int) ((difference4 / (1000 * 60 * 60)) % 24) - 1;
                editText8.setText((hours4 + ":" + minutes4));
            }catch(Exception e)
            {
                System.err.println("ouch!");
            }

editText8将总结editTextH1和editTextH2的总数...如果这两个时间选择器已填满，一切正常。但是现在我想设置editText8也会获得值，即使填充了一次选择器（当只有一个时间选择器被填充时，editText8 = editTextH1的值）...请帮助..我一直在坚持这里超过两天......

Answer 1

如果移动变量以将差异保持在外部范围内，则不需要尝试让它们添加它们。

SimpleDateFormat format = new SimpleDateFormat("HH:mm");
long difference = 0;
long difference1 = 0;
try {
    Date dateb = format.parse(b); //time in
    Date datec = format.parse(c); //time out
    difference = datec.getTime() - dateb.getTime();
    int minutes = (int) ((difference / (1000 * 60)) % 60);
    int hours = (int) ((difference / (1000 * 60 * 60)) % 24) - 1;
    editTextH1.setText((hours + ":" + minutes));
} catch (Exception e) {
     System.err.println("ouch!");
}
try {
    Date dateb1 = format.parse(d);
    Date datec1 = format.parse(e1);
    long difference1 = datec1.getTime() - dateb1.getTime();
    int minutes1 = (int) ((difference1 / (1000 * 60)) % 60);
    int hours1 = (int) ((difference1 / (1000 * 60 * 60)) % 24) - 1;
    editTextH2.setText((hours1 + ":" + minutes1));
 } catch (Exception e) {
    System.err.println("ouch!");
 }
long difference4 = difference + difference1;
int minutes4 = (int) ((difference4 / (1000 * 60)) % 60);
int hours4 = (int) ((difference4 / (1000 * 60 * 60)) % 24) - 1;
editText8.setText((hours4 + ":" + minutes4));

我也会像How to find the duration of difference between two dates in java?

那样使用TimeUnit.MILLISECONDS.toMinutes(difference)和TimeUnit.MILLISECONDS.toHours(difference)

从两个时间选择器获得整体时间

1 个答案: