Question

我希望在特定的时间段内保持for循环。

我正在使用Thread.Sleep(duration)但这是错误的。请让我知道这个问题的任何解决方案。提前感谢您的建议。如果有任何疑虑，请告诉我。

Answer 1

试试这个：

 new Handler().postDelayed(new Runnable() {
            @Override
            public void run() {
               //write here your code
            }
        }, 5000);

Answer 2

您可以使用 Handler

for (int i = 0; i<50 ;i++) {
    new Handler().postDelayed(new Runnable() {

        @Override
        public void run() {
            // perform your task here
        }
    }, 1000 );// delay time in miniseonds
}

Answer 3

要在特定时间内保留Mat img1 = new Mat(); Mat img2 = new Mat(); Utils.bitmapToMat(bmpimg1, img1); Utils.bitmapToMat(bmpimg2, img2); FeatureDetector detector = FeatureDetector.create(FeatureDetector.ORB); DescriptorExtractor descriptor = DescriptorExtractor.create(DescriptorExtractor.ORB);; DescriptorMatcher matcher = DescriptorMatcher.create(DescriptorMatcher.BRUTEFORCE_HAMMING); // First photo Imgproc.cvtColor(img1, img1, Imgproc.COLOR_RGB2GRAY); Mat descriptors1 = new Mat(); MatOfKeyPoint keypoints1 = new MatOfKeyPoint(); detector.detect(img1, keypoints1); descriptor.compute(img1, keypoints1, descriptors1); // Second photo Imgproc.cvtColor(img2, img2, Imgproc.COLOR_RGB2GRAY); Mat descriptors2 = new Mat(); MatOfKeyPoint keypoints2 = new MatOfKeyPoint(); detector.detect(img2, keypoints2); descriptor.compute(img2, keypoints2, descriptors2); // Matching MatOfDMatch matches = new MatOfDMatch(); MatOfDMatch filteredMatches = new MatOfDMatch(); matcher.match(descriptors1, descriptors2, matches); List<DMatch> matchesList = matches.toList(); Double max_dist = 0.0; Double min_dist = 100.0; for(int i = 0;i < matchesList.size(); i++){ Double dist = (double) matchesList.get(i).distance; if (dist < min_dist) min_dist = dist; if ( dist > max_dist) max_dist = dist; } LinkedList<DMatch> good_matches = new LinkedList<DMatch>(); for(int i = 0;i < matchesList.size(); i++){ Log.d("LOG!",(" : matchesList.get(i).distance :" + matchesList.get(i).distance)); Log.d("LOG!",(" : 1.5 * min_dist :" + min_dist)); if (matchesList.get(i).distance <= 1.5 * min_dist) good_matches.addLast(matchesList.get(i)); } MatOfDMatch goodMatches = new MatOfDMatch(); goodMatches.fromList(good_matches); List<DMatch> bestMatches=goodMatches.toList(); Collections.sort(bestMatches,new Comparator<DMatch>() { @Override public int compare(DMatch o1, DMatch o2) { if(o1.distance<o2.distance) return -1; if(o1.distance>o2.distance) return 1; else { Log.d("LOG!",("Equal i value: ")); return 0; } } }); Log.d("LOG!",("bestMatches.size() : "+bestMatches.size() + " : goodMatches size :" + goodMatches.toList().size()));：只需复制上述代码段，然后尝试更改毫秒 1秒= 1000毫秒

for loot

Answer 4

for (int i = 0; i < arrayList.size(); i++)
                    {
                        View view = listview.getChildAt(i);
                        CheckBox dateChk = view.findViewById(R.id.chkDate);

                        if(dateChk.isChecked())
                            {
                               File f = new File(Environment.getExternalStorageDirectory().getAbsolutePath() + "/" + arrayList.get(i).getfilename());
                                  if (f != null && f.exists())
                                    {
                                       f.delete();
                                    }
                            }
                        }

如何在句点的某个特定时间内保持for循环？

4 个答案: