所以我有一个listview,我想按降序排序NumberOfRecords。我有一个自定义数组适配器但我在将数据放入ArrayList之前调用了我的排序类,这是我的Asynctask接收JSON:
public class SampleListTask extends AsyncTask<String, Void, String> {
public ProgressDialog pDialog;
@Override
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(SampleActivity.this);
pDialog.setMessage("Loading...");
pDialog.setCancelable(false);
pDialog.show();
}
@Override
protected String doInBackground(String... path) {
Thread.currentThread().setPriority(Thread.MAX_PRIORITY);
Log.d(Constant.TAG_RANKING, path[0]);
String apiRequestReturn = UtilWebService.getRequest(path[0]);
if (apiRequestReturn.equals("")) {
Log.d(Constant.TAG_SAMPLE, "WebService request is null");
return null;
} else {
Log.d(Constant.TAG_SAMPLE, "WebService request has data");
return apiRequestReturn;
}
}
@Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
if (null != pDialog && pDialog.isShowing()) {
pDialog.dismiss();
}
if (null == result || result.length() == 0) {
application.shortToast("No data found from server");
} else {
try {
JSONObject sampleObject = new JSONObject(result);
JSONArray jsonArray = sampleObject
.getJSONArray(Constant.TAG_SAMPLE);
for (int i = 0; i < jsonArray.length(); i++) {
JSONObject objJson = jsonArray.getJSONObject(i);
sample = new ArraySample();
sample.setId(objJson.getInt(Constant.TAG_SONGID));
sample.setThumbUrl(objJson
.getString(Constant.TAG_IMAGEURL));
sample.setTitle(objJson
.getString(Constant.TAG_NAME));
sample.setArtist(objJson
.getString(Constant.TAG_ARTIST));
sample.setDuration(Utility
.changeStringTimeFormat(objJson
.getString(Constant.TAG_MUSICLENGTH)));
sample.setNumberOfRecords(objJson
.getString(Constant.TAG_NUMBEROFRECORDS));
Collections.sort(sampleList, new SortByRecordNumber()); // This where I call the class
sampleList.add(sample);
}
} catch (JSONException e) {
e.printStackTrace();
}
setAdapterToListview();
}
}
public void setAdapterToListview() {
objRowAdapter = new RowAdapterSample(getApplicationContext(),
R.layout.item_sample, sampleList);
sampleListView.setAdapter(objRowAdapter);
}
}
这是我的排序类:
public class SortByRecordNumber implements Comparator {
public int compare(Object o1, Object o2) {
ArraySample p1 = (ArraySample) o1;
ArraySample p2 = (ArraySample) o2;
return p2.getNumberOfRecords().compareTo(p1.getNumberOfRecords());
}
}
但我得到的结果是:
5
15
14
0
0
我的排序实现错了吗?或者我应该在返回之前将其解析为Integer?
答案 0 :(得分:6)
您可以使用以下代码对整数列表进行降序排序。这里我们重写compare(),以便按降序排序。
//sort list in desc order
Collections.sort(myIntegerList, new Comparator<Integer>() {
public int compare(Integer one, Integer other) {
if (one >= other) {
return -1;
} else {
return 1;
}
}
});
希望它有所帮助。
答案 1 :(得分:2)
好的,所以我通过替换:
解决了这个问题 p2.getNumberOfRecords().compareTo(p1.getNumberOfRecords())
为:
(int) Integer.parseInt(p2.getNumberOfRecords()) - Integer.parseInt(p1.getNumberOfRecords())
因此,String数据类型中的整数的简单比较不会正确结果,而是先解析字符串:
Integer.parseInt(string)
并获取数字字符串的真值。
答案 2 :(得分:2)
尝试使用此比较器。
Comparator objComparator = new Comparator() {
public int compare(Object o1, Object o2) {
int no1 = Integer.parseInt((String) o1);
int no2 = Integer.parseInt((String) o2);
return no1 < no2 ? -1 : no1 == no2 ? 0 : 1;
}
};
Collections.sort(myIntegerList, objComparator);