在Android中排序ArrayLists

  • Z时代
  • 2024-01-10
  • 分类：问答

我正在开发一个应用程序。我想排序ArrayList。我面临的问题是，每当我有一个重复的条目，排序失败，应用程序崩溃。在Android中排序ArrayLists

布局截至目前是：

而我希望有结果以下列方式：

我写下面一段代码分类。

private void sortAlphaNumeric() { 
     /* Take a temporary list to store the original list */ 
     /* 
     * Remove any white spaces in the question number as it might 
     * interfere in sorting 
     */ 
     String[] trimmedArray = new String[m_quesNumber.size()]; 
     for (int i = 0; i < m_quesNumber.size(); i++) { 
      trimmedArray[i] = m_quesNumber.get(i).replace(" ", ""); 
     } 
     m_quesNumber = new ArrayList<String>(); 
     for (int i = 0; i < trimmedArray.length; i++) { 
      m_quesNumber.add(trimmedArray[i]); 
     } 
     ArrayList<String> temp_quesNumber = new ArrayList<String>(
       m_quesNumber); 
     /* Sort the original list */ 
     Collections.sort(m_quesNumber, new AlphanumComparator()); 
     /* Temporary arrays */ 
     String[] temp_prevQuesId = new String[m_quesId.size()]; 
     String[] temp_quesMaxMarks = new String[m_quesMaxScore.size()]; 
     Integer[] temp_quesOptional = new Integer[m_quesOptional.size()]; 
     for (int i = 0; i < m_quesNumber.size(); i++) { 
      int old_position = temp_quesNumber.indexOf(m_quesNumber.get(i)); 
      temp_prevQuesId[i] = m_quesId.get(old_position); 
      temp_quesMaxMarks[i] = m_quesMaxScore.get(old_position); 
      temp_quesOptional[i] = m_quesOptional.get(old_position); 
     } 
     m_quesId = new ArrayList<String>(); 
     m_quesMaxScore = new ArrayList<String>(); 
     m_quesOptional = new ArrayList<Integer>(); 
     for (int i = 0; i < temp_prevQuesId.length; i++) { 
      m_quesId.add(temp_prevQuesId[i]); 
      m_quesMaxScore.add(temp_quesMaxMarks[i]); 
      m_quesOptional.add(temp_quesOptional[i]); 
     } 
    }

回答：

只用这个而不是所有的代码。

ArrayList<ArrayList<String>> data; //then initialize it 
// and pass it here. Sort based on column 0 
Collections.sort(data, new Comparator<ArrayList<String>>() { 
    @Override 
    public int compare(ArrayList<String> row1, ArrayList<String> row2) { 
     return row1.get(0).compareTo(row2.get(0)); 
    } 
});

回答：

做一个模型类像...

class Result { 
    String qNo; 
    int marksScored; 
    int maximumMarks; 
    ... 
}

做出的Result的ArrayList像

List<Result> results = new ArrayList<>();

添加所需的数据，之后您可以通过给定的代码排序。

Collections.sort(results, new Comparator<Result>() { 
      @Override 
      public int compare(Result lhs, Result rhs) { 
       return lhs.qNo.compareToIgnoreCase(rhs.qNo); 
      } 
     });

以上是 在Android中排序ArrayLists 的全部内容， 来源链接： utcz.com/qa/264955.html

回到顶部