如何重構android代碼以選擇佈局？

Question

我有1個頂部佈局和3個底部佈局，用戶從頂部佈局拖動項目到下面3個之一。

下面的代碼：

• 依次把每個圖像中的每個底部佈局成一個列表
• 移除從每個底部佈局

• 把它們放回頂部佈局

  項

  public void onClick(View view) { 
      LinearLayout bottomLinearLayout1 = (LinearLayout) findViewById(R.id.bottom1); 
      LinearLayout bottomLinearLayout2 = (LinearLayout) findViewById(R.id.bottom2); 
      LinearLayout bottomLinearLayout3 = (LinearLayout) findViewById(R.id.bottom3); 
      LinearLayout topLinearLayout = (LinearLayout) findViewById(R.id.topLinear); 
  
      int total_num1 = bottomLinearLayout1.getChildCount(); 
      int total_num2 = bottomLinearLayout2.getChildCount(); 
      int total_num3 = bottomLinearLayout3.getChildCount(); 
  
      View current_image = null; 
      List<View> listOfkids = new ArrayList<>() ; 
  
      //************ REPEATS ************** 
  
      for(int i = 0 ; i < total_num1 ; i++){ 
       current_image = bottomLinearLayout1.getChildAt(i); 
       listOfkids.add(current_image); 
      } 
      bottomLinearLayout1.removeAllViews(); 
  
      for(int i = 0 ; i < listOfkids.size();i++){ 
       topLinearLayout.addView(listOfkids.get(i)); 
      } 
      listOfkids.clear(); 
  
      //************ REPEATS ************** 
  
      for(int i = 0 ; i < total_num2 ; i++){ 
       current_image = bottomLinearLayout2.getChildAt(i); 
       listOfkids.add(current_image); 
      } 
      bottomLinearLayout2.removeAllViews(); 
      for(int i = 0 ; i < listOfkids.size();i++){ 
       topLinearLayout.addView(listOfkids.get(i)); 
      } 
      listOfkids.clear(); 
  
      //************ REPEATS ************** 
  
      for(int i = 0 ; i < total_num3 ; i++){ 
       current_image = bottomLinearLayout3.getChildAt(i); 
       listOfkids.add(current_image); 
      } 
      bottomLinearLayout3.removeAllViews(); 
      for(int i = 0 ; i < listOfkids.size();i++){ 
       topLinearLayout.addView(listOfkids.get(i)); 
      } 
  

基本上，這些循環之間的唯一區別是「bottomLinearLayout」中的最後一位數字;其他代碼只是複製自己！

我能做到這一點：

if(id == R.id.bottom1){ 
String current_layout = "bottomLinearLayout" + 1 ; 
} 
else if(id == R.id.bottom2){ 
current_layout = "bottomLinearLayout" + 2 ; 
} 
else if(id == R.id.bottom3){ 
current_layout = "bottomLinearLayout" + 3 ; 
} 

，然後添加此字符串作爲命令右轉入Java源代碼？

這是可能的嗎？

Answer 1

假設要重構onClick方法，您可以創建佈局ID的數組，並遍歷它們如下：

int bottomLayoutIds[]  = {R.id.bottom1, R.id.bottom2, R.id.bottom3}; 
LinearLayout topLinearLayout = (LinearLayout) findViewById(R.id.topLinear); 

for (int layoutId : bottomLayoutIds){ 
    LinearLayout bottomLinearLayout = (LinearLayout) findViewById(layoutId); 
    int childCount = bottomLinearLayout.getChildCount(); 
    List<View> listOfKids = new ArrayList<>() ; 
    for(int i = 0 ; i < childCount ; i++){ 
     View currentImage = bottomLinearLayout.getChildAt(i); 
     listOfKids.add(currentImage); 
    } 
    bottomLinearLayout.removeAllViews(); 

    for(int i = 0 ; i < listOfKids.size();i++){ 
     topLinearLayout.addView(listOfKids.get(i)); 
    } 
}