Arraylist仅检索最后一个值

时间:2011-04-01 08:46:08

标签: java android arraylist parcelable

您好我正在使用Parcelable将一个arraylist从一个活动传递到另一个活动..

现在的问题是当我点击BILL时,所有行都填充了arraylist的最后一个值。所以我想我无法在调用活动中检索整个arraylist ......

以下是相关代码..

SpinPizza.java

public class SpinPizza extends Activity{
store temp= new store();

     int i=0;

ArrayList<store> B = new ArrayList<store>();


//oncreate method



  int n=Integer.parseInt(edittext.getText().toString());

     temp.setOrder(s.getSelectedItem().toString(), s1.getSelectedItem().toString(),n);

          B.add(temp);

             TextView objText=(TextView) findViewById(R.id.pl);

              TextView objText1=(TextView) findViewById(R.id.pl2);

              objText.setText(B.get(i).getPizzaName());

             objText1.setText(temp.getPizzaSize());

                 i++;




    });

      Button next1 = (Button) findViewById(R.id.bill);    

      next1.setOnClickListener(new View.OnClickListener() {

          public void onClick(View view) {

              Intent myIntent = new Intent(view.getContext(), Bill.class);

                  myIntent.putParcelableArrayListExtra("myclass",B);

              startActivityForResult(myIntent, 0);

          }

      });

}

Bill.java

public class Bill extends Activity {
 @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);

this.setContentView(R.layout.calc);
ArrayList<store> B1 = new ArrayList<store>();

store temp1=new store();

Bundle bj=getIntent().getExtras();


    B1=bj.getParcelableArrayList("myclass");

  TableLayout tl = (TableLayout)findViewById(R.id.myTable);



                      for(int i=0;i<B1.size();i++)
        { 
             TableRow tr = new TableRow(this);
             tr.setId(100+i);
            temp1=B1.get(i);         
             tr.setLayoutParams(new LayoutParams(
                            LayoutParams.FILL_PARENT,
                            LayoutParams.WRAP_CONTENT));



            TextView b = new TextView(this);
        b.setText(temp1.getPizzaName()); 
        b.setId(200+i);

        TextView b1 = new TextView(this);
        b1.setText(temp1.getPizzaSize());
        b1.setId(300+i);

       TextView b2 = new TextView(this);
       b2.setText(String.valueOf(temp1.getQuantity()));
       b2.setId(400+i);


        b.setLayoutParams(new LayoutParams(
                    LayoutParams.FILL_PARENT,
                    LayoutParams.WRAP_CONTENT));
        tr.addView(b);

        b1.setLayoutParams(new LayoutParams(
                LayoutParams.FILL_PARENT,
                LayoutParams.WRAP_CONTENT));
        tr.addView(b1);



        b2.setLayoutParams(new LayoutParams(
                LayoutParams.WRAP_CONTENT,
                LayoutParams.WRAP_CONTENT));

        tr.addView(b2);



tl.addView(tr,new TableLayout.LayoutParams(
          LayoutParams.FILL_PARENT,
          LayoutParams.WRAP_CONTENT));


 }}

}

请帮助。

1 个答案:

答案 0 :(得分:1)

我无法看到temp的创建,也无法在temp修饰符行temp.setOrder(s.getSelectedItem().toString(), s1.getSelectedItem().toString(),n);中看到使用的迭代。您似乎没有使用i,这就是您有相同记录的原因。