在层次视图中获取EditBox

Question

我在EditBox中动态地创建了LinearLayout。这是我的代码：

    final LinearLayout mainLayout = (LinearLayout) findViewById(R.id.main_relative);
    counter = 0;
    addButt.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            counter += 1;
            LinearLayout ll = new LinearLayout(view.getContext());
            ll.setOrientation(LinearLayout.HORIZONTAL);
            ll.setTag(CONSTANT_LAYOUT + counter);
            LinearLayout.LayoutParams layoutParams = new
                    LinearLayout.LayoutParams(LinearLayout.LayoutParams.MATCH_PARENT,
                    LinearLayout.LayoutParams.WRAP_CONTENT);
            layoutParams.setMargins(30, 20, 30, 0);


            EditText etProduct = new EditText(view.getContext());
            etProduct.setLayoutParams(new LinearLayout.LayoutParams(LinearLayout.LayoutParams.WRAP_CONTENT,LinearLayout.LayoutParams.WRAP_CONTENT));
            etProduct.setTag(counter);

            ll.addView(etProduct);
            mainLayout.addView(ll);

      }

就像一种魅力。每当我点击特定的按钮，它就会创建新的线性布局，EditBox在那里和计数器上逐个增加，用户可以在_EditBox_es中增加价值。

现在，我想获得_EditBox_es的价值，我尝试了如下：

    FloatingActionButton fab = (FloatingActionButton) findViewById(R.id.fab);
    fab.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {

            for (int i=0 ; i<counter ; i++) {                    

                View lr = mainLayout.findViewWithTag(CONSTANT_LAYOUT+i);
                EditText et = (EditText)lr.findViewWithTag(counter);
            }

        }
    });
}

但我有个错误：

致命异常:主进程: com.myapp，PID: 15055 java.lang.NullPointerException:尝试在空对象引用上调用虚拟方法的android.view.View java.lang.NullPointerException

有什么建议吗？

Answer 1

请尝试以1而不是0启动for loop，并尝试将标记值大于1或equal给1而不是0。

        for (int i=1 ; i<=counter ; i++) {   // and condition should be i<=counter  = instead of less than               

            View lr = mainLayout.findViewWithTag(CONSTANT_LAYOUT+i);
            EditText et = (EditText)lr.findViewWithTag(counter);
        }

Answer 2

你的循环从零开始。

for (int i=0 ; i<counter ; i++) {
   // No value for Zero index.
}

在设置标记时，您要添加+1，所以第一个值是1，而不是0。

counter += 1; // Default value of counter is zero

Answer 3

您已经在根LinearLayout中添加了新的LinearLayout，也可以将setId()添加到新创建的LinearLayout而不是setTag()中，这对您也很好。

LinearLayout ll = new LinearLayout(view.getContext());
            ll.setOrientation(LinearLayout.HORIZONTAL);
            ll.setId(CONSTANT_LAYOUT + counter);
            LinearLayout.LayoutParams layoutParams = new
                    LinearLayout.LayoutParams(LinearLayout.LayoutParams.MATCH_PARENT,
                    LinearLayout.LayoutParams.WRAP_CONTENT);
            layoutParams.setMargins(30, 20, 30, 0);


            EditText etProduct = new EditText(view.getContext());
            etProduct.setLayoutParams(new LinearLayout.LayoutParams(LinearLayout.LayoutParams.WRAP_CONTENT,LinearLayout.LayoutParams.WRAP_CONTENT));
            etProduct.setId(counter);

            ll.addView(etProduct);
            mainLayout.addView(ll);

和

for (int i=0 ; i<counter ; i++) {                    

                LinearLayout ll = (LinearLayout )mainLayout.findViewById(CONSTANT_LAYOUT+i);
                EditText et = (EditText)ll .findViewById(counter);
            }