当用户把数据输入的EditText检测立即显示的答案（闭合）(Detect when user en

它是如何可能的，如果一个字符被输入到一个EditText这使得它从0到1个字符，然后执行一些动作来检测？

Answer 1:

既然你有一个相当抽象的问题，让我尝试同样通用的答案：

在您onCreate()声明并投出您EditText

EditText editText = (EditText) findViewById(R.id.editText);
editText.addTextChangedListener(filterTextWatcher);

然后， 外面 onCreate()设置了filterTextWatcher是这样的：

private TextWatcher filterTextWatcher = new TextWatcher() {

    @Override
    public void onTextChanged(CharSequence s, int start, int before, int count) {
        // DO THE CALCULATIONS HERE AND SHOW THE RESULT AS PER YOUR CALCULATIONS

        int radius = 0;
        radius = Integer.valueof(s.toString);
        double area = Math.PI * radius * radius;
    }

    @Override
    public void beforeTextChanged(CharSequence s, int start, int count, int after) {

    }

    @Override
    public void afterTextChanged(Editable s) {

    }
};

编辑：

更新后的代码可能出现的计算。 （ 未经测试的代码：我刚刚键入它没有测试修改它在必要时）。

了解更多关于TextWatcher在这里

这里有几个例子让你开始：

1. http://www.android-ever.com/2012/06/android-edittext-textwatcher-example.html
2. http://www.cybernetikz.com/blog/android-textwatcher-example/
3. http://www.allappsdevelopers.com/TopicDetail.aspx?TopicID=22b00052-dad0-4e09-a07e-b74f115ab247
4. http://myandroidsolutions.blogspot.in/2012/06/android-edittext-change-listener.html

Answer 2:

您可以使用TextWatcher您的EditText和处理任何你想要立即与你的用户输入数据的事情。

这里是格式化时立即用户输入数据的示例

final EditText EditTxtFinancialCode = (EditText) findViewById(R.id.edtNewCpFinancialCode);


    EditTxtFinancialCode.addTextChangedListener(new TextWatcher() {

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {

            String a = "";
            boolean flag = true;
            String eachBlock[] = EditTxtFinancialCode.getText().toString().split("-");
            for (int i = 0; i < eachBlock.length; i++) {
                if (eachBlock[i].length() > 4) {
                    flag = false;
                }
            }
            if (flag) {

                EditTxtFinancialCode.setOnKeyListener(new OnKeyListener() {

                    @Override
                    public boolean onKey(View v, int keyCode, KeyEvent event) {

                        if (keyCode == KeyEvent.KEYCODE_DEL)
                            KeyDel = 1;
                        return false;
                    }

                });

                if (KeyDel == 0) {

                    if (((EditTxtFinancialCode.getText().length() + 1) % 5) == 0) {

                        if (EditTxtFinancialCode.getText().toString().split("-").length <= 2) {
                            EditTxtFinancialCode.setText(EditTxtFinancialCode.getText() + "-");
                            EditTxtFinancialCode.setSelection(EditTxtFinancialCode.getText().length());
                        }

                    }
                    a = EditTxtFinancialCode.getText().toString();
                } else {
                    a = EditTxtFinancialCode.getText().toString();
                    KeyDel = 0;
                }

            } else {
                EditTxtFinancialCode.setText(a);

            }

        }

        @Override
        public void beforeTextChanged(CharSequence s, int start, int count,
                int after) {
            // TODO Auto-generated method stub

        }
        @Override
        public void afterTextChanged(Editable arg0) {
            // TODO Auto-generated method stub

        }
    });

Answer 3:

http://developer.android.com/reference/android/widget/TextView.OnEditorActionListener.html

使听者。

   myTextBox.addTextChangedListener(new TextWatcher() {

   public void afterTextChanged(Editable s) {
   }

   public void beforeTextChanged(CharSequence s, int start, 
     int count, int after) {
   }

   public void onTextChanged(CharSequence s, int start, 
     int before, int count) {
   TextView myOutputBox = (TextView) findViewById(R.id.myOutputBox);
   myOutputBox.setText(s);
   }
  });

看到http://www.mysamplecode.com/2012/06/android-edittext-text-change-listener.html

文章来源: Detect when user enters data into edittext immediately shows answer [closed]