How to default an EditText to integer but allow de

问题:

I am using an EditText to allow a user to input a value that is stored in a Double.

However, by default, Doubles look like "0.0" and it's a little annoying for a user to backspace over the extra decimal if it's not used. Is there a way to force-display whole numbers to look like "0" and only show the decimal if the user actually decides to use it?

Current code:

        myEditText = (EditText) view.findViewById(R.id.my_edittext);
        myEditText.setText(myVariable + "");
        myEditText.addTextChangedListener(new TextWatcher() {
            @Override
            public void beforeTextChanged(CharSequence s, int start, int count, int after) {

            }

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

            }

            @Override
            public void afterTextChanged(Editable s) {
                String temp = s.toString();
                if (s.length() > 0){
                    if (OtherMethods.isDouble(temp)) {
                        myVariable = Double.parseDouble(temp);
                    }
                    else {
                        myVariable = 0.0;
                    }
                }
                else {
                    myVariable = 0.0;
                }
            }
        });

The XML:

<EditText
    android:id="@+id/my_edittext"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    android:inputType="numberDecimal"
    android:ems="10"
    android:hint="Input Value"
    android:imeOptions="actionDone"/>

回答1:

To achieve this you can use NumberFormat

    EditText yourEditText = (EditText) findViewById(R.id.editTextID);
    //dummy data, will have user's value
    double aDouble = 4.0;

    //formats to show decimal
    NumberFormat formatter = new DecimalFormat("#0");

    //this will show "4"
    yourEditText.setText(formatter.format(aDouble));

Make sure to validate the user's input. Also, this will only modify what is displayed and not the value itself.

回答2:

1. Parsing Double to String, then String to Int:

   String stringparsed = YourDouble + "";
   int intparsed = Integer.parseInt(stringparsed);
   

2. Using substring to cut the string from a startIndex to finalIndex:

   String stringparsed = YourDouble + "";
   String final = stringparsed.substring(0,1);   //for example, if the double was    0.0, the result is 0