How to delete instantly SPACE from an edittext if a user presses the space?

问题

I have an edittext, and a textwatcher that watches if SPACE arrived or not. If its a SPACE I would like to delete that instantly. Or if its a space I want to make sure it doesnt appear but indicate somehow (seterror, toast) for the user that space is not allowed.

edittext.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) {}
        });

I cannot define onkeydown in the afterTextChaned method, since it gives me an error.

 public boolean onKeyDown(int keyCode, KeyEvent event) {
                super.onKeyDown(keyCode, event);

                if (keyCode == KeyEvent.KEYCODE_SPACE) {

                }
    }

So it is not working (syntax error, misplaced construct for the int keyCode.

Thanks you in advance!

回答1:

The solution is as usually much simpler:

@Override
public void afterTextChanged(Editable s) {
    String result = s.toString().replaceAll(" ", "");
    if (!s.toString().equals(result)) {
         ed.setText(result);
         ed.setSelection(result.length());
         // alert the user
    }
}

This shouldn't have the problems of the previous attempts.

回答2:

setSelection is there to set the cursor again at the end of your EditText:

editText.addTextChangedListener(new TextWatcher() {

                @Override
                public void onTextChanged(CharSequence cs, int arg1, int arg2,
                        int arg3) {}
                @Override
                public void beforeTextChanged(CharSequence s, int arg1, int arg2,
                        int arg3) {}
                @Override
                public void afterTextChanged(Editable arg0) {
                    if(editText.getText().toString().contains(" ")){ editText.setText(editText.getText().toString().replaceAll(" " , ""));
                    editText.setSelection(editText.getText().length());

                    Toast.makeText(getApplicationContext(), "No Spaces Allowed", Toast.LENGTH_LONG).show();
                    }
                }});

回答3:

boolean editclicked =false ;

edittext.addTextChangedListener(new TextWatcher() {
    public void afterTextChanged(Editable s) {
        editclicked = false ;
    }

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

    public void onTextChanged(CharSequence s, int start, int before, int count) {
        editclicked = true;
    });

Put this as a separate function:

public boolean onKeyDown(int keyCode, KeyEvent event) {
    if (editclicked) {
        if (keyCode == KeyEvent.KEYCODE_SPACE) {
            return false
        }
    } else {
        super.onKeyDown(keyCode, event);
    }
}

回答4:

@Override
public void afterTextChanged(Editable s) {
    String result = s.toString().replaceAll("\\s", "");
    if (!s.toString().equals(result)) {
        int pos = editText.getSelectionStart() - (s.length() - result.length());
        editText.setText(result);
        editText.setSelection(Math.max(0,Math.min(pos, result.length())));
        editText.setError("No spaces allowed");
    }
}

\s matches any whitespace character (equal to [\r\n\t\f\v ])

Setting selection like this, allow you to enter or paste text in middle of edittext without loosing cursor position

回答5:

For removing the space instantly you can achieve it by two ways.

One simple solution you can set the digits to your edit text.

android:digits="ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789"

second way you can set a filter

EditText.setFilters(new InputFilter[] { filter });


InputFilter filter = new InputFilter() {
 public CharSequence filter(CharSequence source, int start, int end,
   Spanned dest, int dstart, int dend) {
  for (int i = start; i < end; i++) {
   if (Character.isSpaceChar(source.charAt(i))) {
    return "";
   }
  }
  return null;


      }
     }

回答6:

One more simple way to achieve this using the input Filter

editText.setFilters(new InputFilter[]{new InputFilter() {
        @Override
        public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
            if (source.toString().equalsIgnoreCase(" ")){
                return "";
            }
            return source;
        }
    }});

This will remove the space entered by the user immediately and gives appearance like space is disabled.

回答7:

My relatively simple solution for instant whitespace deletion without removing spannables (styles) in EditText:

Remove at start:

    @Override
public void afterTextChanged(Editable s) {
    int i;
    for (i = 0; i < s.length() && Character.isWhitespace(s.charAt(i)); i++) { ; }
    s.replace(0, i, "");
}

Basically that's it, but you can also do:

Remove at start (without interrupting first input):

    @Override
public void afterTextChanged(Editable s) {
    String text = s.toString();
    if(!text.trim().isEmpty()){
        int i;
        for (i = 0; i < s.length() && Character.isWhitespace(s.charAt(i)); i++) { ; }
        s.replace(0, i, "");
    }
}

Removing at start and end (allow 1 whitespace at end for convinient input):

    @Override
public void afterTextChanged(Editable s) {
    int i;
    //remove at start
    for (i = 0; i < s.length() && Character.isWhitespace(s.charAt(i)); i++) { ; }
    s.replace(0, i, "");
    //remove at end, but allow one whitespace character
    for (i = s.length(); i > 1 && Character.isWhitespace(s.charAt(i-1)) && Character.isWhitespace(s.charAt(i-2)); i--) { ; }
    s.replace(i, s.length(), "");
}

来源：https://stackoverflow.com/questions/9757991/how-to-delete-instantly-space-from-an-edittext-if-a-user-presses-the-space

标签

android

android-edittext

textwatcher