Question

我有一个输入字段我希望它输入密码直到字符串长度为8，之后输入类型应该更改为文本，这意味着如果输入字段有12个字符，则起始8个字符应该是星号和其他字符4应该是文字。我现在使用的方法现在改变了整个edittext的类型，但我不想这样做。

etEmail.addTextChangedListener(new TextWatcher() {
        @Override
        public void beforeTextChanged(CharSequence charSequence, int i, int i1, int i2) {

        }

        @Override
        public void onTextChanged(CharSequence charSequence, int i, int i1, int i2) {
            email = etEmail.getText().toString();

        }

        @Override
        public void afterTextChanged(Editable editable) {
            email = etEmail.getText().toString();
            if(email.length() > 8){
                etEmail.setTransformationMethod(null);
                etEmail.setSelection(etEmail.length());

            }

        }
    });

如果这是字符串：123456789012 输入字段应显示如下：******** 9012

Answer 1

单个EditText中的可见密码和密码都不可能。

如果您想根据字数计数（当其达到8时）使用 TextWatcher onTextChanged 方法更改整个edittext（可见密码或密码）。

注意：editText可能是密码或可见.cant将单个EditText中的属性组合在一起。但是我们可以使用自定义方式将8以上的字符区分开来

editText.addTextChangedListener(textWatcher);


private final TextWatcher  textWatcher= new TextWatcher() {
    public void beforeTextChanged(CharSequence s, int start, int count, int after) {
    }

    public void onTextChanged(CharSequence s, int start, int before, int count) {
     // if(s.length()==8)
     // setInputType(InputType.TYPE_TEXT_VARIATION_VISIBLE_PASSWORD);

      //or replace some text above char 8
       if(s.length()>=8)
       {
       s = s.replace(s.substring(8,s.length()), "replace some special character");
       editText.setText(s);
       }


    }

    public void afterTextChanged(Editable s) {
    }
};

Answer 2

使用下面我尝试过的代码并按照您的建议获得预期结果。

public class MainActivity extends AppCompatActivity {
EditText et;
TextWatcher myTextWatcher;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    et = findViewById(R.id.et);

    myTextWatcher = new MyTextListener();

    et.addTextChangedListener(myTextWatcher);
}

private void updateEditText(String text) {
    if (text.length() < 9) {
        String converted="";
        for (int a=0; a< text.length(); a++) {
            converted = converted + "*";
        }
        et.removeTextChangedListener(myTextWatcher);
        et.setText(converted);
        et.setSelection(text.length());
        et.addTextChangedListener(myTextWatcher);
    }
}

private class MyTextListener implements TextWatcher {

    @Override
    public void beforeTextChanged(CharSequence charSequence, int i, int i1, int i2) {

    }

    @Override
    public void onTextChanged(CharSequence charSequence, int i, int i1, int i2) {
        updateEditText(charSequence.toString());
    }

    @Override
    public void afterTextChanged(Editable editable) {

    }
}

}

对于xml使用xml代码

<?xml version="1.0" encoding="utf-8"?>

<EditText
    android:id="@+id/et"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:hint="Hello World!"
    app:layout_constraintBottom_toBottomOf="parent"
    app:layout_constraintLeft_toLeftOf="parent"
    app:layout_constraintRight_toRightOf="parent"
    app:layout_constraintTop_toTopOf="parent" />

希望能解决你的问题。如果您发现任何问题必须与我分享。

Answer 3

您需要为空例外和字符长度添加案例，并根据您的要求添加特殊情况，但这是您请求的查询的基本实现

etUsername.addTextChangedListener(new TextWatcher() {
            @Override
            public void beforeTextChanged(CharSequence charSequence, int i, int i1, int i2) {

            }

            @Override
            public void onTextChanged(CharSequence charSequence, int i, int i1, int i2) {
                if (charSequence.length() != 0 && charSequence.length() <= 8  && i < charSequence.length() ) {

                    updatedString = etUsername.getText().toString().substring(0,charSequence.length()).replaceAll("[0-9]", "*");
                    etUsername.removeTextChangedListener(this);
                    etUsername.setText(updatedString);
                    etUsername.setSelection(charSequence.length());
                    etUsername.addTextChangedListener(this);
                }
            }

            @Override
            public void afterTextChanged(Editable editable) {
               /* if (etUsername.length() != 0 ){
                    updatedString = etUsername.getText().toString().substring(0,(etUsername.length() > 8 ? 8 : etUsername.length())).replaceAll("[0-9]", "*");
                    etUsername.removeTextChangedListener(this);
                    etUsername.setText(updatedString);
                    etUsername.setSelection(etUsername.length());
                    etUsername.addTextChangedListener(this);
                }*/

            }
        });

我添加的检查以避免空指针异常

updatedString = etUsername.getText().toString().substring(0,charSequence.length()).replaceAll("[0-9]", "*");

在 replaceAll（字符串正则表达式，字符串替换）中，您可以提供任何正则表达式，而且您可以针对性能进行更多优化我只是为您提供基本概念。

Answer 4

试试这个我为你做了并检查了......它运作正常..

etEmail.addTextChangedListener(new TextWatcher() {
   String originalText = "";
   String newCharacter = "";
   String starCharacter = "";
   @Override
   public void beforeTextChanged(CharSequence s, int start, int count, int after) {
   }
   @Override
   public void onTextChanged(CharSequence s, int start, int before, int count) {
      if(s.length()>0) {
         if(originalText.length()<s.length()) {
             newCharacter = s.toString().substring(s.toString().length() - 1);
         }else{
             newCharacter = "";
         }
       }else{
          newCharacter = "";
          originalText = "";
       }
   }
   @Override
   public void afterTextChanged(Editable s) {
       if(s.length()>0){
           //Lock the listener
           etEmail.removeTextChangedListener(this);
           if(s.length()<=8){
               starCharacter = "";
               for (int i = 0; i <s.length() ; i++) {
                   starCharacter += "*";
               }
               etEmail.setText(starCharacter);
               originalText+=newCharacter;
           }else{
               if(!newCharacter.equals("")) {
                   originalText += newCharacter;
               }else{
                   originalText = originalText.substring(0, originalText.length()-1);
               }

                starCharacter = "********";
                etEmail.setText(starCharacter + originalText.substring(8));
            }
                etEmail.setSelection(etEmail.getText().toString().length());
                //Give back the listener
                etEmail.addTextChangedListener(this);
            }else{
                originalText = "";
            }
      }
});

祝你好运..

更改字符串的某个部分的edittext的输入类型

4 个答案: