将EditText密码掩码字符更改为星号(*)

时间:2012-12-27 09:00:04

标签: android android-edittext

有没有办法将密码文本从点(。)更改为星号(*)。

密码以edittext格式输入。

<EditText
        android:id="@+id/passWord1"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:gravity="center"
        android:inputType="number"
        android:password="true"/>

7 个答案:

答案 0 :(得分:55)

在xml文件中插入edittext,

<EditText
    android:id="@+id/passWordEditText"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:gravity="center"
    android:inputType="textPassword"/>

然后你的类文件继续从edittext获取findViewById并为此实现,

EditText edittext = (EditText)findViewById(R.id.passWordEditText);
edittext.setTransformationMethod(new AsteriskPasswordTransformationMethod());

和这个类的实现,

public class AsteriskPasswordTransformationMethod extends PasswordTransformationMethod {
    @Override
    public CharSequence getTransformation(CharSequence source, View view) {
        return new PasswordCharSequence(source);
    }

    private class PasswordCharSequence implements CharSequence {
        private CharSequence mSource;
        public PasswordCharSequence(CharSequence source) {
            mSource = source; // Store char sequence
        }
        public char charAt(int index) {
            return '*'; // This is the important part
        }
        public int length() {
            return mSource.length(); // Return default
        }
        public CharSequence subSequence(int start, int end) {
            return mSource.subSequence(start, end); // Return default
        }
    }
};

答案 1 :(得分:19)

<EditText
        android:id="@+id/passWord1"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:gravity="center"
        android:inputType="textPassword"//here is the change. check it once in your xml
        android:password="true"/>

在eclipse中,当您将光标移动到android:inputType时,单击 Ctrl + Space 时会有提示。然后你可以看到选项列表。在那里你可以选择textPassword

如果您希望*代替.,请查看此Android: Asterisk Password Field

答案 2 :(得分:8)

在Ram kiran发布的链接的帮助下得到答案

text.setTransformationMethod(new AsteriskPasswordTransformationMethod());


public class AsteriskPasswordTransformationMethod extends PasswordTransformationMethod {
@Override
public CharSequence getTransformation(CharSequence source, View view) {
    return new PasswordCharSequence(source);
}

private class PasswordCharSequence implements CharSequence {
    private CharSequence mSource;
    public PasswordCharSequence(CharSequence source) {
        mSource = source; // Store char sequence
    }
    public char charAt(int index) {
        return '*'; // This is the important part
    }
    public int length() {
        return mSource.length(); // Return default
    }
    public CharSequence subSequence(int start, int end) {
        return mSource.subSequence(start, end); // Return default
    }
}
};

答案 3 :(得分:0)

对于android:inputType,有一种密码。

答案 4 :(得分:0)

试试这个

android:inputType="textPassword"       

答案 5 :(得分:0)

以科特林方式:

AsteriskPasswordTransformationMethod类:PasswordTransformationMethod(){

override fun getTransformation(source: CharSequence?, view: View?): CharSequence {
    return super.getTransformation(source, view)
}


 abstract inner class PasswordCharSequence(val source: CharSequence) : CharSequence {
     override val length: Int
         get() = source.length

     override fun get(index: Int): Char =
         '*'

    override fun subSequence(start: Int, end: Int): CharSequence {
        return source.subSequence(start, end) // Return default
    }
}

}`

答案 6 :(得分:0)

科特林

class AsteriskPasswordTransformationMethod : PasswordTransformationMethod() {

    override fun getTransformation(source: CharSequence, view: View): CharSequence {
        return PasswordCharSequence(source)
    }

    inner class PasswordCharSequence (private val source: CharSequence) : CharSequence {

        override val length: Int
            get() = source.length

        override fun get(index: Int): Char = '*'

        override fun subSequence(startIndex: Int, endIndex: Int): CharSequence {
            return source.subSequence(startIndex, endIndex)
        }
        
    }
    
}