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我需要对 EditText 的输入设置一些限制。只允许用户 输入a-z、、、A-Z和。0-9semicolon ;chinese letters

我的布局很简单,一个edittext和一个Button,如果edittext输入符合要求,按钮就会显示并触发事件。

我写了下面的代码,但其中有一些错误,有些功能无法实现,所以我需要有人的帮助,任何帮助将非常感谢。

public class MainActivity extends Activity {
        private static String tag = "MainActivity";
        private Button btn;
        private EditText edt, content;
        private final int[] code = { 8, 13, 32, 48, 49, 50, 51, 52, 53, 54, 55, 56,
                        57, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80,
                        81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 186 };
        private String digits = "0123456789abcdefghijklmnopqrstuvwxyz;";
        private String tmp;
        @Override
        public void onCreate(Bundle savedInstanceState) {
                super.onCreate(savedInstanceState);
                setContentView(R.layout.activity_main);
                btn = (Button) findViewById(R.id.btn_edit);
                btn.setEnabled(false);
                edt = (EditText) findViewById(R.id.edt);                
                edt.addTextChangedListener(edt_watcher);
                // edt.setOnKeyListener(input);
                btn.setOnClickListener(l);        
        }

//My first try:Keyboard to monitor events,Monitoring whether the keyCode is in the button collection of the permit
//The issue is once i push something not in collection,the input content being refreshed
        OnKeyListener input = new OnKeyListener() {
                @Override
                public boolean onKey(View v, int keyCode, KeyEvent event) {
                        // TODO Auto-generated method stub
                        for (int i = 0; i < code.length; i++) {
                                if (keyCode != code) {
                                        edt.setText(tmp);
                                        edt.invalidate();
                                }
                        }
                        return false;
                }
        };
//my second try:Text input to monitor,after the listen i get edt.gettext then do splicing processing,after that i set back to edittext.

//issue:everytime input legal characters will add to the left side.the cursor always keep in left side,any suggestions here? 
        // Text input to monitor
        TextWatcher edt_watcher = new TextWatcher() {

                @Override
                public void onTextChanged(CharSequence s, int start, int before,
                                int count) {
                        String str = s.toString();
                        String edttext = edt.getText().toString();
                        StringBuffer sb = new StringBuffer();
                        // TODO Auto-generated method stub
                        if (s.length() > 0 && str.matches("[a-zA-Z_0-9;]+")
                                        || str.matches("[\u4e00-\u9fa5]+")) {
                                sb = sb.append(s);
                                edt.setText(sb.toString());
                                Log.v(tag, sb.toString()+"======sb");
                                btn.setEnabled(true);
                        } else {
                                edttext = sb.append(tmp).toString();
                                Log.v(tag, edttext + "======edttext");
                                edt.setText(edttext);
                                tmp = str.substring(0,s.toString().length()-2);
                                edt.setText(tmp);
                                btn.setEnabled(false);
                        }        
                        Log.v(tag, s + "======s");
                }

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

                        Log.v(tag, s + "======beforeTextChanged");
                }


//My third try:as comment blow,I don't know how to achieve do all the chinese letters contain in digits
                @Override
                public void afterTextChanged(Editable s) {
                        // TODO Auto-generated method stub
                        Log.v(tag, s + "======afterTextChanged");
                        /*String str = s.toString();
                        if (str.equals(tmp)) {
                                return;
                                // if `tmp==str` return,as i set the result like this,else will be endless loop.
                        }
                        StringBuffer sb = new StringBuffer();
                        for (int i = 0; i < str.length(); i++) {
                                if (digits.indexOf(str.charAt(i)) >= 0) {
                                        // judge input characters allowed or not 
                                        sb.append(str.charAt(i));
                                        //if allowed,add to the result ,else skip it
                                }
                                tmp = sb.toString();// set tmp,next line also can trggier it
                                edt.setText(tmp);// set result
                                edt.invalidate();
                        }
                        if ((str.matches("[a-zA-Z_0-9;]+") || str
                                        .matches("[\u4e00-\u9fa5]+"))) {
                                sb.toString().substring(str.length());
                        }
                        tmp = sb.toString();
                        edt.setText(tmp);
                        edt.invalidate();*/
                }
        };
        //judge if input is Chinese characters,the method not be used
         private boolean isChinese(char c) {
         Character.UnicodeBlock ub = Character.UnicodeBlock.of(c);
         if (ub == Character.UnicodeBlock.CJK_UNIFIED_IDEOGRAPHS
         || ub == Character.UnicodeBlock.CJK_COMPATIBILITY_IDEOGRAPHS
         || ub == Character.UnicodeBlock.CJK_UNIFIED_IDEOGRAPHS_EXTENSION_A
         || ub == Character.UnicodeBlock.GENERAL_PUNCTUATION
         || ub == Character.UnicodeBlock.CJK_SYMBOLS_AND_PUNCTUATION
         || ub == Character.UnicodeBlock.HALFWIDTH_AND_FULLWIDTH_FORMS) {
                 String edttext = edt.getText().toString();
                 StringBuffer sb = new StringBuffer(edttext);
                 sb.append(c);
                 edt.setText(sb);
         return true;
         }
         return false;
         }
        ///judge if input is number or letters or ;,the method not be used
         private boolean isRightData(CharSequence s) {
         if (s.toString().matches("[0-9;]+")
         || s.toString().matches("[a-zA-Z]+")) {
                 String edttext = edt.getText().toString();
                 StringBuffer sb = new StringBuffer(edttext);
                 sb.append(s);
                 edt.setText(sb);
                 edt.invalidate();
         return true;
         }
         return false;
         }

        OnClickListener l = new OnClickListener() {

                @Override
                public void onClick(View v) {
                        // TODO Auto-generated method stub
                        Intent intent = new Intent(MainActivity.this, IldmActivity.class);
                        intent.putExtra("edt", edt.getText().toString());
                        startActivityForResult(intent, 10);
                        Log.v(tag, edt.getText() + "intent");
                }
        };
        @Override
        public boolean onCreateOptionsMenu(Menu menu) {
                // Inflate the menu; this adds items to the action bar if it is present.
                getMenuInflater().inflate(R.menu.main, menu);
                return true;
        }
        public void onActivityResult(int requestCode, int resultCode, Intent data) {
                if (requestCode == 10 && resultCode == RESULT_OK) {
                }
        };
}
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1 回答 1

1

波纹管代码仅接受 0 到 9 位数字,因此可以解决您的问题。

public static void setPhoneNumberFilter(final EditText edt, final int length) { InputFilter[] filters = new InputFilter[1];

    final int len = edt.getText().toString().length();
    Log.e("MIS", "" + edt.getText().toString());
    Log.e("MIS", "" + edt.getText().toString().length());
    if (len > length - 1) {
        return;
    }
    filters[0] = new InputFilter() {

        public CharSequence filter(CharSequence source, int start, int end,
                Spanned dest, int dstart, int dend) {
            // TODO Auto-generated method stub

            try {
                char[] acceptedChars = new char[] { '0', '1', '2', '3',
                        '4', '5', '6', '7', '8', '9' };
                for (int index = 0; index < end; index++) {

                    if (!new String(acceptedChars).contains(String
                            .valueOf(source.charAt(index)))
                            || edt.getText().toString().length() > length - 1) {
                        return "";
                    }
                }

            } catch (Exception e) {
                e.printStackTrace();

            }

            return null;
        }

    };

    edt.setFilters(filters);
}

setPhoneNumberFilter(edtxt,10);

于 2013-05-15T06:41:13.207 回答