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Android EditText字符串返回“”

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我正努力在我的Android应用程序中使用EditText的输入 . 我正在输入十进制值,并且我已经处理了小数点,但是当我删除EditText的所有内容时,我的应用程序崩溃了 . 我在我的onTextChanged函数中为EditText小部件检查了NULL或“”的值,但这已被超越 . 以下是我的相关功能:

private void initializeEditTextListener(){
    final TextView resultView = (TextView)findViewById(R.id.resultView);
    final EditText userInput = (EditText)findViewById(R.id.haveTextField);
    userInput.addTextChangedListener(new TextWatcher()
    {
        @Override
        public void afterTextChanged(Editable s) {

        }

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

        }

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {
            String inputString = userInput.getText().toString();
            if(inputString != null && inputString != "")
            {
                double input = Double.parseDouble(inputString); //<-line causing error
                double result = input * getFactorFromTable();   //specified in logcat
                resultView.setText(Double.toString(result));
            }
            else
            {
                resultView.setText("");
            }
        }
    }
    );
}

我的logcat:

10-19 17:53:59.313 16225-16225 / khandy.application.convertible E / AndroidRuntime:FATAL EXCEPTION:main java.lang.NumberFormatException:无效的double:“”在java.lang.StringToReal.invalidReal(StringToReal.java: 63)at java.lang.StringReal.parseDouble(StringToReal.java:248)at java.lang.Double.parseDouble(Double.java:295)at khandy.application.convertible.EntryActivity $ 1.onTextChanged(EntryActivity.java:237)在android.widget.TextView.sendOnTextChanged(TextView.java:7405)的android.widget.TextView.handleTextChanged(TextView.java:7467)android.widget.TextView $ ChangeWatcher.onTextChanged(TextView.java:9218)在android . android.text.SpannableStringBuilder.delete(SpannableStringBuilder.java:212)在android.text.SpannableStringBuilder.delete的android.text.SpannableStringBuilder.replace(SpannableStringBuilder.java:496)上的text.SpannableStringBuilder.sendTextChanged(SpannableStringBuilder.java:962) (spannableStringBuilder.java:30)在android.text.method.BaseKeyListener.backspaceOrForw ardDelete(BaseKeyListener.java:94)位于android.text.method.By.Ber.Berk.D.Mth .onKeyDown(NumberKeyListener.java:138)在Android.widget.TextView.doKeyDown(TextView.java:5544)的android.widget.TextView.onKeyDown(TextView.java:5356)android.view.KeyEvent.dispatch(KeyEvent . java:2705)在android.view.View.DvatchKeyEvent(View.java:7205)的android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1363)android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1363)at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1363)at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1363)at com.android.internal.policy.impl.PhoneWindow $ DecorView.superDispatchKeyEvent(PhoneWindow.java: 2083)在android.app.Activity.dispatchKeyEvent(Activity.java:2416)的com.android.internal.policy.impl.PhoneWindow.superDispatchKeyEvent(PhoneWindow.java:1459) com.android.internal.policy.impl.PhoneWindow $ DecorView.dispatchKeyEvent(PhoneWindow.java:2010)at android.view.ViewRootImpl.deliverKeyEventPostIme(ViewRootImpl.java:3894)at android.view.ViewRootImpl.deliverKeyEvent(ViewRootImpl.java: 3828)在android.view.ViewRootImpl.dePotInputEvents(ViewRootImpl.java:4483)的android.view.ViewRootImpl.deliverInputEvent(ViewRootImpl.java:4483)android.view.ViewRootImpl.enqueueInputEvent(ViewRootImpl.java:4461) . android.app.ActivityThread上的android.os.Handler.dispatchMessage(Handler.java:99)android.O.Roper.loop(Looper.java:137)上的view.ViewRootImpl $ ViewRootHandler.handleMessage(ViewRootImpl.java:3146) .main(ActivityThread.java:4950)at java.lang.reflect.Method.invokeNative(Native Method)at java.lang.reflect.Method.invoke(Method.java:511)at com.android.internal.os.ZygoteInit位于dalvik.system.NativeStart.main的com.android.internal.os.ZygoteInit.main(ZygoteInit.java:771)上的$ MethodAndArgsCaller.run(ZygoteInit.java:1004) OD)

我担心的是,我被告知无效的双重是“”,但是我正在检查这样的事件......

谢谢您的帮助!

3 回答

  • 1
    String inputString = userInput.getText().toString();
    

    这不是在 onTextChanged 中从 EditText 获取文本的首选方法 . 你的 onTextChanged 函数应该是这样的

    @Override
            public void onTextChanged(CharSequence s, int start, int before, int count) {
                String inputString = s.toString();
                if(inputString != null && inputString != "")
                {
                    double input = Double.parseDouble(inputString); //<-line causing error
                    double result = input * getFactorFromTable();   //specified in logcat
                    resultView.setText(Double.toString(result));
                }
                else
                {
                    resultView.setText("");
                }
            }
    

    onTextChanged 函数 CharSequence s 参数只是来自 EditText 的文本

    你应该捕获由空字符串和包含字母的字符串产生的异常

    @Override
            public void onTextChanged(CharSequence s, int start, int before, int count) {
                String inputString = s.toString();
               try
                {
                    double input = Double.parseDouble(inputString); //<-line causing error
                    double result = input * getFactorFromTable();   //specified in logcat
                    resultView.setText(Double.toString(result));
                }
                catch(NumberFormatException e)
                {
                    resultView.setText("");
                }
    
    
               catch(Exception e)
               {
                    resultView.setText("");
               }
      }
    
  • 1
    inputString != ""
    

    应该

    !inputString.equals("")
    

    当你比较数字或布尔值时,你可以进行比较 . 但是当使用 String 时,你需要 equals 来比较字符串 .

  • 1

    For ease create a method to check wether a String is Null or Empty

    创建一个实用程序类(Say AppUtility .java),并定义一个方法 . 所以现在你可以在程序的任何地方使用这个方法 .

    public class AppUtility{
    
    public static boolean isStringEmpty(String input){
            if(input != null && input.length() == 0){
                return true;
            }
            return false;
        }
    }
    

    And check the condition like that

    String inputString = userInput.getText().toString();
                if(AppUtility.isStringEmpty(inputString)){
                // Do YOUR WORK
        }
    

    希望这会帮助你 .

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