应用程序因 NumberFormatException 而崩溃

Question

代码

import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.widget.EditText;

public class MainActivity extends AppCompatActivity {
    EditText a;

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

        a = (EditText) findViewById(R.id.a);

        int i = Integer.parseInt(a.getText().toString());
  }
}

布局

<?xml version="1.0" encoding="utf-8"?>
<android.support.constraint.ConstraintLayout 
xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    tools:context=".MainActivity">


    <EditText
        android:id="@+id/a"
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:layout_marginBottom="8dp"
        android:layout_marginEnd="8dp"
        android:layout_marginStart="8dp"
        android:layout_marginTop="8dp"
        android:ems="10"
        android:inputType="textPersonName"
        app:layout_constraintBottom_toBottomOf="parent"
        app:layout_constraintEnd_toEndOf="parent"
        app:layout_constraintStart_toStartOf="parent"
        app:layout_constraintTop_toTopOf="parent" />
</android.support.constraint.ConstraintLayout>

异常

java.lang.RuntimeException: Unable to start activity ComponentInfo{com.example.shoaib.demo/com.example.shoaib.demo.MainActivity}: java.lang.NumberFormatException: For input string: "" at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2892) at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:3027) at android.app.servertransaction.LaunchActivityItem.execute(LaunchActivityItem.java:78) at android.app.servertransaction.TransactionExecutor.executeCallbacks(TransactionExecutor.java:101) at android.app.servertransaction.TransactionExecutor.execute(TransactionExecutor.java:73) at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1786) at android.os.Handler.dispatchMessage(Handler.java:106) at android.os.Looper.loop(Looper.java:164) at android.app.ActivityThread.main(ActivityThread.java:6656) at java.lang.reflect.Method.invoke(Native Method) at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:438) at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:823) Caused by: java.lang.NumberFormatException: For input string: "" at java.lang.Integer.parseInt(Integer.java:629) at java.lang.Integer.parseInt(Integer.java:652) at com.example.shoaib.demo.MainActivity.onCreate(MainActivity.java:17) at android.app.Activity.performCreate(Activity.java:7117) at android.app.Activity.performCreate(Activity.java:7108) at android.app.Instrumentation.callActivityOnCreate(Instrumentation.java:1262)

Answer 1

您收到此错误的原因是 EditText 尚未在 UI 上创建。

但是，在调用 Integer.parseInt.

之前，您需要屏幕上的 EditText 以及一些有效数据

您需要使用某种接口，通过它您可以调用 Integer.parseInt。我建议使用按钮。

如果您不想添加按钮，请使用此代码：

a.setOnLongClickListener(new View.OnLongClickListener() {
    @Override
    public boolean onLongClick(View v) {
        int i = Integer.parseInt(a.getText().toString());
        //any operation involving i
        return true;
    }
});

然后，在 a 中输入一些数据后，长按 a，看看魔术！

Answer 2

问题是您正在尝试将非 int（“ ” - 因为 edittext 可能为空）字符转换为 int 字符。您可以尝试下面的代码；

String input = a.getText().toString();

if (TextUtils.isDigitsOnly(input)) {
     i = Integer.parseInt(input);
     // do everything you want with your int here
}

应用程序因 NumberFormatException 而崩溃

application crashed with NumberFormatException

android

runtimeexception

numberformatexception