[英]AlertDialog Force Close with EditText
我得到了錯誤:
E/AndroidRuntime: FATAL EXCEPTION: main
Process: com.crmdev.circuitorlcparalelo, PID: 6778
java.lang.NullPointerException: Attempt to invoke virtual method 'int java.lang.Integer.intValue()' on a null object reference
at com.crmdev.circuitorlcparalelo.MainActivity$3.onClick(MainActivity.java:70)
at android.support.v7.app.AlertController$ButtonHandler.handleMessage(AlertController.java:166)
at android.os.Handler.dispatchMessage(Handler.java:106)
at android.os.Looper.loop(Looper.java:164)
at android.app.ActivityThread.main(ActivityThread.java:6626)
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:811)
對於帶有AlertDialog的以下代碼:
AlertDialog.Builder result = new AlertDialog.Builder(MainActivity.this);
result.setTitle("Iterações");
result.setMessage("Digite o número de iterações desejadas: ");
final EditText iteracoes = new EditText(this);
iteracoes.setInputType(InputType.TYPE_CLASS_NUMBER);
iteracoes.setGravity(9);
iteracoes.setPadding(10,10,10,10);
result.setView(iteracoes);
result.setPositiveButton("Calcular", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
if(iteracoes.getText().equals("")){
Toast.makeText(getApplicationContext(), "Digite algum valor para as iterações!", Toast.LENGTH_LONG).show();
iteracoes.setText("");
}
else{
//Pega o numero de iterações
iteracao = Integer.getInteger(iteracoes.getText().toString());
tensaof = onCalculo(iteracao);
Bundle dad = new Bundle();
dad.putDouble("resultado", tensaof);
Intent intent = new Intent(MainActivity.this, ResultActivity.class);
intent.putExtras(dad);
startActivity(intent);
}
}
});
result.show();
每次我按一下Calcular申請人的“強制關閉”按鈕時,我都會更改onCalculo函數,但會以相同的方式繼續。 當函數onCalculo僅更改TextView時 ,應用程序沒有此問題。 我看不到出了什么問題。 謝謝。
因為您使用錯誤的方法將String
解析為Integer
。 你應該
iteracao = Integer.parseInt(iteracoes.getText().toString());
代替
iteracao = Integer.getInteger(iteracoes.getText().toString());
順便說一句, parseInt
方法可能會引發NumberFormatException
因此您應該放入try/catch
塊中,以防止應用程序崩潰。
Integer iteracao = null;
try {
iteracao = Integer.parseInt(iteracoes.getText().toString());
} catch (NumberFormatException e) {
// TODO: Show a toast to let users know the input value is not valid
}
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