安卓应用程序在onClickListener上崩溃
安卓应用程序在onClickListener上崩溃
提问于 2018-07-10 06:07:52
我有一个应用程序，这是一个简单的彩票系统，但每当我尝试按下其中一个按钮，应用程序冻结，然后崩溃，我不知道为什么。这个程序不会给我错误，所以它真的很令人困惑。我看到了一些关于布局的问题，但我不知道这是什么。我将在问题下的评论中回答页面布局。提前谢谢。
下面是页面脚本：
package com.example.myapplication;
import android.graphics.Color;
import android.icu.text.NumberFormat;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import java.util.*;
import java.math.*;
import org.w3c.dom.Text;
import javax.xml.transform.Result;
public class Lottery extends AppCompatActivity {
Button Check, Random;
EditText User1;
TextView Balance2, ResultText;
int Balance1 = 500;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_lottery);
Check = (Button)findViewById(R.id.CheckNumbers);
Random = (Button)findViewById(R.id.RandomNumber);
User1 = findViewById(R.id.PickNumber);
Balance2 = (TextView)findViewById(R.id.Balance);
ResultText = (TextView)findViewById(R.id.Result);
ResultText.setVisibility(View.GONE);
Balance2.setText("Balance: $500");
Check.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
int User=Integer.parseInt(String.valueOf(User1));
if(User1.toString().trim().equals("")){
ResultText.setBackgroundColor(Color.RED);
ResultText.setText("Please Enter a Number");
ResultText.setVisibility(View.VISIBLE);
}
else if(User > 100 || User < 1){
ResultText.setBackgroundColor(Color.RED);
ResultText.setText("Please Enter a Number between 1 and 100");
ResultText.setVisibility(View.VISIBLE);
}
else {
int answer = (int)((Math.random()*100)+1);
int placeholder = answer % 10;
int AnswerTens = (answer - placeholder)/10;
int AnswerOnes = placeholder;
placeholder = User %10;
int UserTens = (User- placeholder)/10;
int UserOnes = placeholder;
if(User == answer){
ResultText.setBackgroundColor(Color.GREEN);
ResultText.setText("Correct!!! The Number was "+User+" $1000 has been added to your account");
ResultText.setVisibility(View.VISIBLE);
Balance1 += 1000;
Balance2.setText("Balance: $ "+ Balance1);
}
else if((UserTens == AnswerOnes)&&(UserOnes == AnswerTens)){
ResultText.setBackgroundColor(Color.GREEN);
ResultText.setText(" Somewhat Correct! The digit were correct but in the wrong order The answer was "+answer+" $500 has been added to your account");
ResultText.setVisibility(View.VISIBLE);
Balance1 += 500;
Balance2.setText("Balance: $ "+ Balance1);
}
else if((UserTens == AnswerTens)|| (UserTens == AnswerOnes)||(UserOnes == AnswerOnes)||(UserOnes==AnswerTens)){
ResultText.setBackgroundColor(Color.GREEN);
ResultText.setText("Kinda Correct! One digit was correct The answer was "+answer+" $100 has been added to your account");
ResultText.setVisibility(View.VISIBLE);
Balance1 += 100;
Balance2.setText("Balance: $ "+ Balance1);
}
else{
ResultText.setBackgroundColor(Color.RED);
ResultText.setText("Incorrect the Number was "+answer);
ResultText.setVisibility(View.VISIBLE);
}
}
}
});
Random.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
int answer = (int)((Math.random()*100)+1);
User1.setText(answer);
}
});
}
}回答 2
Stack Overflow用户
回答已采纳
发布于 2018-07-10 08:38:59
我看到了导致崩溃的几个可能原因。正如@Siros提到的，您需要在调用setText时使用String.valueOf(answer)
此外，要从EditText中获取字符串，必须调用User1.getText().toString()。调用User1.toString()将返回类似android.support.v7.widget.AppCompatEditText{cab8e45 VFED..CL. ......I. 0,0-0,0 #7f08003e app:id/editText}的内容，它肯定不能被解析为整数。
调用Integer.parseInt(String.valueOf(User1));总是会使用NumberFormatException崩溃(因为它试图解析的字符串与上面的字符串类似)。
当您调用Integer.parseInt(string);时，如果字符串不是有效的整数(例如，空字符串或0.1.2.3)，它将抛出异常。您将看到如下所示的堆栈跟踪：
Process: com.project.testproject, PID: 30357
java.lang.RuntimeException: Unable to start activity ComponentInfo{com.project.testproject/com.project.testproject.MainActivity}: java.lang.NumberFormatException: Invalid int: "android.support.v7.widget.AppCompatEditText{8c2c99a VFED..CL. ......I. 0,0-0,0 #7f08003e app:id/editText}"
at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2416)
at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2476)
at android.app.ActivityThread.-wrap11(ActivityThread.java)
at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1344)
at android.os.Handler.dispatchMessage(Handler.java:102)
at android.os.Looper.loop(Looper.java:148)
at android.app.ActivityThread.main(ActivityThread.java:5417)
at java.lang.reflect.Method.invoke(Native Method)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:726)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:616)
Caused by: java.lang.NumberFormatException: Invalid int: "android.support.v7.widget.AppCompatEditText{8c2c99a VFED..CL. ......I. 0,0-0,0 #7f08003e app:id/editText}"
at java.lang.Integer.invalidInt(Integer.java:138)
at java.lang.Integer.parse(Integer.java:410)
at java.lang.Integer.parseInt(Integer.java:367)
at java.lang.Integer.parseInt(Integer.java:334)
at com.project.testproject.MainActivity.onCreate(MainActivity.java:37)
at android.app.Activity.performCreate(Activity.java:6237)您可能想要使用下面这样的内容：
String UserString = User1.getText().toString();
if( UserString.trim().isEmpty() ) {
//handle empty string
}
else {
try {
int User = Integer.parseInt(UserString);
// Handle valid number cases
}
catch(NumberFormatException ne) {
// handle invalid entry
}
}Stack Overflow用户
发布于 2018-07-10 06:30:35
更改此设置：
User1.setText(answer);要这样做：
User1.setText(String.valueOf(answer));页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/51254689
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