【发布时间】:2015-09-16 03:00:07
【问题描述】:
我创建了一个具有不同级别的问答游戏,每个级别都包含一个问题。我没有用数据库创建它。我只是用字符串。当用户在第一级回答一个问题时,他会被带到第二级,但是当用户返回到第一级时,即使他之前已经解决了,他也必须再次输入答案。无论如何,JAVA 中是否有将答案保留在类型面板中(如果用户解决了它)而无需创建数据库?此外,在类型面板中输入时,用户必须删除“在此处输入...”然后回答。无论如何,当用户点击输入“在此处输入...”时会自动删除吗?
这是我的一级activity.xml:
<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:orientation="vertical"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:background="@drawable/background"
android:weightSum="1"
android:textAlignment="center"
android:id="@+id/level1">
<TextView
android:layout_width="300dp"
android:layout_height="200dp"
android:text="What has 88 keys but cannot open a single door?"
android:id="@+id/que1"
android:width="255dp"
android:textSize="30dp"
android:layout_margin="50dp"
android:textStyle="italic"
android:gravity="center" />
<EditText
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:id="@+id/type1"
android:layout_gravity="center_horizontal"
android:text="Type here..." />
<Button
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:text="Check answer..."
android:id="@+id/check1"
android:layout_gravity="center_horizontal" />
</LinearLayout>
这是我的 Oneactivity.java
package com.golo.user.gaunkhanekatha;
import android.app.Activity;
import android.content.Intent;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;
import android.os.Handler;
public class OneActivity extends Activity {
public SharedPreferences preferences; //ADDED THIS LINE
public Button check;
public EditText typeh;
private Toast toast;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_one);
toast = Toast.makeText(OneActivity.this, "", Toast.LENGTH_SHORT);
check = (Button)findViewById(R.id.check1); //R.id.button is the id on your xml
typeh = (EditText)findViewById(R.id.type1); //this is the EditText id
check.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
// Perform action on click
//Here you must get the text on your EditText
String Answer = (String) typeh.getText().toString(); //here you have the text typed by the user
//You can make an if statement to check if it's correct or not
if(Answer.equals("Piano") || (Answer.equals("Keyboard")))
{
preferences = PreferenceManager.getDefaultSharedPreferences(v.getContext());
SharedPreferences.Editor editor = preferences.edit();
editor.putInt("Level 1 question 1 ", 1); //Depends of the level he have passed.
editor.apply();
///Correct Toast
toast.setText("Correct");
toast.setGravity(Gravity.TOP | Gravity.LEFT, 500, 300);
toast.show();
Intent i = new Intent(OneActivity.this, TwoActivity.class);
startActivity(i);
finish();
}
else{
//It's not the correct answer
toast.setText("Wrong! Try again...");
toast.show();
}
}
});
}
@Override
protected void onDestroy() {
super.onDestroy();
if(toast!= null) {
toast.cancel();
}
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.menu_aboutus, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
//noinspection SimplifiableIfStatement
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
}
此外,toast 会显示在有键盘的地方。有没有办法将吐司屏幕移动到屏幕上清晰可见的位置?
【问题讨论】:
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txt 文件,xml 文件,...