我想让我的应用看起来更专业,所以我决定添加一个启动画面。
我应该如何进行实现呢?
我想让我的应用看起来更专业,所以我决定添加一个启动画面。
我应该如何进行实现呢?
当前回答
启动画面不应该从布局文件加载,加载时可能仍然有一些延迟。
最好的方法是为你的SplashScreenActivity创建一个主题,并设置android:windowBackground为一个可绘制的资源。
https://www.bignerdranch.com/blog/splash-screens-the-right-way/
简而言之:
在manifest中声明你的SplashScreenActivity:
<activity
android:name=".activities.SplashScreenActivity"
android:theme="@style/SplashTheme"
android:screenOrientation="portrait">
<intent-filter>
<action android:name="android.intent.action.MAIN" />
<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
</activity>
在SplashScreenActivity.java中:
@Override
protected void onCreate(@Nullable Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
Intent intent = new Intent(this, MainActivity_.class);
startActivity(intent);
finish();
}
接下来创建主题背景窗口的资源:
<style name="SplashTheme" parent="Theme.Bumpfie.Base">
<item name="android:windowBackground">@drawable/splash</item>
</style>
可绘制文件splash.xml:
<?xml version="1.0" encoding="utf-8"?>
<layer-list xmlns:android="http://schemas.android.com/apk/res/android">
<item android:drawable="@android:color/white"/>
<item>
<bitmap
android:gravity="center"
android:src="@drawable/app_logo"/>
</item>
</layer-list>
其他回答
在我的例子中,我不想创建一个新的活动只是为了显示2秒的图像。当启动我的MainActivity,图像被加载到持有者使用毕加索,我知道这需要大约1秒的加载,所以我决定做以下在我的MainActivity OnCreate:
splashImage = (ImageView) findViewById(R.id.spllll);
this.getWindow().setFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN,WindowManager.LayoutParams.FLAG_FULLSCREEN);
int secondsDelayed = 1;
new Handler().postDelayed(new Runnable() {
public void run() {
getWindow().clearFlags(WindowManager.LayoutParams.FLAG_FULLSCREEN);
splashImage.setVisibility(View.GONE);
}
}, secondsDelayed * 2000);
当启动应用程序时,发生的第一件事是显示ImageView,并通过将窗口标志设置为全屏来删除状态栏。然后我使用Handler运行2秒,2秒后我清除全屏标志,并设置ImageView的可见性为GONE。简单,简单,有效。
它真的很简单,在安卓,我们只是使用处理器的概念来实现启动画面
在SplashScreenActivity java文件中粘贴此代码。
在SplashScreenActivity xml文件中使用imageview放置任何图片。
public void LoadScreen() {
final Handler handler = new Handler();
handler.postDelayed(new Runnable() {
@Override
public void run() {
Intent i = new Intent(SplashScreenActivity.this, AgilanbuGameOptionsActivity.class);
startActivity(i);
}
}, 2000);
}
Create an activity: Splash Create a layout XML file: splash.xml Put UI components in the splash.xml layout so it looks how you want your Splash.java may look like this: public class Splash extends Activity { @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.splash); int secondsDelayed = 1; new Handler().postDelayed(new Runnable() { public void run() { startActivity(new Intent(Splash.this, ActivityB.class)); finish(); } }, secondsDelayed * 1000); } } change ActivityB.class to whichever activity you want to start after the splash screen check your manifest file and it should look like
<活动 android: name = "。屏” android: label = " @string / app_name " > > < /活动 <活动 android: name = "。飞溅” android: label = " @string / title_activity_splash_screen " > <意图过滤器> <action android:name="android.intent.action. main " /> <category android:name="android.intent.category. launcher " /> < /意图过滤器> > < /活动
在Android中,启动画面是一个有点不可用的对象:为了隐藏主活动启动的延迟,它不能尽快加载。使用它有两个原因:广告和网络运营。
实现为对话框使跳跃没有延迟从启动画面到主UI的活动。
public class SplashDialog extends Dialog {
ImageView splashscreen;
SplashLoader loader;
int splashTime = 4000;
public SplashDialog(Context context, int theme) {
super(context, theme);
}
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_splash);
setCancelable(false);
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
cancel();
}
}, splashTime);
}
}
布局:
<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="fill_parent"
android:background="@color/white">
<ImageView
android:id="@+id/splashscreen"
android:layout_width="190dp"
android:layout_height="190dp"
android:background="@drawable/whistle"
android:layout_centerInParent="true" />
</RelativeLayout>
并开始:
public class MyActivity extends ActionBarActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
if (getIntent().getCategories() != null && getIntent().getCategories().contains("android.intent.category.LAUNCHER")) {
showSplashScreen();
}
}
protected Dialog splashDialog;
protected void showSplashScreen() {
splashDialog = new SplashDialog(this, R.style.SplashScreen);
splashDialog.show();
}
...
}
一个超级灵活的启动屏幕如何,可以使用相同的代码,并在AndroidManifest.xml中定义,因此代码永远不需要更改。我通常开发代码库,不喜欢定制代码,因为它很草率。
<activity
android:name=".SplashActivity">
<intent-filter>
<action android:name="android.intent.action.MAIN" />
<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
<meta-data android:name="launch_class" android:value="com.mypackage.MyFirstActivity" />
<meta-data android:name="duration" android:value="5000" />
</activity>
然后SpashActivity本身查找“launch_class”的元数据,然后创建Intent本身。元数据“持续时间”定义了启动画面持续的时间。
public class SplashActivity extends Activity {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle icicle) {
super.onCreate(icicle);
setContentView(R.layout.activity_splash);
ComponentName componentName = new ComponentName(this, this.getClass());
try {
Bundle bundle = null;
bundle = getPackageManager().getActivityInfo(componentName, PackageManager.GET_META_DATA).metaData;
String launch_class = bundle.getString("launch_class");
//default of 2 seconds, otherwise defined in manifest
int duration = bundle.getInt("duration", 2000);
if(launch_class != null) {
try {
final Class<?> c = Class.forName(launch_class);
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
Intent intent = new Intent(SplashActivity.this, c);
startActivity(intent);
finish();
}
}, duration);
} catch (ClassNotFoundException e) {
e.printStackTrace();
}
}
} catch (PackageManager.NameNotFoundException e) {
e.printStackTrace();
}
}
}