安卓：振动法（？如果手机没有震动） [英] Android: vibrator method (if phone has no vibrator?)

问题描述

我想用振动器方法在我的应用程序，和我得到它的工作在我的手机里面有一个振动器这是伟大的。然而电话不具有一个振动器会发生什么。它不工作？它停止的应用程序工作？或者它不是在市场上露面呢？或者我要问的手机，如果它有一个振动器？

I want to use the vibrator method in my app, and i have got it working on my phone which has a vibrator which is great. however phones that don't have a vibrator what happens. does it not work at all? does it stop the app working? or does it not show up in the market at all? or do i have to ask the phone if it has a vibrator?

我也想知道，如果这code是好还是需要的任何调整？这里是我的code ..

I would also like to know if this code is good or needs any adjustments? here is my code..

Vibrator vi;

vi = (Vibrator) getSystemService(Context.VIBRATOR_SERVICE);

vi.vibrate(100);

<uses-permission android:name="android.permission.VIBRATE" /> (In manifest)

谢谢，任何帮助将是巨大的。

Thanks, any help would be great.

推荐答案

检查文档，<一个href="http://developer.android.com/reference/android/os/Vibrator.html">http://developer.android.com/reference/android/os/Vibrator.html

所有你需要做的是检查是否有震动的手机，像这样的present：

all you need to do is check if a vibrator is present on the phone like so:

 Vibrator vi;

 vi = (Vibrator) getSystemService(Context.VIBRATOR_SERVICE);

 if(vi.hasVibrator()){
     vi.vibrate(100);
 }

的震动权限的Andr​​oid市场由于的可能会过滤的您的应用程序只是手机用振动。为了避免这种情况，你可以使用标签与属性要求=假

Because of the vibrate permission Android market may filter your app to just phones with a vibrate. To avoid this you can use the tag with the attribute of required="false"

 <uses-permission android:name="android.permission.VIBRATE" />
 <uses-feature android:name="there.isnt.a.vibrate.feature" android:required="false" />

这一切都记录在这里：

It's all documented here:

<一个href="http://developer.android.com/guide/topics/manifest/uses-permission-element.html">http://developer.android.com/guide/topics/manifest/uses-permission-element.html

<一个href="http://developer.android.com/guide/topics/manifest/uses-feature-element.html">http://developer.android.com/guide/topics/manifest/uses-feature-element.html

然而

没有一个振动特征串，因此Android电子市场的不会过滤您的应用程序，因为你正在使用的振动许可。所以，你确定只使用用途，权限和做在Java code检查。

There is not a Vibrate feature string, therefore Android Market Will Not filter your app because you are using the vibrate permission. So your ok to just use uses-permission and do the check in the Java code.

设备需要一个振动器与Android市场兼容的，但当然，这并不去亚马逊和其他应用程序市场（巴恩斯＆安培; Noble Nook等不具有VIB）

Devices need a vibrator to be compatible with the android market, but of course this doesn't go for the amazon and other app markets (Barnes & Noble Nook doesn't have a vib).

这是支持由黛安Hackthorn（在谷歌的Andr​​oid牵头开发）回复此主题： <一href="http://groups.google.com/group/android-developers/browse_thread/thread/7713e796ea2d0f5f">http://groups.google.com/group/android-developers/browse_thread/thread/7713e796ea2d0f5f

This is backed up by Dianne Hackthorn (Android lead dev at Google's) reply to this thread: http://groups.google.com/group/android-developers/browse_thread/thread/7713e796ea2d0f5f

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