这是我的code：

VideoView vd;
vd = (VideoView) findViewById(R.id.videoview2);
String path = "android.resource://" + getPackageName() + "/"
                        + R.raw.video1;
                vd.setVideoURI(Uri.parse(path));
                vd.start();

这工作，但我想R.raw.video1是一个字串becaus我有很多的视频播放。

This works, but i want the R.raw.video1 to be a string becaus i have a lot of videos to play.

所以我想是这样的：

String videoResource = "R.raw.video1"
 String path = "android.resource://" + getPackageName() + "/"
                            + videoResource;

不幸的是这不工作，我如何得到它的工作？

Unfortunately this doesn't work, how do i get it to work?

推荐答案

您需要的资源，要工作，可以通过它的名字用检索的标识符：

You need the identifier of the resource for that to work, which can be retrieved by its name using:

int id = getResources().
    getIdentifier("name_of_resource", "id", getPackageName());

所以，新的code将变成：

So your new code would become:

int videoResource = getResources().
    getIdentifier("video1", "raw", getPackageName());
String path = "android.resource://" + getPackageName() + "/" + videoResource;