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JAXB - 如何在没有标头的情况下编组 java 对象

我正在尝试封送一个 java 对象,但我想删除 Jaxb 引入的标头。

目的:

@XmlRootElement
public class FormElement implements Serializable {
    private String id;
    private String name;
    private Integer order;
}
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预期输出:

<element>
    <id>asd</id>
    <name>asd</name>
    <order>1</order>
</element>
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我的输出:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<formElement>
  <id>asd</id>
  <name>asd</name>
  <order>1</order>
</formElement>
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我的元帅方法:

public String marshal() {
        JAXBContext context;
        try {
            context = JAXBContext.newInstance(FormElement.class);
            Marshaller marshaller = context.createMarshaller();
            StringWriter stringWriter = new StringWriter();
            marshaller.marshal(this, stringWriter);
            return stringWriter.toString();
        } catch (JAXBException e) {
        }
        return null;
    }
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我怎样才能删除它?

提前致谢。

java xml jaxb marshalling

4
推荐指数
1
解决办法
6815
查看次数

java.io.FileNotFoundException:output.mp4(只读文件系统)

我正在尝试将我的图像转换为视频.我找到了示例代码并用于我的应用程序.但我得到的错误就像

java.io.FileNotFoundException: output.mp4 (Read-only file system)

这是我的代码的片段:

    public class Main2Activity extends AppCompatActivity {

        AndroidSequenceEncoder encoder;
        @RequiresApi(api = Build.VERSION_CODES.N)
        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.activity_main2);

            ArrayList<Bitmap> arrayList = new ArrayList<>();
            Bitmap b = BitmapFactory.decodeResource(getResources(),R.drawable.empty_collage);
            Bitmap a = BitmapFactory.decodeResource(getResources(),R.drawable.images);

            arrayList.add(a);
            arrayList.add(b);

            File file = new File("output.mp4");

            SeekableByteChannel out = null;
            try {
                out = (SeekableByteChannel) NIOUtils.writableFileChannel(String.valueOf(file));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
                Log.e("msg",e.toString());
            }

            try {
                 encoder = new AndroidSequenceEncoder((org.jcodec.common.io.SeekableByteChannel) out, Rational.R(25,1));
            } catch (IOException e) {
                e.printStackTrace();
            }

            Toast.makeText(this, …
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animation android

1
推荐指数
1
解决办法
711
查看次数

标签 统计

android ×1

animation ×1

java ×1

jaxb ×1

marshalling ×1

xml ×1