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我正在使用 msgpack 序列化数据。我有一些代码可以很好地序列化数据。

public void testJackson() throws Exception {
    ByteArrayOutputStream out = new ByteArrayOutputStream();

    String data1 = "test data";
    int data2 = 10;
    List<String> data3 = new ArrayList<String>();
    data3.add("list data1");
    data3.add("list data1");

    ObjectMapper mapper = new ObjectMapper();
    mapper.writeValue(out, data1);
    mapper.writeValue(out, data2);
    mapper.writeValue(out, data3);

    // TODO: How to deserialize?

}

但现在我不知道如何反序列化数据。我在任何地方都找不到任何解决方案。如果有人可以帮助如何进行，那将是很好的。

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3 回答 3

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@Japu_D_Cret 谢谢你这么详细的回答！其实我想用msgpack来传输数据，我用msgpack让它工作了，这是我的代码

    ByteArrayOutputStream out = new ByteArrayOutputStream();

    String data1 = "test data";
    int data2 = 10;
    List<String> data3 = new ArrayList<String>();
    data3.add("list data1");
    data3.add("list data1");

    MessagePack packer = new MessagePack();

    packer.write(out, data1);
    packer.write(out, data2);
    packer.write(out, data3);

    // TODO: How to deserialize?

    BufferUnpacker unpacker = packer.createBufferUnpacker(out.toByteArray());
    System.out.println(unpacker.readString());
    System.out.println(unpacker.readInt());
    System.out.println(unpacker.read(Templates.tList(Templates.TString)));

然后我在 msgpack 网站上找到了 jackson-databind，它也支持 msgpack 格式。我对这两个做了一些测试，发现jackson的serialize性能比msgpack要好，所以我想用jackson代替msgpack。

于 2017-03-30T06:30:14.963 回答
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问题

我尝试了很多readValue方法，但我只能得到第一个String，关于第二个和第三个值我不知道

问题是，Jackson 总是读取第一个数据，因为数据既没有从 中删除，也没有明确告诉 Jackson 下一个数据是从位置 A 到位置 B

解决方案

此示例有效并且与您的代码相似，但不是很优雅。在这里我明确告诉杰克逊我的数据在哪里，但我必须知道它是如何写入的，这是一个过于具体的解决方案

File dataFile = new File("jackson.txt");
if(!dataFile.exists())
  dataFile.createNewFile();
FileOutputStream fileOut = new FileOutputStream(dataFile);
ByteArrayOutputStream out = new ByteArrayOutputStream();
FileInputStream fileIn = new FileInputStream(dataFile);

String writeData1 = "test data";
int writeData2 = 10;
List<String> writeData3 = new ArrayList<String>();
writeData3.add("list data1");
writeData3.add("list data1");

ObjectMapper mapper = new ObjectMapper();
byte[] writeData1Bytes = mapper.writeValueAsBytes(writeData1);
out.write(writeData1Bytes);
byte[] writeData2Bytes = mapper.writeValueAsBytes(writeData2);
out.write(writeData2Bytes);
byte[] writeData3Bytes = mapper.writeValueAsBytes(writeData3);
out.write(writeData3Bytes);
out.writeTo(fileOut);

// TODO: How to deserialize?
int pos = 0;
byte[] readData = new byte[1000];
fileIn.read(readData);
String readData1 = mapper.readValue(readData, pos, writeData1Bytes.length, String.class);
pos += writeData1Bytes.length;
Integer readData2 = mapper.readValue(readData, pos, writeData2Bytes.length, Integer.class);
pos += writeData2Bytes.length;
ArrayList readData3 = mapper.readValue(readData, pos, writeData3Bytes.length, ArrayList.class);
pos += writeData3Bytes.length;

System.out.printf("readData1 = %s%n", readData1);
System.out.printf("readData2 = %s%n", readData2);
System.out.printf("readData3 = %s%n", readData3);

文件看起来像这样

"test data"10["list data1","list data1"]

如何正确地做

一种更优雅的方法是将您的数据封装在一个对象中，该对象可以转换为有效的 JSON 字符串，并且杰克逊不需要任何更多信息

public class JacksonTest {
  public static class DataNode {
    @JsonProperty("data1")
    private String data1;
    @JsonProperty("data2")
    private int data2;
    @JsonProperty("data3")
    private List<String> data3;

    //needed for Jackson
    public DataNode() {
    }

    public DataNode(String data1, int data2, List<String> data3) {
      this.data1 = data1;
      this.data2 = data2;
      this.data3 = data3;
    }
  }

  public static void main(String[] args) throws Exception {
    File dataFile = new File("jackson.txt");
    if(!dataFile.exists())
      dataFile.createNewFile();
    FileOutputStream fileOut = new FileOutputStream(dataFile);
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    FileInputStream fileIn = new FileInputStream(dataFile);

    String writeData1 = "test data";
    int writeData2 = 10;
    List<String> writeData3 = new ArrayList<String>();
    writeData3.add("list data1");
    writeData3.add("list data1");

    DataNode writeData = new DataNode(writeData1, writeData2, writeData3);

    ObjectMapper mapper = new ObjectMapper();
    mapper.writeValue(out, writeData);
    out.writeTo(fileOut);

    // TODO: How to deserialize?
    DataNode readData = mapper.readValue(fileIn, DataNode.class);

    System.out.printf("readData1 = %s%n", readData.data1);
    System.out.printf("readData2 = %s%n", readData.data2);
    System.out.printf("readData3 = %s%n", readData.data3);
  }
}

文件的内容看起来像这样

{"data1":"test data","data2":10,"data3":["list data1","list data1"]}
于 2017-03-30T05:34:26.710 回答
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您将需要使用ObjectMapperreadValue中的一种方法- 可能具有 Reader 或 InputStream 作为第一个参数的方法。

于 2017-03-30T04:52:32.313 回答