使用 spark 将 POST 正文解析为 java 对象

Question

我不久前从 spring 迁移到 spark，现在我停留在一些基本的东西上。

当我发出 POST 请求在正文中发送数据时，我想让 JAVA 对象回到控制器中..

在spring我曾经做过

@RequestBody User user

它是自动 "filled"..

现在有了 spark，我有了方法：

request.body();

但这给了我一个像这样的序列化字符串：

id=7&name=Pablo+Mat%C3%ADas&lastname=Gomez&githubUsername=pablomatiasgomez

那么如何获取用户 DTO？

当然，用户 class 具有属性

id
名字
姓氏
github用户名

Answer 1

据我所知，Spark 不提供此功能。当我将它用于小型宠物项目时，我编写了一些小型实用方法来将 URL 编码的字符串解析为 POJO，如下所示：

import com.google.gson.Gson;
import java.io.UnsupportedEncodingException;
import java.net.URLDecoder;
import java.util.LinkedHashMap;
import java.util.Map;

public class Test {

  private static final Gson GSON = new Gson();

  public static <T> T convert(String urlencoded, Class<T> type) {
    try {
      Map<String, Object> map = asMap(urlencoded);
      String json = GSON.toJson(map);
      return GSON.fromJson(json, type);
    }
    catch (Exception e) {
      e.printStackTrace(); // TODO log
      return null;
    }
  }

  public static Map<String, Object> asMap(String urlencoded) throws UnsupportedEncodingException {
    return asMap(urlencoded, "UTF-8");
  }

  @SuppressWarnings("unchecked")
  public static Map<String, Object> asMap(String urlencoded, String encoding) throws UnsupportedEncodingException {

    Map<String, Object> map = new LinkedHashMap<>();

    for (String keyValue : urlencoded.trim().split("&")) {

      String[] tokens = keyValue.trim().split("=");
      String key = tokens[0];
      String value = tokens.length == 1 ? null : URLDecoder.decode(tokens[1], encoding);

      String[] keys = key.split("\.");
      Map<String, Object> pointer = map;

      for (int i = 0; i < keys.length - 1; i++) {

        String currentKey = keys[i];
        Map<String, Object> nested = (Map<String, Object>) pointer.get(keys[i]);

        if (nested == null) {
          nested = new LinkedHashMap<>();
        }

        pointer.put(currentKey, nested);
        pointer = nested;
      }

      pointer.put(keys[keys.length - 1], value);
    }

    return map;
  }

  public static void main(String[] args) {
    String payload = "id=7&name=Pablo+Mat%C3%ADas&lastname=Gomez&githubUsername=pablomatiasgomez";
    User user = convert(payload, User.class);
    System.out.println(user);
  }
}

class User {

  long id;
  String name;
  String lastname;
  String githubUsername;

  @Override
  public String toString() {
    return "User{" +
        "id=" + id +
        ", name='" + name + '\'' +
        ", lastname='" + lastname + '\'' +
        ", githubUsername='" + githubUsername + '\'' +
        '}';
  }
}

运行此 Test class 将在您的控制台上打印以下内容：

User{id=7, name='Pablo Matías', lastname='Gomez', githubUsername='pablomatiasgomez'}

请注意，当 User 中有嵌套结构时，这也适用，例如，由几个其他字段组成的 Address。只需用 "." 分隔字段，如下所示：

public class Test {

  // ... same code ...

  public static void main(String[] args) {
    String payload = "id=7&name=Pablo+Mat%C3%ADas&lastname=Gomez&githubUsername=pablomatiasgomez&" +
        "address.street=Coolsingel&address.number=42a&address.city=Rotterdam";
    User user = convert(payload, User.class);
    System.out.println(user);
  }
}

class User {

  long id;
  String name;
  String lastname;
  String githubUsername;
  Address address;

  @Override
  public String toString() {
    return "User{" +
        "\n  id=" + id +
        "\n  name='" + name + '\'' +
        "\n  lastname='" + lastname + '\'' +
        "\n  githubUsername='" + githubUsername + "'" +
        "\n  address=" + address + "\n" +
        '}';
  }
}

class Address {


  String street;
  String number;
  String city;

  @Override
  public String toString() {
    return "Address{" +
        "street='" + street + '\'' +
        ", number='" + number + '\'' +
        ", city='" + city + '\'' +
        '}';
  }
}

这将打印：

User{
  id=7
  name='Pablo Matías'
  lastname='Gomez'
  githubUsername='pablomatiasgomez'
  address=Address{street='Coolsingel', number='42a', city='Rotterdam'}
}

编辑

如果负载包含一个列表，比如 Users，你可以这样做：

public class Test {

  private static final Gson GSON = new Gson();

  public static <T> T convert(String urlencoded, Type type) {
    try {
      Map<String, Object> map = asMap(urlencoded);
      String json = GSON.toJson(containsList(map) ? map.values() : map);
      return GSON.fromJson(json, type);
    }
    catch (Exception e) {
      e.printStackTrace();
      return null;
    }
  }

  private static boolean containsList(Map<String, Object> map) {
    return !map.isEmpty() && new ArrayList<>(map.keySet()).get(0).contains("[");
  }

  public static Map<String, Object> asMap(String urlencoded) throws UnsupportedEncodingException {
    return asMap(urlencoded, "UTF-8");
  }

  @SuppressWarnings("unchecked")
  public static Map<String, Object> asMap(String urlencoded, String encoding) throws UnsupportedEncodingException {

    Map<String, Object> map = new LinkedHashMap<>();

    for (String keyValue : urlencoded.trim().split("&")) {

      String[] tokens = keyValue.trim().split("=");
      String key = tokens[0];
      String value = tokens.length == 1 ? null : URLDecoder.decode(tokens[1], encoding);

      String[] keys = key.split("\.");
      Map<String, Object> pointer = map;

      for (int i = 0; i < keys.length - 1; i++) {

        String currentKey = keys[i];
        Map<String, Object> nested = (Map<String, Object>) pointer.get(keys[i]);

        if (nested == null) {
          nested = new LinkedHashMap<>();
        }

        pointer.put(currentKey, nested);
        pointer = nested;
      }

      pointer.put(keys[keys.length - 1], value);
    }

    return map;
  }

  public static void main(String[] args) throws Exception {

    String payload = "id=7&name=Pablo Mat%C3%ADas";
    User user = convert(payload, User.class);
    System.out.println("single user   -> " + user);

    payload = "users[0].id=7&users[0].name=Pablo Mat%C3%ADas&users[1].id=42&users[1].name=Bart";
    List<User> users = convert(payload, new TypeToken<List<User>>(){}.getType());
    System.out.println("list of users -> : " + users);
  }
}

class User {

  long id;
  String name;

  @Override
  public String toString() {
    return "User{" +
        "id=" + id +
        ", name='" + name + '\'' +
        '}';
  }
}

这将打印：

single user   -> User{id=7, name='Pablo Matías'}
list of users -> : [User{id=7, name='Pablo Matías'}, User{id=42, name='Bart'}]

Answer 2

我找到了一种更简单的方法，它不涉及 URL 编码。

在客户端，将您的 javascript 对象转换为 JSON 字符串并使用它设置查询参数 (yourObject)：

var obj = null;
obj = {
  yourObject: JSON.stringify(currentObject)
};
   
$.ajax({
  type: "GET",
  url: "saveAnObject",
  data: obj, 
  contentType: "application/json; charset=utf-8",
  dataType: "json",
  success: function(data) {
    console.log('saveAnObject result: ' + data + ".");

  },
  error: function() {
    
  },
  cache: false
});

然后在 Spark 中：

 get("/saveAnObject", (req, res) - > {
     
     String yourObjectStr = "" + req.queryParams("yourObject");
   
     // Convert the JSON string to a POJO obj
     Gson gson = new GsonBuilder().create();
     YourObject pojoObj = gson.fromJson(yourObjectStr , YourObject.class);

     // do something with your pojoObj object.

使用 spark 将 POST 正文解析为 java 对象

Parse POST body to java object using spark

java

spring

spark-java

编辑