Question

我有一个具有以下值的字符串：

{
  "keyA": {
    "ID": "123",
    "Name": "TESTA",
    "Mobile": "1111"
  },
  "keyB": {
    "ID": "456",
    "Name": "TESTB",
    "Mobile": "2222"
  }
}

如何将json转换为列表和类格式

我想将它解析为List

请告知如何使用Jackson ObjectMapper？

Answer 1

您可以尝试进行测试：

public static void main(String[] args) {
    String jsonString = "{\n" +
                "  \"keyA\": {\n" +
                "    \"ID\": \"123\",\n" +
                "    \"Name\": \"TESTA\",\n" +
                "    \"Mobile\": \"1111\"\n" +
                "  },\n" +
                "  \"keyB\": {\n" +
                "    \"ID\": \"456\",\n" +
                "    \"Name\": \"TESTB\",\n" +
                "    \"Mobile\": \"2222\"\n" +
                "  }\n" +
                "}";
    List<CustomObject> customObjects = new ArrayList<CustomObject>();
    ObjectMapper mapper = new ObjectMapper();
    try {
        ClassA myClassAObject= mapper.readValue(jsonString, ClassA.class);
        customObjects.add(myClassAObject.getKeyA());
        customObjects.add(myClassAObject.getKeyB());
    }
    catch(Exception e) {
        e.printStackTrace();
    }
    System.out.println(customObjects.size());
}

我的其他模特课CustomObject：

public class CustomObject {

    private String id;
    private String name;
    private String mobile;

    @JsonProperty("ID")
    public String getId() {
        return id;
    }

    @JsonProperty("ID")
    public void setId(String id) {
        this.id = id;
    }

    @JsonProperty("Name")
    public String getName() {
        return name;
    }

    @JsonProperty("Name")
    public void setName(String name) {
        this.name = name;
    }

    @JsonProperty("Mobile")
    public String getMobile() {
        return mobile;
    }

    @JsonProperty("Mobile")
    public void setMobile(String mobile) {
        this.mobile = mobile;
    }
}

和ClassA

public class ClassA {

    private CustomObject keyA;
    private CustomObject keyB;

    public CustomObject getKeyA() {
        return keyA;
    }

    public void setKeyA(CustomObject keyA) {
        this.keyA = keyA;
    }

    public CustomObject getKeyB() {
        return keyB;
    }

    public void setKeyB(CustomObject keyB) {
        this.keyB = keyB;
    }
}

如果您遇到问题，这是我在Maven中使用的库版本：

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.4.1.3</version>
</dependency>

如何使用Jackson将JSON字符串解析为列表

1 个答案: