使用Jackson将JSON字符串转换为漂亮的打印JSON输出

这是我的JSON字符串:

{"attributes":[{"nm":"ACCOUNT","lv":[{"v":{"Id":null,"State":null},"vt":"java.util.Map","cn":1}],"vt":"java.util.Map","status":"SUCCESS","lmd":13585},{"nm":"PROFILE","lv":[{"v":{"Party":null,"Ads":null},"vt":"java.util.Map","cn":2}],"vt":"java.util.Map","status":"SUCCESS","lmd":41962}]}

我需要将上面的JSON字符串转换为漂亮的打印JSON输出(使用Jackson)，如下所示:

{
    "attributes": [
        {
            "nm": "ACCOUNT",
            "lv": [
                {
                    "v": {
                        "Id": null,
                        "State": null
                    },
                    "vt": "java.util.Map",
                    "cn": 1
                }
            ],
            "vt": "java.util.Map",
            "status": "SUCCESS",
            "lmd": 13585
        },
        {
            "nm": "PROFILE
            "lv": [
                {
                    "v": {
                        "Party": null,
                        "Ads": null
                    },
                    "vt": "java.util.Map",
                    "cn": 2
                }
            ],
            "vt": "java.util.Map",
            "status": "SUCCESS",
            "lmd": 41962
        }
    ]
}

谁能根据我上面的例子给我一个例子?如何实现这个场景?我知道有很多例子，但我不能正确地理解它们。任何帮助都将感谢一个简单的例子。

更新:

下面是我正在使用的代码:

ObjectMapper mapper = new ObjectMapper();
System.out.println(mapper.defaultPrettyPrintingWriter().writeValueAsString(jsonString));

但这并不适用于我需要输出的方式如上所述。

这是我在上面JSON中使用的POJO:

public class UrlInfo implements Serializable {

    private List<Attributes> attribute;

}

class Attributes {

    private String nm;
    private List<ValueList> lv;
    private String vt;
    private String status;
    private String lmd;

}


class ValueList {
    private String vt;
    private String cn;
    private List<String> v;
}

有人能告诉我，我是否为JSON得到了正确的POJO吗?

更新:

String result = restTemplate.getForObject(url.toString(), String.class);

ObjectMapper mapper = new ObjectMapper();
Object json = mapper.readValue(result, Object.class);

String indented = mapper.defaultPrettyPrintingWriter().writeValueAsString(json);

System.out.println(indented);//This print statement show correct way I need

model.addAttribute("response", (indented));

下面一行输出如下内容:

System.out.println(indented);


{
  "attributes" : [ {
    "nm" : "ACCOUNT",
    "error" : "null SYS00019CancellationException in CoreImpl fetchAttributes\n java.util.concurrent.CancellationException\n\tat java.util.concurrent.FutureTask$Sync.innerGet(FutureTask.java:231)\n\tat java.util.concurrent.FutureTask.",
    "status" : "ERROR"
  } ]
}

这就是我需要的表现方式。但当我像这样把它添加到模型中:

model.addAttribute("response", (indented));

然后显示在一个结果表单jsp页面，如下所示:

    <fieldset>
        <legend>Response:</legend>
            <strong>${response}</strong><br />

    </fieldset>

我得到的结果是这样的:

{ "attributes" : [ { "nm" : "ACCOUNT", "error" : "null    
SYS00019CancellationException in CoreImpl fetchAttributes\n 
java.util.concurrent.CancellationException\n\tat 
java.util.concurrent.FutureTask$Sync.innerGet(FutureTask.java:231)\n\tat 
java.util.concurrent.FutureTask.", "status" : "ERROR" } ] }

我不需要。我需要上面打印出来的方式。有人能告诉我为什么会这样吗?

当前回答

如果你格式化字符串并返回像RestApiResponse< string >这样的对象，你会得到不需要的字符，如转义等:\n， \"。解决方案是将您的JSON-string转换为Jackson JsonNode对象并返回RestApiResponse<JsonNode>:

ObjectMapper mapper = new ObjectMapper();
JsonNode tree = objectMapper.readTree(jsonString);
RestApiResponse<JsonNode> response = new RestApiResponse<>();
apiResponse.setData(tree);
return response;

2021-02-23 09:29:44

其他回答

这看起来可能是你问题的答案。它说它正在使用Spring，但我认为这对您的情况仍然有帮助。让我在这里内联代码，这样更方便:

import java.io.FileReader;

import org.codehaus.jackson.map.ObjectMapper;
import org.codehaus.jackson.map.ObjectWriter;

public class Foo
{
  public static void main(String[] args) throws Exception
  {
    ObjectMapper mapper = new ObjectMapper();
    MyClass myObject = mapper.readValue(new FileReader("input.json"), MyClass.class);
    // this is Jackson 1.x API only: 
    ObjectWriter writer = mapper.defaultPrettyPrintingWriter();
    // ***IMPORTANT!!!*** for Jackson 2.x use the line below instead of the one above: 
    // ObjectWriter writer = mapper.writer().withDefaultPrettyPrinter();
    System.out.println(writer.writeValueAsString(myObject));
  }
}

class MyClass
{
  String one;
  String[] two;
  MyOtherClass three;

  public String getOne() {return one;}
  void setOne(String one) {this.one = one;}
  public String[] getTwo() {return two;}
  void setTwo(String[] two) {this.two = two;}
  public MyOtherClass getThree() {return three;}
  void setThree(MyOtherClass three) {this.three = three;}
}

class MyOtherClass
{
  String four;
  String[] five;

  public String getFour() {return four;}
  void setFour(String four) {this.four = four;}
  public String[] getFive() {return five;}
  void setFive(String[] five) {this.five = five;}
}

2013-01-25 06:03:07

要缩进任何旧JSON，只需将其绑定为Object，如下所示:

Object json = mapper.readValue(input, Object.class);

然后缩进写出来:

String indented = mapper.writerWithDefaultPrettyPrinter().writeValueAsString(json);

这避免了必须定义实际的POJO来映射数据。

或者你也可以使用JsonNode (JSON树)。

2013-01-26 00:25:45

使用Jackson 1.9+的新方法如下:

Object json = OBJECT_MAPPER.readValue(diffResponseJson, Object.class);
String indented = OBJECT_MAPPER.writerWithDefaultPrettyPrinter()
                               .writeValueAsString(json);

输出将被正确格式化!

2014-01-14 02:49:06

任何使用POJO、DDO或响应类返回JSON的人都可以使用spring.jackson.serialization。在它们的属性文件中缩进-输出=true。它自动格式化响应。

2022-11-17 08:41:54

最简单也是最紧凑的解决方案(适用于v2.3.3):

ObjectMapper mapper = new ObjectMapper();
mapper.enable(SerializationFeature.INDENT_OUTPUT);
mapper.writeValueAsString(obj)

2014-06-01 17:32:09

使用Jackson将JSON字符串转换为漂亮的打印JSON输出

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