Question

我试图通过指定该数组的路径来弄清楚如何更改JSON数组中的现有对象。

问题是我不知道如何指定数组的路径以便可以更改它。我尝试过的每种方法都没有关于如何更改ObjectNode中的数组的示例。

这是我现在的代码，在其中我尝试通过指针表示法指定数组的路径来获取和更改数组，但是，这总是不返回数组：

构建POJO类不是一种选择。

我的json结构中的数组如下：

{
  RESULTS: {
     ATTACHMENTS: {
       ATTACHMENT: [{object}]
  }
 }
}

代码：

   private JsonNode attachmentConversion(JsonNode object){
        final String ATTACHMENT_POINTER = "/RESULTS/ATTACHMENTS/ATTACHMENT"; //Path to the array
        ObjectNode OUTPUT = object.deepCopy();
        OUTPUT.remove("DETAILS");

        //Validate is an array - It properly fetches the array 
        if(object.at(ATTACHMENT_POINTER).isArray()){
            int counter=0;

            //Loop through attachment array - It properly fethes each object in the array
            for(final JsonNode objNode : object.at(ATTACHMENT_POINTER)){
                ObjectNode objectNode = objNode.deepCopy();
                OUTPUT.withArray(ATTACHMENT_POINTER) //HERE IS THE PROBLEM - How do I specify the path to the array I want to replace with a new object in the withArray api?  This always returns back as an empty array, even though the loop above properly fetches the array elements.
                        .set(counter
                                , objectNode
                                .replace("ATTACH_CONTENT"
                                        , xmlToJson.parseXmlToJson(objNode.get("ATTACH_CONTENT")
                                        .asText())));
                counter = counter+1;
            }
        }
        return new ObjectMapper()
                .createObjectNode()
                .set("customertopology", OUTPUT);
    }

Answer 1

我认为你不想改变传递给方法的对象，所以你做了一个深层复制。并且您想要返回一个新的JSON对象。

到目前为止我是对的吗？如果是，那么当您使用深层副本时，请使用它，遍历您关注的数组的每个元素，并将Node值替换为您想要的值。

private static JsonNode attachmentConversion(JsonNode object){



    final String ATTACHMENT_POINTER = "/RESULTS/ATTACHMENTS/ATTACHMENT"; //Path to the array
    ObjectNode OUTPUT = object.deepCopy();
    OUTPUT.remove("DETAILS");

    //Validate is an array - It properly fetches the array
    if(OUTPUT.at(ATTACHMENT_POINTER).isArray()){
        int counter=0;

        //Loop through attachment array - It properly fethes each object in the array
        for(final JsonNode objNode : OUTPUT.at(ATTACHMENT_POINTER)){
            ((ObjectNode)objNode).replace("ATTACH_CONTENT", xmlToJson.parseXmlToJson(objNode.get("ATTACH_CONTENT"));
        }

    }
    return new ObjectMapper()
            .createObjectNode()
            .set("customertopology", OUTPUT);
}

Answer 2

您可以执行

，而不是提供整个路径

let topViewLabelConstraints: [NSLayoutConstraint] = [
    NSLayoutConstraint(item: topViewLabel, attribute: .centerX, relatedBy: .equal, toItem: redView, attribute: .centerX, multiplier: 1, constant: 0),
    NSLayoutConstraint(item: topViewLabel, attribute: .centerY, relatedBy: .equal, toItem: redView, attribute: .centerY, multiplier: 1, constant: 0)
]
redView.addConstraints(topViewLabelConstraints)

这应该返回数组。使用fasterxml的完整代码

OUTPUT.get("RESULTS").get("ATTACHMENTS").get("ATTACHMENT");

以下代码打印为我。

String json = "{\n" +
            "  \"RESULTS\": {\n" +
            "     \"ATTACHMENTS\": {\n" +
            "       \"ATTACHMENT\": [{\"key\": 1}, {\"key\": 2}]\n" +
            "  }\n" +
            " }\n" +
            "}";

ObjectMapper mapper = new ObjectMapper();
JsonNode node = mapper.readTree(json);
node.get("RESULTS").get("ATTACHMENTS").get("ATTACHMENT").forEach(obj -> {
        System.out.println(obj.get("key"));
    });

使用Jackson更改嵌套JSON对象中的数组

2 个答案: