如何避免Jackson在Java中转义双引号? [英] How to avoid Jackson escaping double quotes in Java?

问题描述

我正在一个项目上，想使用ObjectMapper重写一些用Gson编写的代码给Jackson.因此，我尝试使用以下一些属性来创建JSON字符串:

I am working on a project and wanted to rewrite some code written in Gson to Jackson using ObjectMapper. So I am trying to create a JSON string using some properties as below:

ObjectMapper objectMapper = new ObjectMapper();
ObjectNode objNode= objectMapper.createObjectNode();
objNode.put("identifierVal", UUID.randomUUID().toString());
objNode.put("version", "two");
List<String> namesList= new ArrayList<>();
namesList.add("test");
objNode.put("namesList", namesList.toString());
String requestObject = objectMapper.writeValueAsString(objNode.toString());

预期结果:

{
   "identifierVal":1234,
   "version":"two",
   "namesList":[
      "test"
   ]
}

实际:

"{  
        "\identifierVal\": 1234,   
        "\version\":"\two\",
"\namesList\": ["\test\"]

    }"

因此，一旦我使用杰克逊(Jackson)创建了JSON字符串，事实证明它是在转义字段名称和值的双引号，并在前导和尾随空格处添加\.因此，由于转义，服务呼叫失败.

So once I create a JSON String using Jackson, it turns out it is escaping double quotes for field names and values and adding a \ at leading and trailing spaces. So service call fails due to the escaping.

我已经阅读了一些文档，并且我可以理解杰克逊正在转义双引号.但是有一种方法可以避免转义双引号以及添加前导和尾随双引号.

I have went through some documentation and I can understand Jackson is escaping double quotes. But is there a way to avoid escaping double quotes and adding of leading and trailing double quotes.

感谢您的帮助.顺便说一句，我点击了以下链接:

Any help is appreciated. BTW, I followed the links below:

为什么ObjectNode在Json字符串中添加反斜杠

Jackson在json中添加反斜杠

推荐答案

问题是您正在将JSON对象通过其toString()方法转换为字符串，然后再将其传递给objectMapper.更改此:

The problem is you are converting your JSON object to a String via its toString() method before passing it to the objectMapper. Change this:

String requestObject = objectMapper.writeValueAsString(objNode.toString());

对此:

String requestObject = objectMapper.writeValueAsString(objNode);

您还需要更改此内容:

List<String> namesList= new ArrayList<>();
namesList.add("test");
objNode.put("namesList", namesList.toString());

对此:

ArrayNode namesNode = objNode.arrayNode();
namesNode.add("test");
objNode.set("namesList", namesNode);

它会按预期工作.

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