如何仅使用 Jackson 将 XML 转换为 JSON？

我从服务器收到 XML 格式的响应。但我需要以 JSON 格式显示它。

有没有办法不用第三方API就可以转换呢？我使用了 Jackson，但为此我需要创建 POJO。

服务器的响应是这样的：

<?xml version='1.0'?>
<errors><error><status>400</status><message>The field 'quantity' is invalid.</message><details><invalid_reason>The quantity specified is greater than the quantity of the product that is available to ship.</invalid_reason><available_quantity>0</available_quantity><order_product_id>12525</order_product_id></details></error></errors>

使用杰克逊 2.x

您可以使用 Jackson 来做到这一点，并且不需要 POJO：

String xml = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n" +
             "<errors>\n" +
             "  <error>\n" +
             "    <status>400</status>\n" +
             "    <message>The field 'quantity' is invalid.</message>\n" +
             "    <details>\n" +
             "      <invalid_reason>The quantity specified is greater than the quantity of the product that is available to ship.</invalid_reason>\n" +
             "      <available_quantity>0</available_quantity>\n" +
             "      <order_product_id>12525</order_product_id>\n" +
             "    </details>\n" +
             "  </error>\n" +
             "</errors>";

XmlMapper xmlMapper = new XmlMapper();
JsonNode node = xmlMapper.readTree(xml.getBytes());

ObjectMapper jsonMapper = new ObjectMapper();
String json = jsonMapper.writeValueAsString(node);

需要以下依赖项：

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-core</artifactId>
    <version>2.8.2</version>
</dependency>
<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.8.2</version>
</dependency>
<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-annotations</artifactId>
    <version>2.8.2</version>
</dependency>
<dependency>
    <groupId>com.fasterxml.jackson.dataformat</groupId>
    <artifactId>jackson-dataformat-xml</artifactId>
    <version>2.8.2</version>
</dependency>

请注意XmlMapper中规定的限制文档 https://github.com/FasterXML/jackson-dataformat-xml:

树模型仅以有限的方式支持：具体来说，Java 数组和Collections可以写入，但不能读取，因为在没有附加信息的情况下无法区分数组和对象。

正如很好地强调的那样杰克逊作者在评论中 https://stackoverflow.com/questions/39493173/how-to-convert-xml-to-json-using-only-jackson/39493394?noredirect=1#comment120312236_39493394, 杰克逊2.12最后改进的 XML 处理 https://cowtowncoder.medium.com/jackson-2-12-improved-xml-b9487889a23f，以便保留重复项if using JsonNode or Object作为目标类型。

使用 JSON.org

您还可以使用 JSON.org 来完成此操作：

String xml = "<?xml version=\"1.0\" encoding=\"UTF-8\"?>\n" +
             "<errors>\n" +
             "  <error>\n" +
             "    <status>400</status>\n" +
             "    <message>The field 'quantity' is invalid.</message>\n" +
             "    <details>\n" +
             "      <invalid_reason>The quantity specified is greater than the quantity of the product that is available to ship.</invalid_reason>\n" +
             "      <available_quantity>0</available_quantity>\n" +
             "      <order_product_id>12525</order_product_id>\n" +
             "    </details>\n" +
             "  </error>\n" +
             "</errors>";

String json = XML.toJSONObject(xml).toString();

需要以下依赖：

<dependency>
    <groupId>org.json</groupId>
    <artifactId>json</artifactId>
    <version>20160810</version>
</dependency>