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我正在编写一个将 JSON 返回给客户端的 JAX-RS Web 服务，并且我正在尝试使用这些org.codehaus.jackson库来处理 JSON 对象。我遇到的问题是我JsonNode在返回给客户端之前没有得到正确的序列化。响应中有一堆无关的属性。我想要JsonNode返回的最简单的 JSON 表示。

这是一个（人为的）示例：

@GET
@Path("user")
@Produces(MediaType.APPLICATION_JSON)
public JsonNode getUser() {
    ObjectNode node = JsonNodeFactory.instance.objectNode();
    node.put("user", "jDoe");
    return node;
}

对此的 JSON 响应非常难看：

{"object":true,"elements":[{"textual":true,"textValue":"jDoe","binaryValue":"jDoe","valueAsText":"jDoe","valueNode":true,"containerNode":false,"missingNode":false,"array":false,"object":false,"pojo":false,"number":false,"integralNumber":false,"floatingPointNumber":false,"int":false,"long":false,"double":false,"bigDecimal":false,"bigInteger":false,"boolean":false,"null":false,"binary":false,"booleanValue":false,"intValue":0,"longValue":0,"doubleValue":0.0,"decimalValue":0,"bigIntegerValue":0,"valueAsInt":0,"valueAsLong":0,"valueAsDouble":0.0,"valueAsBoolean":false,"elements":[],"fieldNames":[],"fields":[]}],"fieldNames":["user"],"fields":[{"key":"user","value":{"textual":true,"textValue":"jDoe","binaryValue":"jDoe","valueAsText":"jDoe","valueNode":true,"containerNode":false,"missingNode":false,"array":false,"object":false,"pojo":false,"number":false,"integralNumber":false,"floatingPointNumber":false,"int":false,"long":false,"double":false,"bigDecimal":false,"bigInteger":false,"boolean":false,"null":false,"binary":false,"booleanValue":false,"intValue":0,"longValue":0,"doubleValue":0.0,"decimalValue":0,"bigIntegerValue":0,"valueAsInt":0,"valueAsLong":0,"valueAsDouble":0.0,"valueAsBoolean":false,"elements":[],"fieldNames":[],"fields":[]}}],"containerNode":true,"valueNode":false,"missingNode":false,"array":false,"pojo":false,"number":false,"integralNumber":false,"floatingPointNumber":false,"int":false,"long":false,"double":false,"bigDecimal":false,"bigInteger":false,"textual":false,"boolean":false,"null":false,"binary":false,"booleanValue":false,"intValue":0,"longValue":0,"doubleValue":0.0,"decimalValue":0,"bigIntegerValue":0,"valueAsInt":0,"valueAsLong":0,"valueAsDouble":0.0,"valueAsBoolean":false}

我希望它简单

{"user":"jDoe"}

我错过了什么吗？我不明白为什么将简单的 JsonNode 序列化为 JSON 会出现问题。

对于它的价值，JSONObject从 JSON4J 库返回 a 按预期工作，但我更喜欢使用 Jackson 的库。

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1 回答 1

3

这应该按原样工作，所以我怀疑可能存在版本不兼容。由于 Jersey 使用 Jackson 1.x（1.8 或 1.9），因此您需要使用相同的版本；2.0 类位于不同的 Java 包中，以允许 1.x 和 2.x 共存（避免强制升级）。

因此，请确保您使用与 Jersey 相同的版本。

于 2012-09-27T00:11:56.250 回答