Java Jackson：忽略 Json 配置值
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Hymanson: Ignore Json configuration value
提问by tzulberti
I have the following json file:
我有以下 json 文件：
{
"segments": {
"externalId": 123,
"name": "Tomas Zulberti",
"shouldInform": true,
"id": 4
}
}
But the java model is as follows:
但是java模型如下：
public class Segment {
private String id;
private String name;
private boolean shouldInform;
// getter and setters here...
}
When Hymanson is parsing it raises an exception becuase there is no getter or setter for the field "externalId". It there a decorator that can be used to ignore a json field?
当 Hymanson 解析它时会引发异常，因为“externalId”字段没有 getter 或 setter。是否有可用于忽略 json 字段的装饰器？
采纳答案by StaxMan
You can use annotation @JsonIgnoreProperties; if it's just one value you want to skip, something like:
您可以使用注释@JsonIgnoreProperties；如果它只是您想跳过的一个值，例如：
@JsonIgnoreProperties({"externalId"})
or to ignore anything that can't be used:
或者忽略任何不能使用的东西：
@JsonIgnoreProperties(ignoreUnknown=true)
There are other ways to do it too, for rest check out FasterXML Hymanson wiki.
还有其他方法可以做到这一点，其余的请查看FasterXML Hymanson wiki。
回答by SamDJava
Also we can use mapper.enable(DeserializationFeature.FAIL_ON_IGNORED_PROPERTIES); instead @JsonIgnoreProperties(ignoreUnknown=true)
此外，我们可以使用mapper.enable（DeserializationFeature。FAIL_ON_IGNORED_PROPERTIES）; 相反@JsonIgnoreProperties(ignoreUnknown=true)
but for particular property we can use
但是对于特定的属性，我们可以使用
@JsonIgnoreProperties({"externalId"})
public class Segment {
private String id;
private String name;
private boolean shouldInform;
// getter and setters here...
}
