忽略杰克逊序列化的特定字段

发布于 2024-12-24 19:14:55 字数 535 浏览 0 评论 0原文

我正在使用杰克逊图书馆。

我想在序列化/反序列化时忽略特定字段，因此例如：

public static class Foo {
    public String foo = "a";
    public String bar = "b";

    @JsonIgnore
    public String foobar = "c";
}

应该给我：

{
foo: "a",
bar: "b",
}

但我得到：

{
foo: "a",
bar: "b",
foobar: "c"
}

我正在使用以下代码序列化对象：

ObjectMapper mapper = new ObjectMapper();
String out = mapper.writeValueAsString(new Foo());

我的类中字段的真实类型是Log4J 记录器类。我做错了什么？

原文

I'm using the Jackson library.

I want to ignore a specific field when serializing/deserializing, so for example:

public static class Foo {
    public String foo = "a";
    public String bar = "b";

    @JsonIgnore
    public String foobar = "c";
}

Should give me:

{
foo: "a",
bar: "b",
}

But I'm getting:

{
foo: "a",
bar: "b",
foobar: "c"
}

I'm serializing the object with this code:

ObjectMapper mapper = new ObjectMapper();
String out = mapper.writeValueAsString(new Foo());

The real type of the field on my class is an instance of the Log4J Logger class. What am I doing wrong?

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始于初秋 2024-12-31 19:14:55

好吧，由于某种原因，我错过了这个答案。

以下代码按预期工作：

@JsonIgnoreProperties({"foobar"})
public static class Foo {
    public String foo = "a";
    public String bar = "b";

    public String foobar = "c";
}

//Test code
ObjectMapper mapper = new ObjectMapper();
Foo foo = new Foo();
foo.foobar = "foobar";
foo.foo = "Foo";
String out = mapper.writeValueAsString(foo);
Foo f = mapper.readValue(out, Foo.class);

Ok, so for some reason I missed this answer.

The following code works as expected:

@JsonIgnoreProperties({"foobar"})
public static class Foo {
    public String foo = "a";
    public String bar = "b";

    public String foobar = "c";
}

//Test code
ObjectMapper mapper = new ObjectMapper();
Foo foo = new Foo();
foo.foobar = "foobar";
foo.foo = "Foo";
String out = mapper.writeValueAsString(foo);
Foo f = mapper.readValue(out, Foo.class);

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滴情不沾 2024-12-31 19:14:55

另外值得注意的是，此解决方案使用 DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES： https://stackoverflow.com/a/18850479/1256179

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过度放纵 2024-12-31 19:14:55

参考我如何告诉 Jackson 忽略我无法控制源代码的属性？

您可以使用 Jackson Mixins。例如：

class YourClass {
  public int ignoreThis() { return 0; }    
}

使用这个 Mixin

abstract class MixIn {
  @JsonIgnore abstract int ignoreThis(); // we don't need it!  
}

使用这个：

objectMapper.addMixIn(YourClass.class, MixIn.class);

Reference from How can I tell jackson to ignore a property for which I don't have control over the source code?

You can use Jackson Mixins. For example:

class YourClass {
  public int ignoreThis() { return 0; }    
}

With this Mixin

abstract class MixIn {
  @JsonIgnore abstract int ignoreThis(); // we don't need it!  
}

With this:

objectMapper.addMixIn(YourClass.class, MixIn.class);

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迟月 2024-12-31 19:14:55

尝试忽略 get 方法，而使用 set 方法。例如

@JsonIgnore
public String getPassword() {
    return _password;
}

@JsonProperty(JSON.PASSWORD)
public void setPassword(String value) {
    _password = value;
}

在我的测试用例中工作。

这个想法是反序列化器可以写入该值，但序列化器不应该看到该值。

try to ignore the get method, and use the set method. e.g.

@JsonIgnore
public String getPassword() {
    return _password;
}

@JsonProperty(JSON.PASSWORD)
public void setPassword(String value) {
    _password = value;
}

Worked in my test case.

The idea is the deserializer can write the value, but the serializer should not see the value.

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