Java Jackson无法识别字段,即使它为';她在班上

Java Jackson无法识别字段,即使它为';她在班上,java,json,file,jackson,Java,Json,File,Jackson,我有这门课: public static class Person { public Person(){ } public String get_id() { return _id; } public int getIndex() { return index; } public String getGuid() { r

我有这门课:

public static class Person {

    public Person(){

    }

        public String get_id() {
            return _id;
        }

        public int getIndex() {
            return index;
        }

        public String getGuid() {
            return guid;
        }

        public boolean isActive() {
            return isActive;
        }

        public String getBalance() {
            return balance;
        }

        public String getPicture() {
            return picture;
        }

        public int getAge() {
            return age;
        }

        public String getEyeColor() {
            return eyeColor;
        }

        public String getName() {
            return name;
        }

        public String getGender() {
            return gender;
        }

        public String getCompany() {
            return company;
        }

        public String getEmail() {
            return email;
        }

        public String getPhone() {
            return phone;
        }

        public String getAddress() {
            return address;
        }

        public String getAbout() {
            return about;
        }

        public String getRegistered() {
            return registered;
        }

        public double getLatitude() {
            return latitude;
        }

        public double getLongitude() {
            return longitude;
        }

        public List<String> getTags() {
            return tags;
        }

        public List<App.Friend> getFriends() {
            return friends;
        }

        public String getGreeting() {
            return greeting;
        }

        public String getFavoriteFruit() {
            return favoriteFruit;
        }

        private boolean isActive;
        private String _id;
        private int index;
        private String guid;
        private String balance;
        private String picture;
        private int age;
        private String eyeColor;
        private String name;
        private String gender;
        private String company;
        private String email;
        private String phone;
        private String address;
        private String about;
        private String registered;
        private double latitude;
        private double longitude;
        private List<String> tags;
        private List<App.Friend> friends;
        private String greeting;
        private String favoriteFruit;

    @Override
    public String toString() {
        return "Person{" +
                "_id='" + _id + '\'' +
                ", index=" + index +
                ", guid='" + guid + '\'' +
                ", isActive=" + isActive +
                ", balance='" + balance + '\'' +
                ", picture='" + picture + '\'' +
                ", age=" + age +
                ", eyeColor='" + eyeColor + '\'' +
                ", name='" + name + '\'' +
                ", gender='" + gender + '\'' +
                ", company='" + company + '\'' +
                ", email='" + email + '\'' +
                ", phone='" + phone + '\'' +
                ", address='" + address + '\'' +
                ", about='" + about + '\'' +
                ", registered='" + registered + '\'' +
                ", latitude=" + latitude +
                ", longitude=" + longitude +
                ", tags=" + tags +
                ", friends=" + friends +
                ", greeting='" + greeting + '\'' +
                ", favoriteFruit='" + favoriteFruit + '\'' +
                '}';
    }
}

因此,我排除了注释中建议的格式错误的json文件的可能性,解决方案非常简单,我只需将
isActive
字段的getter和setter方法的名称更改为
getIsActive
setIsActive
(IntelliJ IDEA使用不同的名称自动创建了这些方法

当您使用IDE的generate选项创建getter和setter时,它将使用不同的名称创建getter和setter

这个问题可能是在我们使用isActive时产生的,isDeleted变量中的名称类型

手动为此类型的字段创建getter setter以解决此问题

    getIsActive(); 
    setIsActive();

访问器方法的预期名称可能是
getIsActive
setIsActive
。是的,现在它可以工作了,非常感谢@Berger:)如果您无法修改该类,请参见我的回答:
boolean isActive;

public boolean isActive() {
    return isActive;
}

public void setActive(boolean isActive) {
    this.isActive = isActive;
}
    getIsActive(); 
    setIsActive();