javajackson对象Map器在正确的变量名上失败

mbyulnm0  于 2021-07-12  发布在  Java
关注(0)|答案(2)|浏览(385)

注意:我知道变量的名称不正确。我正在清理另一个开发人员的旧代码,他将这些代码列为成员变量。他们将被正确命名,是的,我知道我需要注解,但我在这里要找的是一个解释,为什么变量是employeeid,它在寻找employeeid,但找到一个不存在的变量memployeeid。顺便说一句,这种情况发生在服务器上,而不仅仅是在ide中。
出于某种未知的原因-在修改变量名、保存、重新启动、使缓存失效等之后-jackson无法Map到propper变量名。为此,我使用intellij和java。
错误:

Exception in badge service getting by badgeId: Unrecognized field 
"EmployeeId" (class com.companyName.attendance.entity.DTOs.badgeservice.BadgeDTO), not marked as ignorable (6 known properties: "mBadgeId", "mEmployeeId", "mLanId", "mFirstName", "mEmail", "mLastName"]) 
at [Source: (String)"{"backoff":null,"error_id":null,"error_message":null,"error_name":null,"has_more":false,"items":[{"EmployeeId":"888888","LanId":"NTest","FirstName":"Name","MiddleName":null,"LastName":"Test","Email":null,"Location":null,"Title":null,"Phone":null,"DepartmentId":null,"DepartmentName":null,"DepartmentNumber":null,"Groups":null,"Found":false,"BadgeId":"222222","JobTitle":null,"Picture":null,"OrgUrl":null,"Manager":null,"Coworkers":null,"DirectReports":null}],"page":null,"page_size":null,"quo"[truncated 61 chars]; line: 1, column: 113] (through reference chain: com.companyName.attendance.entity.DTOs.badgeservice.BadgeServiceResponseDTO["items"]->java.util.ArrayList[0]->com.companyName.attendance.entity.DTOs.badgeservice.BadgeDTO["EmployeeId"])

如您所见,它没有在dto中找到employeeid,而是说期望的feld是memployeeid。但是,下面是我的dto变量声明:

public class BadgeDTO {
//TODO: Convert member variables to proper practice names
String BadgeId;
String EmployeeId;
String FirstName;
String LanId;
String LastName;
String Email;

public BadgeDTO(String BadgeId, String EmployeeId, String FirstName, 
String LanId, String LastName, String Email) {
    super();
    this.BadgeId = BadgeId;
    this.EmployeeId = EmployeeId;
    this.FirstName = FirstName;
    this.LanId = LanId;
    this.LastName = LastName;
    this.Email = Email;
}
public BadgeDTO() {
    super();
}
public String getmBadgeId() {
    return BadgeId;
}
public void setmBadgeId(String BadgeId) {
    this.BadgeId = BadgeId;
}
public String getmEmployeeId() {
    return EmployeeId;
}
public void setmEmployeeId(String EmployeeId) {
    this.EmployeeId = EmployeeId;
}
public String getmFirstname() {
    return FirstName;
}
public void setmFirstName(String FirstName) {
    this.FirstName = FirstName;
}
public String getmLanId() {
    return LanId;
}
public void setmLanId(String LanId) {
    this.LanId = LanId;
}
public String getmLastName() {
    return LastName;
}
public void setmLastName(String LastName) {
    this.LastName = LastName;
}
public String getmEmail() {
    return Email;
}
public void setmEmail(String Email) {
    this.Email = Email;
}
@Override
public String toString() {
    return "BadgeDTO [BadgeId=" + BadgeId + ", EmployeeId=" + EmployeeId + 
", Firstname=" + FirstName
            + ", LanId=" + LanId + ", LastName=" + LastName + ", Email=" + 
Email + "]";
}
}

现在最疯狂的是,做以下工作:

@JsonProperty("BadgeId")
String BadgeId;
@JsonProperty("EmployeeId")
String EmployeeId;
@JsonProperty("FirstName")
String FirstName;

json码:

{
"backoff": null,
"error_id": null,
"error_message": null,
"error_name": null,
"has_more": false,
"items": [
    {
        "EmployeeId": "888888",
        "LanId": "TName",
        "FirstName": "Test",
        "MiddleName": null,
        "LastName": "Name",
        "Email": null,
        "Location": null,
        "Title": null,
        "Phone": null,
        "DepartmentId": null,
        "DepartmentName": null,
        "DepartmentNumber": null,
        "Groups": null,
        "Found": false,
        "BadgeId": "222222",
        "JobTitle": null,
        "Picture": null,
        "OrgUrl": null,
        "Manager": null,
        "Coworkers": null,
        "DirectReports": null
    }
],
"page": null,
"page_size": null,
"quota_max": null,
"quota_remaining": null,
"total": null,
"type": null
}

因此,我已经使用上面的jsonproperty注解解决了这个问题,但是为什么它仍然在寻找memployeeid,而它在我的整个代码中不存在呢?我想失效/重启可以解决这个问题,但它没有

6qftjkof

6qftjkof1#

答案通常是:注意你写的代码。
getter和setter将字段公开为m*(其中*是实际的字段名)。
看看这个:

getmEmployeeId(
   ^
   |

这是字母“m”。
因此,这将字段名公开为“memployeeid”
更多详细信息:
方法返回的值与getter公开的值的名称无关。java要求getter名称的格式为“getfieldname”,setter nanes的格式为“setfieldname”,其中“fieldname”是任何不是方法名的“get”或“set”部分的值。
这被称为“javabean命名约定”,如果您使用java编写代码并使用任何第三方java库,那么您必须理解并遵守它。

ff29svar

ff29svar2#

匹配 JSON 钥匙 POJO 属性 Jackson 使用一种叫做 PropertyNamingStrategy . 在你的 JSON 我们至少可以找到两种策略:
蛇壳( page_size , error_message 等)
上\u Camel \u箱( EmployeeId , DepartmentId 等)
从另一边 POJO 类提供第三种策略:
“m”+大写\u( mEmployeeId , mDepartmentId 等)
这就是为什么 JSON 不匹配 POJO . 要使其发挥作用,您需要实施如下所示的新战略:

class MNamingStrategy extends PropertyNamingStrategy {

    @Override
    public String nameForSetterMethod(MapperConfig<?> config, AnnotatedMethod method,
        String defaultName) {
        return defaultName.substring(1); // remove first `m` letter
    }
}

您可以这样使用它:

@JsonNaming(MNamingStrategy.class)
class Clazz {

    private int Id = 11;

    public int getmId() {
        return Id;
    }

    public void setmId(int id) {
        this.Id = id;
    }

    @Override
    public String toString() {
        return "Clazz{" +
            "Id=" + Id +
            '}';
    }
}

从现在起,您可以反序列化上面的内容 JSON 给予 POJO .
当您添加 @JsonProperty 带有已告知的属性名称的注解 Jackson 使用自定义Map。
另请参见:
Spring Jackson酒店命名策略
物业管理策略
更多Jackson注解

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