Java 将当前时区设置为 @JsonFormat 时区值
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Set Current TimeZone to @JsonFormat timezone value
提问by Vishnu K
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy", timezone = "Asia/Kolkata")
private Date activationDate;
Hi Guys,
嗨，大家好，
From the above java code, i want to set timezone value as Current System timezone using below TimeZone.getDefault().getID()- it returns value as "Asia/Kolkata"
从上面的 java 代码中，我想使用下面的TimeZone.getDefault().getID()将时区值设置为当前系统时区 - 它返回值作为“亚洲/加尔各答”
But if i set this code to json format
但是如果我将此代码设置为 json 格式
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy", timezone = TimeZone.getDefault().getID())
I am getting error like "The value for annotation attribute JsonFormat.timezone must be a constant expression"
我收到类似“注释属性 JsonFormat.timezone 的值必须是常量表达式”之类的错误
Pls help me to solve this issue.
请帮我解决这个问题。
Thanks in advance, Vishnu
提前致谢，毗湿奴
回答by Raju Sharma
You cannot assign timezonevalue a dynamic or a runtime value. It should be constant or a compile time value and enumstoo accepted.
您不能为时区值分配动态值或运行时值。它应该是常量或编译时值，并且枚举也被接受。
So you should assign a constant to timezone. like below.
所以你应该为timezone分配一个常量。像下面。
private static final String MY_TIME_ZONE="Asia/Kolkata";
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy", timezone = MY_TIME_ZONE);
回答by 0__1
You can use enumeration in order to possibly enrich you time zones that you would use. A solution using enumeration is the following enumeration class implementation.
您可以使用枚举来丰富您将使用的时区。使用枚举的解决方案是以下枚举类实现。
package <your package goes here>;
import java.util.TimeZone;
public enum TimeZoneEnum {
DEFAULT(TimeZone.getDefault()),
ASIA_KOLKATA = (TimeZone.getTimeZone("Africa/Abidjan")),
//other timezones you maybe need
...
private final TimeZone tz;
private TimeZoneEnum(final TimeZone tz)
{
this.tz = tz;
}
public final TimeZone getTimeZone()
{
return tz;
}
}
Then you can utilize you enumeration like below:
然后你可以使用你的枚举，如下所示：
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy", timezone = TimeZoneEnum.ASIA_KOLKATA )
回答by xonya
You can use JsonFormat.DEFAULT_TIMEZONE, after properly configuring the ObjectMapper:
您可以使用JsonFormat.DEFAULT_TIMEZONE，经过正确配置ObjectMapper：
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy", timezone = JsonFormat.DEFAULT_TIMEZONE)
From the docs:
从文档：
Value that indicates that default TimeZone (from deserialization or serialization context) should be used: annotation does not define value to use.
NOTE: default here does NOT mean JVM defaults but Hymanson databindings default, usually UTC, but may be changed on ObjectMapper.
指示应使用默认 TimeZone（来自反序列化或序列化上下文）的值：注释未定义要使用的值。
注意：这里的默认值并不意味着 JVM 默认值，而是 Hymanson 数据绑定默认值，通常是 UTC，但可能会在 ObjectMapper 上更改。
In order to configure the ObjectMapper:
为了配置ObjectMapper：
@Configuration
public class MyApp {
@Autowired
public void configureHymanson(ObjectMapper objectMapper) {
objectMapper.setTimeZone(TimeZone.getDefault());
}
}
To set the default TimeZone on your application use this JVM property:
要在您的应用程序上设置默认时区，请使用此 JVM 属性：
-Duser.timezone=Asia/Kolkata
