blocks|key|213765|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|213766|我需要使用什么格式来执行与杰克逊的转换？Date是一个很好的字段类型吗？|blockquote|213767|213768|213769|在这方面，Date是一个很好的字段类型。通过使用ObjectMapper.setDateFormat，可以非常容易地对JSON进行解析|offset|length|style|CODE|213770|DateFormat+df+=+new+SimpleDateFormat("yyyy-MM-dd+HH:mm+a+z");
myObjectMapper.setDateFormat(df);|code-block|syntax|javascript|213771|一般来说，在变量被|213772|映射到对象成员之前，有没有一种方法来处理它们？例如，更改格式、计算等。|213773|213774|213775|是。您有几个选择，包括实现自定义的JsonDeserializer，例如扩展JsonDeserializer<Date>。This是一个很好的开始。|213776|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/FasterXML/jackson-docs/wiki/JacksonHowToCustomSerializers^0|0|0|0|0|5|4|O|Q|0|0|0|0|0|0|H|G|12|M|1P|4|0|0^^$0|@$1|2|3|-4|4|5|6|16|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|17|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|18|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|19|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|1A|7|@$H|1B|I|1C|J|K]|$H|1D|I|1E|J|K]]|8|@]|9|$]]|$1|L|3|M|4|N|6|1F|7|@]|8|@]|9|$O|P]]|$1|Q|3|R|4|5|6|1G|7|@]|8|@]|9|$]]|$1|S|3|T|4|C|6|1H|7|@]|8|@]|9|$]]|$1|U|3|-4|4|5|6|1I|7|@]|8|@]|9|$]]|$1|V|3|-4|4|5|6|1J|7|@]|8|@]|9|$]]|$1|W|3|X|4|5|6|1K|7|@$H|1L|I|1M|J|K]|$H|1N|I|1O|J|K]]|8|@$H|1P|I|1Q|1|1R]]|9|$]]|$1|Y|3|-4|4|5|6|1S|7|@]|8|@]|9|$]]]|Z|$10|$4|11|12|13|9|$14|15]]]]

<blockquote>
 What is the formatting I need to use to carry out conversion with Jackson? Is Date a good field type for this?
</blockquote>

<code>Date</code> is a fine field type for this. You can make the JSON parse-able pretty easily by using <code>ObjectMapper.setDateFormat</code>:

<pre><code>DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm a z");
myObjectMapper.setDateFormat(df);
</code></pre>

<blockquote>
 In general, is there a way to process the variables before they get mapped to Object members by Jackson? Something like, changing the format, calculations, etc.
</blockquote>

Yes. You have a few options, including implementing a custom <code>JsonDeserializer</code>, e.g. extending <code>JsonDeserializer&lt;Date&gt;</code>. <a href="https://github.com/FasterXML/jackson-docs/wiki/JacksonHowToCustomSerializers" rel="noreferrer">This</a> is a good start.

blocks|key|2856986|text|从Jackson+v2.0开始，你可以直接在对象成员上使用@JsonFormat注解；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2856987|@JsonFormat(shape+=+JsonFormat.Shape.STRING,+pattern+=+"yyyy-MM-dd+HH:mm+a+z")
private+Date+date;|code-block|syntax|javascript|2856988|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Since Jackson v2.0, you can use @JsonFormat annotation directly on Object members; 

<pre><code>@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm a z")
private Date date;
</code></pre>

blocks|key|213810|text|当然，还有一种称为序列化和反序列化的自动化方法，您也可以使用pb2q提到的特定注释(@JsonSerialize，@JsonDeserialize)来定义它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|213811|您可以同时使用java.util.Date和java.util.Calendar+...也许JodaTime也是如此。|213812|+@JsonFormat注解对我不起作用，因为我想要(它已经将时区调整为不同的值)在反序列化期间(序列化工作完美)：|213813|@JsonFormat(locale+=+"hu",+shape+=+JsonFormat.Shape.STRING,+pattern+=+"yyyy-MM-dd+HH:mm",+timezone+=+"CET")

@JsonFormat(locale+=+"hu",+shape+=+JsonFormat.Shape.STRING,+pattern+=+"yyyy-MM-dd+HH:mm",+timezone+=+"Europe/Budapest")|code-block|syntax|javascript|213814|如果您想要预测的结果，您需要使用自定义序列化程序和自定义反序列化程序，而不是@格式注释。我在这里找到了非常好的教程和解决方案http://www.baeldung.com/jackson-serialize-dates|213815|这里有Date字段的示例，但我需要日历字段，所以here是我的实现|213816|序列化程序类：|213817|public+class+CustomCalendarSerializer+extends+JsonSerializer<Calendar>+{

++++public+static+final+SimpleDateFormat+FORMATTER+=+new+SimpleDateFormat("yyyy-MM-dd+HH:mm");
++++public+static+final+Locale+LOCALE_HUNGARIAN+=+new+Locale("hu",+"HU");
++++public+static+final+TimeZone+LOCAL_TIME_ZONE+=+TimeZone.getTimeZone("Europe/Budapest");

++++@Override
++++public+void+serialize(Calendar+value,+JsonGenerator+gen,+SerializerProvider+arg2)
++++++++++++throws+IOException,+JsonProcessingException+{
++++++++if+(value+==+null)+{
++++++++++++gen.writeNull();
++++++++}+else+{
++++++++++++gen.writeString(FORMATTER.format(value.getTime()));
++++++++}
++++}
}|213818|deserializer类：|213819|public+class+CustomCalendarDeserializer+extends+JsonDeserializer<Calendar>+{

++++@Override
++++public+Calendar+deserialize(JsonParser+jsonparser,+DeserializationContext+context)
++++++++++++throws+IOException,+JsonProcessingException+{
++++++++String+dateAsString+=+jsonparser.getText();
++++++++try+{
++++++++++++Date+date+=+CustomCalendarSerializer.FORMATTER.parse(dateAsString);
++++++++++++Calendar+calendar+=+Calendar.getInstance(
++++++++++++++++CustomCalendarSerializer.LOCAL_TIME_ZONE,+
++++++++++++++++CustomCalendarSerializer.LOCALE_HUNGARIAN
++++++++++++);
++++++++++++calendar.setTime(date);
++++++++++++return+calendar;
++++++++}+catch+(ParseException+e)+{
++++++++++++throw+new+RuntimeException(e);
++++++++}
++++}
}|213820|和上述类的使用：|213821|public+class+CalendarEntry+{

++++@JsonSerialize(using+=+CustomCalendarSerializer.class)
++++@JsonDeserialize(using+=+CustomCalendarDeserializer.class)
++++private+Calendar+calendar;

++++//+...+additional+things+...
}|213822|使用此实现时，序列化和反序列化过程的执行将连续产生原始值。|213823|只使用@JsonFormat注释，反序列化给出了不同的结果，我认为，因为库的内部时区默认设置，你不能用注释参数改变(这也是我使用Jackson库2.5.3和2.6.3版本的经验)。|213824|entityMap|0|LINK|mutability|MUTABLE|url|http://www.baeldung.com/jackson-serialize-dates^0|16|E|1L|G|0|0|0|Q|U|3|15|5|0|0|13|5|1Q|1B|0|0|3|4|H|2|O|9|0|0|5|0|0|0|C|0|0|0|7|0|0|0|T|0|0|2I|0^^$0|@$1|2|3|4|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1L|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1M|8|@$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]]|D|@]|E|$]]|$1|J|3|K|5|L|7|1T|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1U|8|@$9|1V|A|1W|B|C]]|D|@$9|1X|A|1Y|1|1Z]]|E|$]]|$1|Q|3|R|5|6|7|20|8|@$9|21|A|22|B|C]|$9|23|A|24|B|C]|$9|25|A|26|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|27|8|@$9|28|A|29|B|C]]|D|@]|E|$]]|$1|U|3|V|5|L|7|2A|8|@]|D|@]|E|$M|N]]|$1|W|3|X|5|6|7|2B|8|@$9|2C|A|2D|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|L|7|2E|8|@]|D|@]|E|$M|N]]|$1|10|3|11|5|6|7|2F|8|@$9|2G|A|2H|B|C]]|D|@]|E|$]]|$1|12|3|13|5|L|7|2I|8|@]|D|@]|E|$M|N]]|$1|14|3|15|5|6|7|2J|8|@$9|2K|A|2L|B|C]]|D|@]|E|$]]|$1|16|3|17|5|6|7|2M|8|@$9|2N|A|2O|B|C]]|D|@]|E|$]]|$1|18|3|-4|5|6|7|2P|8|@]|D|@]|E|$]]]|19|$1A|$5|1B|1C|1D|E|$1E|1F]]]]

Of course there is an automated way called serialization and deserialization and you can define it with specific annotations (@JsonSerialize,@JsonDeserialize) as mentioned by pb2q as well.

You can use both java.util.Date and java.util.Calendar
... and probably JodaTime as well.

The @JsonFormat annotations not worked for me as I wanted (it has adjusted the timezone to different value) during deserialization (the serialization worked perfect):

<pre><code>@JsonFormat(locale = "hu", shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm", timezone = "CET")

@JsonFormat(locale = "hu", shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd HH:mm", timezone = "Europe/Budapest")
</code></pre>

You need to use custom serializer and custom deserializer instead of the @JsonFormat annotation if you want predicted result. I have found real good tutorial and solution here <a href="http://www.baeldung.com/jackson-serialize-dates">http://www.baeldung.com/jackson-serialize-dates</a>

There are examples for Date fields but I needed for Calendar fields so here is my implementation:

The serializer class:

<pre><code>public class CustomCalendarSerializer extends JsonSerializer&lt;Calendar&gt; {

 public static final SimpleDateFormat FORMATTER = new SimpleDateFormat("yyyy-MM-dd HH:mm");
 public static final Locale LOCALE_HUNGARIAN = new Locale("hu", "HU");
 public static final TimeZone LOCAL_TIME_ZONE = TimeZone.getTimeZone("Europe/Budapest");

 @Override
 public void serialize(Calendar value, JsonGenerator gen, SerializerProvider arg2)
 throws IOException, JsonProcessingException {
 if (value == null) {
 gen.writeNull();
 } else {
 gen.writeString(FORMATTER.format(value.getTime()));
 }
 }
}
</code></pre>

The deserializer class:

<pre><code>public class CustomCalendarDeserializer extends JsonDeserializer&lt;Calendar&gt; {

 @Override
 public Calendar deserialize(JsonParser jsonparser, DeserializationContext context)
 throws IOException, JsonProcessingException {
 String dateAsString = jsonparser.getText();
 try {
 Date date = CustomCalendarSerializer.FORMATTER.parse(dateAsString);
 Calendar calendar = Calendar.getInstance(
 CustomCalendarSerializer.LOCAL_TIME_ZONE, 
 CustomCalendarSerializer.LOCALE_HUNGARIAN
 );
 calendar.setTime(date);
 return calendar;
 } catch (ParseException e) {
 throw new RuntimeException(e);
 }
 }
}
</code></pre>

and the usage of the above classes:

<pre><code>public class CalendarEntry {

 @JsonSerialize(using = CustomCalendarSerializer.class)
 @JsonDeserialize(using = CustomCalendarDeserializer.class)
 private Calendar calendar;

 // ... additional things ...
}
</code></pre>

Using this implementation the execution of the serialization and deserialization process consecutively results the origin value.

Only using the @JsonFormat annotation the deserialization gives different result I think because of the library internal timezone default setup what you can not change with annotation parameters (that was my experience with Jackson library 2.5.3 and 2.6.3 version as well).

blocks|key|3070474|text|在@miklov-kriven非常有用的答案的基础上，我希望这两个额外的考虑点能证明对某些人有用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3070475|(1)我发现将序列化程序和反序列化程序作为静态内部类包含在同一个类中是一个很好的想法。注意，使用ThreadLocal实现SimpleDateFormat的线程安全。|3070476|public+class+DateConverter+{

++++private+static+final+ThreadLocal<SimpleDateFormat>+sdf+=+
++++++++ThreadLocal.<SimpleDateFormat>withInitial(
++++++++++++++++()+->+{return+new+SimpleDateFormat("yyyy-MM-dd+HH:mm+a+z");});

++++public+static+class+Serialize+extends+JsonSerializer<Date>+{
++++++++@Override
++++++++public+void+serialize(Date+value,+JsonGenerator+jgen+SerializerProvider+provider)+throws+Exception+{
++++++++++++if+(value+==+null)+{
++++++++++++++++jgen.writeNull();
++++++++++++}
++++++++++++else+{
++++++++++++++++jgen.writeString(sdf.get().format(value));
++++++++++++}
++++++++}
++++}

++++public+static+class+Deserialize+extends+JsonDeserializer<Date>+{
++++++++@Overrride
++++++++public+Date+deserialize(JsonParser+jp,+DeserializationContext+ctxt)+throws+Exception+{
++++++++++++String+dateAsString+=+jp.getText();
++++++++++++try+{
++++++++++++++++if+(Strings.isNullOrEmpty(dateAsString))+{
++++++++++++++++++++return+null;
++++++++++++++++}
++++++++++++++++else+{
++++++++++++++++++++return+new+Date(sdf.get().parse(dateAsString).getTime());
++++++++++++++++}
++++++++++++}
++++++++++++catch+(ParseException+pe)+{
++++++++++++++++throw+new+RuntimeException(pe);
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|3070477|(2)作为在每个单独的类成员上使用@JsonSerialize和@JsonSerialize注释的替代方法，您还可以考虑通过在应用程序级别应用自定义序列化来覆盖Jackson的默认序列化，也就是说，Jackson将使用这个自定义序列化来序列化Date类型的所有类成员，而不在每个字段上显式注释。例如，如果您使用的是Spring+Boot，一种方法如下：|3070478|@SpringBootApplication
public+class+Application+{
++++public+static+void+main(String[]+args)+{
++++++++SpringApplication.run(Application.class,+args);
++++}

++++@Bean
++++public+Module+customModule()+{
++++++++SimpleModule+module+=+new+SimpleModule();
++++++++module.addSerializer(Date.class,+new+DateConverter.Serialize());
++++++++module.addDeserializer(Date.class,+new+Dateconverter.Deserialize());
++++++++return+module;
++++}
}|3070479|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Building on @miklov-kriven's very helpful answer, I hope these two additional points of consideration prove helpful to someone:

(1) I find it a nice idea to include serializer and de-serializer as static inner classes in the same class. NB, using ThreadLocal for thread safety of SimpleDateFormat.

<pre><code>public class DateConverter {

 private static final ThreadLocal&lt;SimpleDateFormat&gt; sdf = 
 ThreadLocal.&lt;SimpleDateFormat&gt;withInitial(
 () -&gt; {return new SimpleDateFormat("yyyy-MM-dd HH:mm a z");});

 public static class Serialize extends JsonSerializer&lt;Date&gt; {
 @Override
 public void serialize(Date value, JsonGenerator jgen SerializerProvider provider) throws Exception {
 if (value == null) {
 jgen.writeNull();
 }
 else {
 jgen.writeString(sdf.get().format(value));
 }
 }
 }

 public static class Deserialize extends JsonDeserializer&lt;Date&gt; {
 @Overrride
 public Date deserialize(JsonParser jp, DeserializationContext ctxt) throws Exception {
 String dateAsString = jp.getText();
 try {
 if (Strings.isNullOrEmpty(dateAsString)) {
 return null;
 }
 else {
 return new Date(sdf.get().parse(dateAsString).getTime());
 }
 }
 catch (ParseException pe) {
 throw new RuntimeException(pe);
 }
 }
 }
}
</code></pre>

(2) As an alternative to using @JsonSerialize and @JsonDeserialize annotations on each individual class member you could also consider overriding Jackson's default serialization by applying the custom serialization at an application level, that is all class members of type Date will be serialized by Jackson using this custom serialization without explicit annotation on each field. If you are using Spring Boot for example one way to do this would as follows:

<pre><code>@SpringBootApplication
public class Application {
 public static void main(String[] args) {
 SpringApplication.run(Application.class, args);
 }

 @Bean
 public Module customModule() {
 SimpleModule module = new SimpleModule();
 module.addSerializer(Date.class, new DateConverter.Serialize());
 module.addDeserializer(Date.class, new Dateconverter.Deserialize());
 return module;
 }
}
</code></pre>

blocks|key|3070509|text|如果有人在使用java.sql.Date的自定义日期格式时遇到问题，这是最简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3070510|ObjectMapper+mapper+=+new+ObjectMapper();
SimpleModule+module+=+new+SimpleModule();
module.addSerializer(java.sql.Date.class,+new+DateSerializer());
mapper.registerModule(module);|code-block|syntax|javascript|3070511|(这个差强人意的回答省去了我很多麻烦：https://stackoverflow.com/a/35212795/3149048+)|offset|length|3070512|杰克逊默认使用java.sql.Date的SqlDateSerializer，但是目前，这个序列化程序没有考虑日期格式，请参阅这个问题：https://github.com/FasterXML/jackson-databind/issues/1407。解决方法是为java.sql.Date注册一个不同的序列化程序，如代码示例所示。|3070513|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/35212795/3149048|1|https://github.com/FasterXML/jackson-databind/issues/1407^0|0|0|J|18|0|0|1W|1L|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|K|3|L|5|6|7|12|8|@]|9|@$I|13|J|14|1|15]]|A|$]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]|U|$5|P|Q|R|A|$S|V]]]]

If anyone has problems with using a custom dateformat for java.sql.Date, this is the simplest solution:

<pre><code>ObjectMapper mapper = new ObjectMapper();
SimpleModule module = new SimpleModule();
module.addSerializer(java.sql.Date.class, new DateSerializer());
mapper.registerModule(module);
</code></pre>

(This SO-answer saved me a lot of trouble: <a href="https://stackoverflow.com/a/35212795/3149048">https://stackoverflow.com/a/35212795/3149048</a> )

Jackson uses the SqlDateSerializer by default for java.sql.Date, but currently, this serializer doesn't take the dateformat into account, see this issue: <a href="https://github.com/FasterXML/jackson-databind/issues/1407" rel="nofollow noreferrer">https://github.com/FasterXML/jackson-databind/issues/1407</a> . 
The workaround is to register a different serializer for java.sql.Date as shown in the code example.

blocks|key|215964|text|这是一个使用RFC3339+datetime格式的spring+boot应用程序的完整示例|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|215965|package+bj.demo;

import+com.fasterxml.jackson.databind.ObjectMapper;
import+org.springframework.beans.factory.annotation.Autowired;
import+org.springframework.boot.SpringApplication;
import+org.springframework.boot.autoconfigure.SpringBootApplication;
import+org.springframework.boot.context.event.ApplicationReadyEvent;
import+org.springframework.context.ApplicationListener;

import+java.text.SimpleDateFormat;

/**
+*+Created+by+BaiJiFeiLong@gmail.com+at+2018/5/4+10:22
+*/
@SpringBootApplication
public+class+BarApp+implements+ApplicationListener<ApplicationReadyEvent>+{

++++public+static+void+main(String[]+args)+{
++++++++SpringApplication.run(BarApp.class,+args);
++++}

++++@Autowired
++++private+ObjectMapper+objectMapper;

++++@Override
++++public+void+onApplicationEvent(ApplicationReadyEvent+applicationReadyEvent)+{
++++++++objectMapper.setDateFormat(new+SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssXXX"));
++++}
}|code-block|syntax|javascript|215966|entityMap^0|6|7|P|B|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|S|8|@]|E|@]|F|$J|K]]|$1|L|3|-4|5|6|7|T|8|@]|E|@]|F|$]]]|M|$]]

Just a complete example for spring boot application with <code>RFC3339</code> datetime format

<pre><code>package bj.demo;

import com.fasterxml.jackson.databind.ObjectMapper;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.context.event.ApplicationReadyEvent;
import org.springframework.context.ApplicationListener;

import java.text.SimpleDateFormat;

/**
 * Created by BaiJiFeiLong@gmail.com at 2018/5/4 10:22
 */
@SpringBootApplication
public class BarApp implements ApplicationListener&lt;ApplicationReadyEvent&gt; {

 public static void main(String[] args) {
 SpringApplication.run(BarApp.class, args);
 }

 @Autowired
 private ObjectMapper objectMapper;

 @Override
 public void onApplicationEvent(ApplicationReadyEvent applicationReadyEvent) {
 objectMapper.setDateFormat(new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssXXX"));
 }
}
</code></pre>

blocks|key|211192|text|我想指出的是，像另一个答案中描述的那样设置SimpleDateFormat只对java.util.Date有效，我假设这是问题中的意思。但对于java.sql.Date，格式化程序不起作用。在我的例子中，格式化程序不能工作的原因不是很明显，因为在应该序列化的模型中，字段实际上是一个java.utl.Date，但实际的对象最终是一个java.sql.Date。这是可能的，因为|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|211193|public+class+java.sql+extends+java.util.Date|code-block|syntax|javascript|211194|所以这实际上是有效的|211195|java.util.Date+date+=+new+java.sql.Date(1542381115815L);|211196|因此，如果您想知道为什么日期字段的格式不正确，请确保该对象确实是一个java.util.Date。|211197|还提到了为什么不添加处理java.sql.Date的Here。|211198|211199|这会破坏变化，我不认为这是有根据的。如果我们从头开始，我会同意这种改变，但事情并没有那么严重。|blockquote|211200|211201|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/FasterXML/jackson-databind/pull/1412^0|L|G|13|E|1Z|D|3X|D|4M|D|0|0|0|0|Y|E|0|C|D|Q|4|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1H|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@$9|1P|A|1Q|1|1R]]|E|$]]|$1|S|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]|$1|T|3|U|5|V|7|1T|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1V|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]]]

I want to point out that setting a <code>SimpleDateFormat</code> like described in the other answer only works for a <code>java.util.Date</code> which I assume is meant in the question.
But for <code>java.sql.Date</code> the formatter does not work.
In my case it was not very obvious why the formatter did not work because in the model which should be serialized the field was in fact a <code>java.utl.Date</code> but the actual object ended up beeing a <code>java.sql.Date</code>.
This is possible because

<pre><code>public class java.sql extends java.util.Date
</code></pre>

So this is actually valid

<pre><code>java.util.Date date = new java.sql.Date(1542381115815L);
</code></pre>

So if you are wondering why your Date field is not correctly formatted make sure that the object is really a <code>java.util.Date</code>.

<a href="https://github.com/FasterXML/jackson-databind/pull/1412" rel="nofollow noreferrer">Here</a> is also mentioned why handling <code>java.sql.Date</code> will not be added.

<blockquote>
 This would then be breaking change, and I don't think that is warranted. If we were starting from scratch I would agree with the change, but as things are not so much.
</blockquote>

blocks|key|213893|text|为我工作。SpringBoot。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|213894|+import+com.alibaba.fastjson.annotation.JSONField;

+@JSONField(format+=+"yyyy-MM-dd+HH:mm:ss")++
+private+Date+createTime;|code-block|syntax|javascript|213895|输出：|213896|{+
+++"createTime":+"2019-06-14+13:07:21"
}|213897|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Working for me. SpringBoot.

<pre><code> import com.alibaba.fastjson.annotation.JSONField;

 @JSONField(format = "yyyy-MM-dd HH:mm:ss") 
 private Date createTime;
</code></pre>

output:

<pre><code>{ 
 "createTime": "2019-06-14 13:07:21"
}
</code></pre>

blocks|key|214088|text|在日期中添加字符，如T和Z|type|unstyled|depth|inlineStyleRanges|entityRanges|data|214089|@JsonFormat(shape+=+JsonFormat.Shape.STRING,+pattern+=+"yyyy-MM-dd'T'HH:mm:ss'Z'")
private+Date+currentTime;|code-block|syntax|javascript|214090|输出|214091|{
++++"currentTime":+"2019-12-11T11:40:49Z"
}|214092|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

To add characters such as T and Z in your date

<pre><code>@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss'Z'")
private Date currentTime;
</code></pre>

output 

<pre><code>{
 "currentTime": "2019-12-11T11:40:49Z"
}
</code></pre>

I have a Date format coming from API like this:

<pre><code>"start_time": "2015-10-1 3:00 PM GMT+1:00"
</code></pre>

Which is YYYY-DD-MM HH:MM am/pm GMT timestamp. 
I am mapping this value to a Date variable in POJO. Obviously, its showing conversion error.

I would like to know 2 things:

<ol>
<li>What is the formatting I need to use to carry out conversion with Jackson? Is Date a good field type for this?</li>
<li>In general, is there a way to process the variables before they get mapped to Object members by Jackson? Something like, changing the format, calculations, etc.</li>
</ol>

Date format Mapping to JSON Jackson

我有一个来自API的日期格式，格式如下："start_time": "2015-10-1 3:00 PM GMT+1:00"这是YYYY-DD-MM HH:MM am/pm GMT时间戳。我将这个值映射到POJO中的一个日期变量。显然，它显示了转换错误。我想知道两件事：我需要使用什么格式来执行与杰克逊的转换？Date是...

问到JSON Jackson的日期格式映射
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