使用@DateTimeFormat，客户端发送的请求在语法上不正确 [英] The request sent by the client was syntactically incorrect using @DateTimeFormat

问题描述

我有一个JSON格式的字符串，我使用HTTP-PUT将其发送到具有Spring MVC和Hibernate的服务器。

$ b

Controller < code $：
$ b $ pre $ @RequestMapping（value =/，method = RequestMethod.PUT）
public ResponseEntity< Map< String，Object>> myTest（
@RequestHeader（a）字符串a，
@RequestBody MyTestClass b）{...}


JSON ：

  {
number：123，
test：11/14
} 
  

test 是一个java.util.Date（MySQL - > date），并且我注释了POJO：



  @Column（name =TEST）
 @DateTimeFormat（pattern =MM / yy）
私人日期测试; 
  

因此 test 应格式化为月/年。但是我用Firefox RESTClient试过了，我总是得到
客户端发送的请求在语法上是不正确的。删除 test
看起来，用 @DateTimeFormat（pattern =MM / yy）看起来，code>

解决方案

因为您使用的是 RequestBody application / json 内容类型，Spring将使用它的 MappingJackson2HttpMessageConverter 将JSON转换为一个对象你的类型。但是，您提供的日期字符串 11/14 与任何预先配置的日期模式都不匹配，因此无法正确解析。执行这项工作的 MappingJackson2HttpMessageConverter ，或者更确切地说是 ObjectMapper 并不知道有关 @DateTimeFormat ，一个Spring注释。

您需要告诉Jackson您要使用哪个日期模式。您可以使用自定义日期反序列化器来完成这项工作。

  public class CustomDateDeserializer extends JsonDeserializer< Date> {
 @Override 
 public Date deserialize（JsonParser jp，DeserializationContext ctxt）
 throws IOException，JsonProcessingException {
 SimpleDateFormat format = new SimpleDateFormat（MM / yy）; 
 String date = jp.getText（）; 
 
尝试{
 return format.parse（date）; 
} catch（ParseException e）{
 throw new JsonParseException（e）; 
 
 
 
 
 
 
 $ b $ p然后简单地注释你的字段杰克逊知道如何反序列化它。
  @JsonDeserialize（使用= CustomDateDeserializer.class）
私人日期测试; 
  
您可以使用 @DateTimeFormat 使用带有 @ModelAttribute 的url编码表单参数。 Spring注册了一些转换器，可以将字符串值从请求参数转换为 Date 对象。 这在deocumentation中有描述。  
 
I have a JSON formatted String which I send with a HTTP-PUT to a server with Spring MVC and Hibernate.

Controller:
@RequestMapping(value = "/", method = RequestMethod.PUT)
public ResponseEntity<Map<String, Object>> myTest(
        @RequestHeader("a") String a,
        @RequestBody MyTestClass b) { … }
JSON:
{
 "number":"123",
 "test":"11/14"
}
test is a java.util.Date (MySQL -> date) and I annotated the POJO like this:
@Column(name = "TEST")
@DateTimeFormat(pattern = "MM/yy")
private Date test;
So test should be formatted as month/year. But I tried it with Firefox RESTClient and I always get 
The request sent by the client was syntactically incorrect. Removing test, everything is okay and works as expected.

So it seems, that with @DateTimeFormat(pattern = "MM/yy") is something wrong?
解决方案
Because you are using RequestBody with an application/json content type, Spring will use its MappingJackson2HttpMessageConverter to convert your JSON to an object of your type. However the date string you provide, 11/14 does not match any of the pre-configured date patterns and therefore it can't parse it correctly. The MappingJackson2HttpMessageConverter, or more specifically the ObjectMapper that does the job, doesn't know anything about @DateTimeFormat, a Spring annotation.

You will need to tell Jackson which date pattern you want to use. You can do so with a custom date deserializer 
public class CustomDateDeserializer extends JsonDeserializer<Date> {
    @Override
    public Date deserialize(JsonParser jp, DeserializationContext ctxt)
            throws IOException, JsonProcessingException {
        SimpleDateFormat format = new SimpleDateFormat("MM/yy");
        String date = jp.getText();

        try {
            return format.parse(date);
        } catch (ParseException e) {
            throw new JsonParseException(e);
        }
    }
}
Then simply annotate your field so that Jackson knows how to deserialize it.
@JsonDeserialize(using = CustomDateDeserializer.class)
private Date test;
You could use @DateTimeFormat if you were using url-encoded form parameters with @ModelAttribute. Spring registers some converters that can convert a String value from a request parameters into a Date object. This is described in the deocumentation.

                        
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