Cannot deserialize value of type `java.util.Date` from String

Question

Using Spring 1.5.8.RELEASE Jackson mapper giving the following exception.

Cannot deserialize value of type `java.util.Date` from String "2018-09-04T10:44:46": expected format "yyyy-MM-dd'T'HH:mm:ss.SSS"

at [Source: UNKNOWN; line: -1, column: -1] (through reference chain: com.copart.conversationapi.calldisposition.model.vo.CallLogEntity["callEndTime"])

CallEntity.java

@JsonProperty("callEndTime")
@Column(name = "call_end_ts")
@JsonFormat(pattern="yyyy-MM-dd'T'HH:mm:ss.SSS")
private Date callEndTime;

DAO.java

ObjectMapper mapper = new ObjectMapper();
HashMap<String, Object> finalHashMap;
finalHashMap = convertMultiToString(requestMap);
mapper.enable(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY);
CallLogEntity callLogEntity = mapper.convertValue(finalHashMap, CallEntity.class);

pom.xml

<dependency>
     <groupId>com.fasterxml.jackson.core</groupId>
     <artifactId>jackson-databind</artifactId>
     <version>2.9.0</version>
     <exclusions>
        <exclusion>
           <groupId>com.fasterxml.jackson.core</groupId>
           <artifactId>jackson-core</artifactId>
        </exclusion>
        <exclusion>
           <groupId>com.fasterxml.jackson.core</groupId>
           <artifactId>jackson-annotations</artifactId>
        </exclusion>
     </exclusions>
  </dependency>

mkhanoyan · Accepted Answer · 2019-03-18 18:19:38Z

79

Change your @JsonFormat line to this.

@JsonFormat(pattern="yyyy-MM-dd'T'HH:mm:ss")

The format pattern you have right now expects the sting to have millisecond values - but your example string doesn't have them.

answered Mar 18, 2019 at 18:19

mkhanoyan

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2 Comments

Utkarsh Gandhi Over a year ago

Even I pass milliseconds, then also it throws an exception. Though milliseconds are not so important to me and this annotation worked perfectly. Thanks!

dokam_scotland Over a year ago

Then you can define any format that you want like on my case - @JsonFormat(pattern="yyyy-MM-dd HH:mm:ss")

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