Exception deserializing JSON String to java.sql.Timestamp w Jackson

Question

I get this error when attempting to deserialize but I feel there is a deserializing interface i need to configure.

The error:

com.fasterxml.jackson.databind.exc.InvalidFormatException: Cannot deserialize value of type `java.sql.Timestamp` from String "2022-01-13 23:57:58": not a valid representation (error: Failed to parse Date value '2022-01-13 23:57:58': Cannot parse date "2022-01-13 23:57:58": while it seems to fit format 'yyyy-MM-dd'T'HH:mm:ss.SSSX', parsing fails (leniency? null))
 at [Source: (String)"{"endTime": "2022-01-13 23:57:58", "startTime": "2022-01-13 23:57:56", "numResults": 6, "numFailures": 0, "sampleResults": {"0": {"id": 0, "enabled": false, "resultsMap": {"tst": "FAIL", "tst2": "PASS"}, "containerId": null, "criteriaMap": {}}, "1": {"id": 1, "enabled": false, "resultsMap": {"tst": "FAIL", "tst2": "PASS"}, "containerId": null, "criteriaMap": {}}, "2": {"id": 2, "enabled": true, "resultsMap": {}, "containerId": null, "criteriaMap": {"ICCI_VE_OSW_B_Z01": "N_A", "ICCI_VE_OSW_B_Z02""[truncated 464 chars]; line: 1, column: 13] (through reference chain: filtec.test.TestRun["endTime"])
    at com.fasterxml.jackson.databind.exc.InvalidFormatException.from(InvalidFormatException.java:67)
    at com.fasterxml.jackson.databind.DeserializationContext.weirdStringException(DeserializationContext.java:1991)
    at com.fasterxml.jackson.databind.DeserializationContext.handleWeirdStringValue(DeserializationContext.java:1219)
    at com.fasterxml.jackson.databind.deser.std.StdDeserializer._parseDate(StdDeserializer.java:1261)
    ...
    at com.fasterxml.jackson.databind.deser.DefaultDeserializationContext.readRootValue(DefaultDeserializationContext.java:322)
    at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4675)
    at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3630)
    at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3598)
    at filtec.db.TestRunEntity.getData(TestRunEntity.java:79)

the calling method:

public TestRun getData() {
    TestRun data = null;
    try {
        data = MAPPER.readValue(jsonDoc, TestRun.class);
    } catch (JsonProcessingException e) {
        ApplicationManager.logStackTrace(e);
    }
    return data;
}

my serializer:

public class JsonTimestampSerializer extends JsonSerializer<Timestamp> {
    @Override
    public void serialize(Timestamp value, JsonGenerator gen, SerializerProvider serializers) throws IOException {
        try {
            SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
            String s = sdf.format(value);
            gen.writeString(s);
        } catch (DateTimeParseException e) {
            System.err.println(e);
            gen.writeString("");
        }
    }
}

and my TestRun POJO fields:

public class TestRun {
    //...
    /** Internal state of a test run. */
    enum State {
        CREATED,
        IN_PROGRESS,
        COMPLETE
    }
    
    /** State of the test session. */
    State state = State.CREATED;
    
    /** The start time of the run. */
    @JsonSerialize(using = JsonTimestampSerializer.class)
    Timestamp startTime;
    
    /** The end time of the test run. */
    @JsonSerialize(using = JsonTimestampSerializer.class)
    Timestamp endTime;

Where's your de-serializer? Also, sdf.format(value) can't throw DateTimeParseException... have you confused serialization and de-serialization? — Jim Garrison
– Jim Garrison, Commented Jan 18, 2022 at 1:05
Do I need to add a deserializer with a class such as: fasterxml.github.io/jackson-databind/javadoc/2.7/com/fasterxml/… ? — simgineer
– simgineer, Commented Jan 18, 2022 at 1:07
Basic principle: If you serialized something yourself, unless you duplicated an existing built-in serializer (why would you do that), you will have to provide the corresponding de-serializer. DateTimeParserException, cannot occur here, but could occur in a deserializer. — Jim Garrison
– Jim Garrison, Commented Jan 18, 2022 at 1:19

simgineer · Accepted Answer · 2022-01-18 02:19:31Z

2

Looks like a solution was to add a deserializer but an even more ideal solution per Jim's suggestion is to use a built-in serializer which works for the string format I am using.

@JsonFormat(shape=JsonFormat.Shape.STRING, pattern="yyyy-MM-dd HH:mm:ss")
Timestamp startTime;

edited Jan 18, 2022 at 2:19

answered Jan 18, 2022 at 1:45

simgineer

1,9262 gold badges24 silver badges54 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Exception deserializing JSON String to java.sql.Timestamp w Jackson

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related