How to use jackson to parse string convert to json in JAVA?

Question

I am trying to convert string into json using jackson.

I have used ObjectMapper().readValue() and deserialize into a DragDropJsonDto class

qList.getHeader() is like this "<p>{
 "RC" : ["N" , "Raj" , "USA"], 
"LC" : [ 
"Enter Country Name :" , 
"Enter State Name :", 
"Enter City Name :" ]
 }</p>"

public class DragDropJsonDto {

private List<String> RC;
private List<String> LC;

public List<String> getRC() {
    return RC;
}

public void setRC(List<String> RC) {
    this.RC = RC;
}

public List<String> getLC() {
    return LC;
}

public void setLC(List<String> LC) {
    this.LC = LC;
}

}

DragDropJsonDto dragDropJson = new ObjectMapper().readValue(qList.getHeader(), DragDropJsonDto.class);

I am not able to convert into json exception arises Error Unrecognized field "RC" (class com.visataar.lt.web.dto.DragDropJsonDto), not marked as ignorable (2 known properties: "rc", "lc"])

Does this answer your question? Does Jackson deserialise the second character to lowercase for a property — Joe
– Joe, Commented Jun 17, 2021 at 11:50

josejuan · Accepted Answer · 2021-06-17 11:54:56Z

1

By default jackson use lowercase, if you need RC and LC use:

private class DragDropJsonDto {
    ...
    @JsonProperty("RC")
    private List<String> RC;
    @JsonProperty("LC")
    private List<String> LC;
    ...
}

(or in getters)

answered Jun 17, 2021 at 11:54

josejuan

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sanjeevRm · Accepted Answer · 2021-06-17 11:57:51Z

0

You can set Fail on unknown properties as false to avoid this issue

ObjectMapper objectMapper = getObjectMapper();
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

answered Jun 17, 2021 at 11:57

sanjeevRm

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Comments

Gaurav Jeswani · Accepted Answer · 2021-06-17 12:37:19Z

0

As the qList.getHeader() response String Object Deserialization has 2 issues:

i) Response JSON String contains <p> and </p>. So you need to remove these sub-strings. As below:

String objectJson = qList.getHeader().replace("<p>", "").replace("</p>", "");
DragDropJsonDto f = m.readValue(objectJson, DragDropJsonDto.class);

ii) JSON String is having Keys in capital letters. As by default jackson uses small letters you need to use JsonProperty in DTO. As below:

  @JsonProperty("RC") 
  private List<String> RC;
  @JsonProperty("LC") 
  private List<String> LC;

answered Jun 17, 2021 at 12:37

Gaurav Jeswani

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