CODEWITHSUNDEEP
javaspring-bootjackson
Alex
Alex

Reputation: 131

How to deserialize JSON string when field name and/or value contain white space leading/tailing by using jackson

Input json: 

{"id   ": "   1"}

java class defineded:

class ID {
    String id;
}

how to correctly get the ID object with "id = 1" instead of "id = null"

Upvotes: 1

Views: 2711

Answers (2)

Alice Cooper
Alice Cooper

Reputation: 1

In my opinion, it is better to edit JSON beforehand and remove extra spaces.

var json = "{\"id   \": \"   1\"}";
json = json.replaceAll("\\s+", "");

You'll get: {"id":"1"} And then your POJO class will be correct.

After you can try to deserialize this json via Jackson. For example:

var id = mapper.readValue(json, ID.class);

Upvotes: 0

Artem Namednev
Artem Namednev

Reputation: 417

You may use @JsonProperty annotation like this:

class ID {

    @JsonProperty("id   ")
    String id;

    public String getId() {
        return id;
    }
}

And then:

String id = myIdClassInstance.getId().trim();

But this will work only in cases where the number of spaces is fixed.

Also, you can access the fields of the object only at clearly defined names.

If you really want that "id" and "id " to point to the same field, you will need to format the JSON manually before sending it.

Upvotes: 3

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