Searched on Google with the first line of a JAVA stack trace?

We can recommend more relevant solutions and speed up debugging when you paste your entire stack trace with the exception message. Try a sample exception.

Recommended solutions based on your search

Samebug tips

  1. ,

    you should add @JsonIgnoresProperties, This allows properties not defined in your POJO to be ignored. Needed if you are just trying to gamble your way through few properties in JSON. you do this by @JsonIgnoreProperties(property=true)

  2. ,

    You need to define another class for the information within ResObj {"ClientNum":"12345","ServerNum":"78945","IdNum":"020252"}. Otherwise jackson cannot determine how to deserialize.

Solutions on the web

via GitHub by tjwilson90
, 2 years ago
Unrecognized field "bar" (class ImmutableFoo$Json), not marked as ignorable (one known property: "isBar"]) at [Source: {"bar": true}; line: 1, column: 13] (through reference chain: Json["bar"])
via GitHub by ellisjoe
, 2 years ago
Unrecognized field "test" (class com.ImmutableTestClass$Json), not marked as ignorable (one known property: "isTest"]) at [Source: {"test":true}; line: 1, column: 13] (through reference chain: com.Json["test"])
via GitHub by mbuechner
, 2 years ago
Unrecognized field "claims" (class org.wikidata.wdtk.datamodel.json.jackson.JacksonPropertyDocument), not marked as ignorable (5 known properties: "datatype", "descriptions", "id", "aliases", "labels"]) at [Source: java.util.zip.GZIPInputStream@53a4
Empty query.