Create a JsonProcessingException

Question

I'm trying to create a JsonProcessingException to be thrown by a mock object.

when(mapper.writeValueAsString(any(Object.class))).thenThrow(new JsonProcessingException("Error"));

However I'm unable to create a JsonProcessingException object as all the constructors are protected. How do I get around this?

why don't you create a new class namely `MyJsonProcessingException` and inherit `JsonProcessingException` class? — R9J, May 27 '15 at 06:32

score 65 · Accepted Answer · answered Nov 13 '17 at 23:05

65

how about you create an anonymous exception of type JsonProcessingException

when(mapper.writeValueAsString(any(Object.class))).thenThrow(new JsonProcessingException("Error"){});

The {} braces does the trick. This is much better since it is not confusing to the reader of the test code.

answered Nov 13 '17 at 23:05

rajan.jana

5

this will do the trick (code will compile), but at the same time will result in `Checked exception is invalid for this method!`. Real solution will involve extending ObjectMapper in your test method and passing that to the SUT – senseiwu Dec 10 '18 at 15:09
1

im a bit confused, what is the backets doing? – Robbo_UK Jul 11 '19 at 09:50
2

@Robbo_UK It creates an anonymous class which extends `JsonProcessingException`. https://stackoverflow.com/questions/28073047/anonymous-initialization-of-class-with-protected-constructor – xli Apr 28 '20 at 23:17
1

While it might do the same thing behind the scenes, the `doThrow/when` syntax suggested @albin-p-babu is cleaner. – ithinkisam Oct 29 '20 at 19:10
I'm seeing the following error when I try this: `unreported exception com.fasterxml.jackson.core.JsonProcessingException; must be caught or declared to be thrown`. Any ideas how to fix this? – Beau Forest Dec 04 '22 at 04:04

score 20 · Answer 2 · answered May 27 '15 at 18:11

20

How about throwing one of the known direct subclasses instead?

Direct Known Subclasses:
JsonGenerationException, JsonMappingException, JsonParseException

Direct Known Subclasses:
JsonGenerationException, JsonParseException

answered May 27 '15 at 18:11

Jose Martinez

Works Perfectly! Thanks. – pez May 28 '15 at 00:03
this does not work at all, nor should it. JsonProcessingException is a checked exception, meaning that all subclasses of JsonProcessingException are checked exceptions as well – DanDan Aug 05 '19 at 10:11

score 14 · Answer 3 · edited Feb 25 '19 at 21:33

14

This one worked for me which allowed to throw JsonProcessingException itself

doThrow(JsonProcessingException.class).when(mockedObjectMapper).writeValueAsString(Mockito.any());

edited Feb 25 '19 at 21:33

PLASMA chicken

answered Feb 25 '19 at 19:25

Albin

score 8 · Answer 4 · edited Feb 27 '20 at 14:18

8

I came across this issue today also. The simplest and cleanest solution I came up with was to create and throw a mock JsonProcessingException

when(mapper.writeValueAsString(any(Object.class)))
    .thenThrow(mock(JsonProcessingException.class));

edited Feb 27 '20 at 14:18

spottedmahn

answered Jan 03 '20 at 23:56

Aaron McGhie

score 4 · Answer 5 · answered Feb 23 '20 at 15:36

4

Faced the same issue today. You can use:

Mockito.when(mockObjectMapper.writeValueAsString(Mockito.any())).thenThrow(JsonProcessingException.class);

answered Feb 23 '20 at 15:36

StoneCold

score -1 · Answer 6 · answered Jul 18 '19 at 03:41

-1

Try to use thenAnswer and create an anonymous class from JsonProcessingException

when(mapper.writeValueAsString(any(Object.class))).thenAnswer(x-> {throw new JsonProcessingException(""){};});

answered Jul 18 '19 at 03:41

yNuhp7cV0uPCbA

6 Answers6