java 创建一个 JsonProcessingException
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Create a JsonProcessingException
提问by pez
I'm trying to create a JsonProcessingException to be thrown by a mock object.
我正在尝试创建一个由模拟对象抛出的 JsonProcessingException。
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?
但是我无法创建 JsonProcessingException 对象，因为所有构造函数都受到保护。我该如何解决这个问题？
回答by rajan.jana
how about you create an anonymous exception of type JsonProcessingException
你如何创建一个 JsonProcessingException 类型的匿名异常
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.
{} 大括号可以解决问题。这要好得多，因为它不会让测试代码的读者感到困惑。
回答by Jose Martinez
回答by ALBIN P BABU
This one worked for me which allowed to throw JsonProcessingException itself
这个对我有用，它允许自己抛出 JsonProcessingException
doThrow(JsonProcessingException.class).when(mockedObjectMapper).writeValueAsString(Mockito.any());
回答by Aaron McGhie
I came across this issue today also. The simplest and cleanest solution I came up with was to create and throw a mock JsonProcessingException
我今天也遇到了这个问题。我想出的最简单、最干净的解决方案是创建并抛出一个模拟 JsonProcessingException
when(mapper.writeValueAsString(any(Object.class)))
.thenThrow(mock(JsonProcessingException.class));
回答by StoneCold
Faced the same issue today. You can use:
今天遇到了同样的问题。您可以使用：
Mockito.when(mockObjectMapper.writeValueAsString(Mockito.any())).thenThrow(JsonProcessingException.class);
回答by Giau Ngo
Try to use thenAnswerand create an anonymous class from JsonProcessingException
尝试使用thenAnswer和创建一个匿名类JsonProcessingException
when(mapper.writeValueAsString(any(Object.class))).thenAnswer(x-> {throw new JsonProcessingException(""){};});
