Jackson and JSON in Java, finally learn with a coding-first approach:
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1. Overview
In this short tutorial, weβll shed light on how to fix the exception JsonMappingException: Can not deserialize instance of java.util.ArrayList from Object value (token `JsonToken.START_OBJECT`).
First, weβll highlight the main cause of the exception. Then weβll demonstrate how to reproduce it in practice, and finally, how to solve it.
2. Understanding the Exception
Typically, Jackson throws JsonMappingException to signal a fatal mapping error when deserializing a JSON string. Therefore, the stack trace βCan not deserialize instance of java.util.ArrayListβ indicates that Jackson failed to map a JSON property into an instance of ArrayList.
In short, the most common cause of this exception is using curly braces β{β¦}β to denote a collection instead of brackets β[β¦]β.
3. Reproducing the Exception
Now that we know what causes Jackson to throw JsonMappingException, letβs see how to reproduce it using a practical example.
Letβs consider the Country class:
public class Country {
private String name;
private List<String> cities;
// standard getters and setters
}
As we can see, a country is defined by a name and a list of cities.
Next, letβs pretend to use curly braces to define cities in a JSON string:
{
"name": "Netherlands",
"cities": {"Amsterdam", "Tamassint"}
}
Now trying to deserialize the JSON string into an object of Country type will result in the following:
Cannot deserialize value of type `java.util.ArrayList<java.lang.String>` from Object value (token `JsonToken.START_OBJECT`)
at [Source: (String)"{"name":"Netherlands","cities":{"Amsterdam", "Tamassint"}}"; line: 1, column: 32]
(through reference chain: com.baeldung.mappingexception.Country["cities"])
...
Finally, weβll create a test case to confirm this:
@Test
public final void givenJsonWithInvalidList_whenDeserializing_thenThrowException() throws JsonParseException, IOException {
String json = "{\"name\":\"Netherlands\",\"cities\":{\"Amsterdam\", \"Tamassint\"}}";
ObjectMapper mapper = new ObjectMapper();
Exception exception = assertThrows(JsonMappingException.class, () -> mapper.reader()
.forType(Country.class)
.readValue(json));
assertTrue(exception.getMessage()
.contains("Cannot deserialize value of type `java.util.ArrayList<java.lang.String>`"));
}
As shown above, Jackson fails with βCannot deserialize value of type `java.util.ArrayList<java.lang.String>`β.
The main reason here is that we used curly braces to represent a list of cities. For Jackson, {βAmsterdamβ, βTamassintβ} isnβt a JSON array.
4. Fixing the Exception
The easiest way to avoid the exception is to use brackets instead of curly braces to define a collection of elements. So to solve the exception, we need to fix our JSON string first:
{
"name": "Netherlands",
"cities": ["Amsterdam", "Tamassint"]
}
Now letβs verify that everything works as excepted using a test case:
@Test
public final void givenJsonWithValidList_whenDeserializing_thenCorrect() throws JsonParseException, IOException {
String json = "{\"name\":\"Netherlands\",\"cities\":[\"Amsterdam\", \"Tamassint\"]}";
ObjectMapper mapper = new ObjectMapper();
Country country = mapper.reader()
.forType(Country.class)
.readValue(json);
assertEquals("Netherlands", country.getName());
assertEquals(Arrays.asList("Amsterdam", "Tamassint"), country.getCities());
}
As we can see, the new JSON string is successfully deserialized into a Country object.
5. Conclusion
In this brief article, we discussed the main cause of the JsonMappingException: Can not deserialize instance of java.util.ArrayList from Object value (token `JsonToken.START_OBJECT`). Then we showcased how to produce the exception, and how to solve it.
