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Mapping a json string to an object with jackson will throw MismatchedInputException
How do I efficiently iterate over each entry in a Java Map?Sort a Map<Key, Value> by valuesHow do I read / convert an InputStream into a String in Java?Jackson with JSON: Unrecognized field, not marked as ignorableIgnoring new fields on JSON objects using JacksonHow do I convert a String to an int in Java?How to use Jackson to deserialise an array of objectsWhy is char[] preferred over String for passwords?How to tell Jackson to ignore a field during serialization if its value is null?Jackson @JsonProperty(required=true) doesn't throw an exception
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I have a simple class
public class AuthenticationToken
public String token;
public AuthenticationToken(String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
With jackson I am trying to map an string to this object like this
private String input = ""token":"adf"";
@Test
public void whenJsonString_ThenCreateAuthenticationObject() throws IOException
ObjectMapper jsonMapper = new ObjectMapper();
AuthenticationToken tokenObject = jsonMapper.readValue(input, AuthenticationToken.class);
assertThat(tokenObject).isNotNull();
But it throws the following exception
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `foo.AuthenticationToken` (although at least one Creator exists): cannot deserialize from Object value (no delegate- or property-based Creator)
at [Source: (String)""token":"adf""; line: 1, column: 2]
I tried to annotate the property in my AuthenticationToken as a @JsonProperty but this also resulted in in this exception.
java jackson
add a comment |
I have a simple class
public class AuthenticationToken
public String token;
public AuthenticationToken(String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
With jackson I am trying to map an string to this object like this
private String input = ""token":"adf"";
@Test
public void whenJsonString_ThenCreateAuthenticationObject() throws IOException
ObjectMapper jsonMapper = new ObjectMapper();
AuthenticationToken tokenObject = jsonMapper.readValue(input, AuthenticationToken.class);
assertThat(tokenObject).isNotNull();
But it throws the following exception
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `foo.AuthenticationToken` (although at least one Creator exists): cannot deserialize from Object value (no delegate- or property-based Creator)
at [Source: (String)""token":"adf""; line: 1, column: 2]
I tried to annotate the property in my AuthenticationToken as a @JsonProperty but this also resulted in in this exception.
java jackson
2
You need to define a default constructor.
– BackSlash
Mar 27 at 13:25
Jackson doesn't know how to call the constructor. You probably want to annotate it using@JsonCreator
and its argument(s) uisng@JsonProperty
, i.e.@JsonCreator AuthenticationToken(@JsonProperty("token") String token)
– Thomas
Mar 27 at 13:25
add a comment |
I have a simple class
public class AuthenticationToken
public String token;
public AuthenticationToken(String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
With jackson I am trying to map an string to this object like this
private String input = ""token":"adf"";
@Test
public void whenJsonString_ThenCreateAuthenticationObject() throws IOException
ObjectMapper jsonMapper = new ObjectMapper();
AuthenticationToken tokenObject = jsonMapper.readValue(input, AuthenticationToken.class);
assertThat(tokenObject).isNotNull();
But it throws the following exception
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `foo.AuthenticationToken` (although at least one Creator exists): cannot deserialize from Object value (no delegate- or property-based Creator)
at [Source: (String)""token":"adf""; line: 1, column: 2]
I tried to annotate the property in my AuthenticationToken as a @JsonProperty but this also resulted in in this exception.
java jackson
I have a simple class
public class AuthenticationToken
public String token;
public AuthenticationToken(String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
With jackson I am trying to map an string to this object like this
private String input = ""token":"adf"";
@Test
public void whenJsonString_ThenCreateAuthenticationObject() throws IOException
ObjectMapper jsonMapper = new ObjectMapper();
AuthenticationToken tokenObject = jsonMapper.readValue(input, AuthenticationToken.class);
assertThat(tokenObject).isNotNull();
But it throws the following exception
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of `foo.AuthenticationToken` (although at least one Creator exists): cannot deserialize from Object value (no delegate- or property-based Creator)
at [Source: (String)""token":"adf""; line: 1, column: 2]
I tried to annotate the property in my AuthenticationToken as a @JsonProperty but this also resulted in in this exception.
java jackson
java jackson
asked Mar 27 at 13:22
Al PhabaAl Phaba
2,5149 gold badges35 silver badges61 bronze badges
2,5149 gold badges35 silver badges61 bronze badges
2
You need to define a default constructor.
– BackSlash
Mar 27 at 13:25
Jackson doesn't know how to call the constructor. You probably want to annotate it using@JsonCreator
and its argument(s) uisng@JsonProperty
, i.e.@JsonCreator AuthenticationToken(@JsonProperty("token") String token)
– Thomas
Mar 27 at 13:25
add a comment |
2
You need to define a default constructor.
– BackSlash
Mar 27 at 13:25
Jackson doesn't know how to call the constructor. You probably want to annotate it using@JsonCreator
and its argument(s) uisng@JsonProperty
, i.e.@JsonCreator AuthenticationToken(@JsonProperty("token") String token)
– Thomas
Mar 27 at 13:25
2
2
You need to define a default constructor.
– BackSlash
Mar 27 at 13:25
You need to define a default constructor.
– BackSlash
Mar 27 at 13:25
Jackson doesn't know how to call the constructor. You probably want to annotate it using
@JsonCreator
and its argument(s) uisng @JsonProperty
, i.e. @JsonCreator AuthenticationToken(@JsonProperty("token") String token)
– Thomas
Mar 27 at 13:25
Jackson doesn't know how to call the constructor. You probably want to annotate it using
@JsonCreator
and its argument(s) uisng @JsonProperty
, i.e. @JsonCreator AuthenticationToken(@JsonProperty("token") String token)
– Thomas
Mar 27 at 13:25
add a comment |
2 Answers
2
active
oldest
votes
Jackson will by default expect an "empty" constructor and will automatically fill your Object by the getters and setters that are provided for each field.
So removing the arguments of your constructor will already solve your problem:
public class AuthenticationToken
public String token;
public AuthenticationToken()
public String getToken()
return token;
public void setToken(String token)
this.token = token;
You could also just add an additional empty constructor if you want to keep your current one as it is. Tested both options for your Test Case, both work fine.
add a comment |
Annotate the class constructor with @JsonCreator
Marker annotation that can be used to define constructors and factory
methods as one to use for instantiating new instances of the
associated class.
public class AuthenticationToken
public String token;
@JsonCreator
public AuthenticationToken(@JsonProperty("token") final String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Jackson will by default expect an "empty" constructor and will automatically fill your Object by the getters and setters that are provided for each field.
So removing the arguments of your constructor will already solve your problem:
public class AuthenticationToken
public String token;
public AuthenticationToken()
public String getToken()
return token;
public void setToken(String token)
this.token = token;
You could also just add an additional empty constructor if you want to keep your current one as it is. Tested both options for your Test Case, both work fine.
add a comment |
Jackson will by default expect an "empty" constructor and will automatically fill your Object by the getters and setters that are provided for each field.
So removing the arguments of your constructor will already solve your problem:
public class AuthenticationToken
public String token;
public AuthenticationToken()
public String getToken()
return token;
public void setToken(String token)
this.token = token;
You could also just add an additional empty constructor if you want to keep your current one as it is. Tested both options for your Test Case, both work fine.
add a comment |
Jackson will by default expect an "empty" constructor and will automatically fill your Object by the getters and setters that are provided for each field.
So removing the arguments of your constructor will already solve your problem:
public class AuthenticationToken
public String token;
public AuthenticationToken()
public String getToken()
return token;
public void setToken(String token)
this.token = token;
You could also just add an additional empty constructor if you want to keep your current one as it is. Tested both options for your Test Case, both work fine.
Jackson will by default expect an "empty" constructor and will automatically fill your Object by the getters and setters that are provided for each field.
So removing the arguments of your constructor will already solve your problem:
public class AuthenticationToken
public String token;
public AuthenticationToken()
public String getToken()
return token;
public void setToken(String token)
this.token = token;
You could also just add an additional empty constructor if you want to keep your current one as it is. Tested both options for your Test Case, both work fine.
edited Mar 27 at 13:54
answered Mar 27 at 13:48
T AT A
8021 gold badge9 silver badges17 bronze badges
8021 gold badge9 silver badges17 bronze badges
add a comment |
add a comment |
Annotate the class constructor with @JsonCreator
Marker annotation that can be used to define constructors and factory
methods as one to use for instantiating new instances of the
associated class.
public class AuthenticationToken
public String token;
@JsonCreator
public AuthenticationToken(@JsonProperty("token") final String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
add a comment |
Annotate the class constructor with @JsonCreator
Marker annotation that can be used to define constructors and factory
methods as one to use for instantiating new instances of the
associated class.
public class AuthenticationToken
public String token;
@JsonCreator
public AuthenticationToken(@JsonProperty("token") final String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
add a comment |
Annotate the class constructor with @JsonCreator
Marker annotation that can be used to define constructors and factory
methods as one to use for instantiating new instances of the
associated class.
public class AuthenticationToken
public String token;
@JsonCreator
public AuthenticationToken(@JsonProperty("token") final String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
Annotate the class constructor with @JsonCreator
Marker annotation that can be used to define constructors and factory
methods as one to use for instantiating new instances of the
associated class.
public class AuthenticationToken
public String token;
@JsonCreator
public AuthenticationToken(@JsonProperty("token") final String token)
this.token = token;
public String getToken()
return token;
public void setToken(String token)
this.token = token;
answered Mar 27 at 13:26
LppEddLppEdd
10.5k3 gold badges20 silver badges52 bronze badges
10.5k3 gold badges20 silver badges52 bronze badges
add a comment |
add a comment |
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2
You need to define a default constructor.
– BackSlash
Mar 27 at 13:25
Jackson doesn't know how to call the constructor. You probably want to annotate it using
@JsonCreator
and its argument(s) uisng@JsonProperty
, i.e.@JsonCreator AuthenticationToken(@JsonProperty("token") String token)
– Thomas
Mar 27 at 13:25