Hooli Hooli - 6 months ago 8
Java Question

How do I serialize & deserialize CSV properly?

I've been trying to serialize an object to a CSV

String
but the object contains a
List
and
@JsonUnwrapped
doesn't work on
List
objects.

Expected sample output:

color,part.name\n
red,gearbox\n
red,door\n
red,bumper


Actual output:

com.fasterxml.jackson.core.JsonGenerationException: Unrecognized column 'name':


Here is my code: (Most of it is the 2 POJO's)

import com.fasterxml.jackson.annotation.JsonAutoDetect;
import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.annotation.JsonPropertyOrder;
import com.fasterxml.jackson.annotation.JsonRootName;
import com.fasterxml.jackson.dataformat.csv.CsvMapper;
import com.fasterxml.jackson.dataformat.csv.CsvSchema;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlElementWrapper;
import java.io.IOException;
import static java.util.Arrays.asList;
import java.util.List;
import java.util.logging.Level;
import java.util.logging.Logger;

public class NestedWrapping {

@JsonRootName("Car")
@JsonInclude(JsonInclude.Include.NON_DEFAULT)
@JsonAutoDetect(fieldVisibility = JsonAutoDetect.Visibility.ANY, getterVisibility = JsonAutoDetect.Visibility.NONE, setterVisibility = JsonAutoDetect.Visibility.NONE)
@JsonPropertyOrder({"color"})
public static class Car {

@JsonProperty("color")
private String color;

@JsonFormat(shape = JsonFormat.Shape.STRING)
@JacksonXmlElementWrapper(useWrapping = false)
private List<Part> parts;

public String getColor() {
return color;
}

public void setColor(String color) {
this.color = color;
}

public List<Part> getParts() {
return parts;
}

public void setParts(List<Part> parts) {
this.parts = parts;
}

}

@JsonInclude(JsonInclude.Include.NON_DEFAULT)
@JsonAutoDetect(fieldVisibility = JsonAutoDetect.Visibility.ANY, getterVisibility = JsonAutoDetect.Visibility.NONE, setterVisibility = JsonAutoDetect.Visibility.NONE)
@JsonPropertyOrder({
"name"
})
public static class Part {

@JsonProperty("name")
private String name;

public String getName() {
return name;
}

public void setName(String name) {
this.name = name;
}

}

public static void main(String args[]) {
try {
Car car = new Car();
car.setColor("red");
Part part1 = new Part();
part1.setName("geabox");
Part part2 = new Part();
part2.setName("door");
Part part3 = new Part();
part3.setName("bumper");
car.setParts(asList(part1, part2, part3));
System.out.println("serialized: " + serialize(car, Car.class, true));
} catch (IOException ex) {
Logger.getLogger(NestedWrapping.class.getName()).log(Level.SEVERE, null, ex);
}
}

public static final synchronized String serialize(final Object object, final Class type, final Boolean withHeaders) throws IOException {
CsvMapper csvMapper = new CsvMapper();
CsvSchema csvSchema;
if (withHeaders) {
csvSchema = csvMapper.schemaFor(type).withHeader();
} else {
csvSchema = csvMapper.schemaFor(type).withoutHeader();
}
return csvMapper.writer(csvSchema).writeValueAsString(object);
}

}


Nothing I try seems to work, I've read every post on stackoverflow and github about the topic but I can't find a working solution.

Sorry about any pointless annotations that I've left behind for no reason and if you answer with code, please feel free to remove them.

Answer

From the error, I would like to believe that it has something to do with your schema for a Car, which has the columns of {"color"} taken from @JsonPropertyOrder on Car and not a "name" value.

You probably want to add "parts" in there, but you would get the same error that "name" is not part of that schema.

After a few changes to your code, I was able to serialize and deserialize a Car object.

Part

Here, after some other changes it needed a constructor with a single String value, so add that

@JsonPropertyOrder({"name"})
public static class Part {
    @JsonProperty("name")
    private String name;

    public Part() {
        this("");
    }

    public Part(String partJSON) {
        // TODO: Unserialize the parameter... it is a serialized Part string... 
        this.name = partJSON;
    }

Car

Here, you will need to implement a method that will convert the List<Part> into a CSV-readable format manually.

Such a method would look like this

@JsonGetter("parts")
public String getPartString() {
    String separator = ";";
    StringBuilder sb = new StringBuilder();

    Iterator<Part> iter = this.parts.iterator();
    while (iter.hasNext()) {
        Part p = iter.next();
        sb.append(p.getName());

        if (iter.hasNext())
            sb.append(separator);
    }

    return sb.toString();
} 

Also, don't forget to fix the schema at the top of the class

@JsonPropertyOrder({"color", "parts"})
public static class Car {

    @JsonProperty("color")
    private String color;
    @JsonProperty("parts")
    private List<Part> parts;

    public Car() {
        this.parts = new ArrayList<>();
    }

serialize

You can change your serialize method to take the type of the class as a generic type parameter instead of an explicit Class like so.

public static final synchronized <T> String serialize(final T object, final Boolean withHeaders) throws IOException {
    CsvMapper csvMapper = new CsvMapper();
    CsvSchema csvSchema = csvMapper.schemaFor(object.getClass());

    if (withHeaders) {
        csvSchema = csvSchema.withHeader();
    } else {
        csvSchema = csvSchema.withoutHeader();
    }

    return csvMapper.writer(csvSchema).writeValueAsString(object);
}

main - writer

Now, if you serialize a Car, you should see

color,parts
red,gearbox;door;bumper

main - reader

And reading that CSV string and looping over the Car.getParts()

Car car = mapper.readerFor(Car.class).with(csvSchema).readValue(csv);

for (Part p : car.getParts()) {
    System.out.println(p.getName());
}
gearbox
door
bumper
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