数组的 Jackson-CSV 架构

Mat*_*ski 6 java csv schema jackson deserialization

我有CSV需要解析的文件。该文件的架构如下:

姓名、联系人 - 其中联系人是每个人的许多字符串,其中该联系人列的数量不规则。

例如:

john, john@wick.com, 123 123 123, fb/john.wick
mike, 123 0303 11
dave,
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我正在尝试为我的 bean创建一个CsvSchemawith :Jacskon CSV

public class Person {
    private String name;
    private String[] contacts;
}
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通过创建自定义架构:

CsvSchema schema = CsvSchema.builder()
    .addColumn("name")
    .addArrayColumn("contacts", ",")
    .build(); 
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但我得到这个:

com.fasterxml.jackson.dataformat.csv.CsvMappingException: Too many entries: expected at most 2
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如何解决Jackson CSV这样的问题?

Java代码:

CsvMapper mapper = new CsvMapper();
CsvSchema schema = CsvSchema.builder()
    .addColumn("name")
    .addArrayColumn("contacts", ",")
    .build();
MappingIterator<Person> it = mapper.readerFor(Person.class).with(schema)
    .readValues(csvString);
List<Person> all = it.readAll();
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Mic*_*ber 4

您可以使用CsvParser.Feature.WRAP_AS_ARRAY功能并将整行读取为List<String>. 在构造函数中你可以转换ListPerson对象。请参见下面的示例:

import com.fasterxml.jackson.databind.MappingIterator;
import com.fasterxml.jackson.dataformat.csv.CsvMapper;
import com.fasterxml.jackson.dataformat.csv.CsvParser;

import java.io.File;
import java.util.List;
import java.util.stream.Collectors;

public class CsvApp {

    public static void main(String[] args) throws Exception {
        File csvFile = new File("./resource/test.csv").getAbsoluteFile();

        CsvMapper csvMapper = new CsvMapper();
        csvMapper.enable(CsvParser.Feature.WRAP_AS_ARRAY);

        MappingIterator<List<String>> rows = csvMapper.readerFor(List.class).readValues(csvFile);
        List<Person> persons = rows.readAll().stream()
                .filter(row -> !row.isEmpty())
                .map(Person::new)
                .collect(Collectors.toList());

        persons.forEach(System.out::println);
    }
}

class Person {

    private String name;
    private List<String> contacts;

    public Person(List<String> row) {
        this.name = row.remove(0);
        this.contacts = row;
    }

    public String getName() {
        return name;
    }

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

    public List<String> getContacts() {
        return contacts;
    }

    public void setContacts(List<String> contacts) {
        this.contacts = contacts;
    }

    @Override
    public String toString() {
        return "Person{" +
                "name='" + name + '\'' +
                ", contacts=" + contacts +
                '}';
    }
}
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对于您的输入,上面的代码打印:

Person{name='john', contacts=[john@wick.com, 123 123 123, fb/john.wick]}
Person{name='mike', contacts=[123 0303 11]}
Person{name='dave', contacts=[]}
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