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Java convert list of string with comma separated values to list of objects

I'd like to convert a json array of person

persons: [
"1, Franck, 1980-01-01T00:00:00",
"2, Marc, 1981-01-01T00:00:00"
]


To a list of Person object using this class:

class Person
private Integer id;
private String name;
private Date dateOfBirth;

// getter and setter



Would it be possible to use a converter and Java 8 to do it?

public Person convert(String from)
String data = from.split(",");
return new Person(Integer.parseInt(data[0]), data[1], new Date(data[2]));



string

json

2 Answers
2

You can do it like so,

Pattern idNumber = Pattern.compile("\d+");
List<Person> persons = Arrays.stream(from.split("}")).filter(s -> idNumber.matcher(s).find())
.map(s -> s.substring(s.indexOf("") + 1)).map(s -> s.split(","))
.map(a -> new Person(Integer.parseInt(a[0].replaceAll(""", "")), a[1],
LocalDateTime.parse(a[2].trim().replaceAll(""", ""))))
.collect(Collectors.toList());


First split each string using "" character, and then filter out invalid tokens, which does not contain a digit. Notice that each valid payload should contain Id number which is a digit. Finally remove any spurious trailing characters occur before the Id digit and map each resulting String to a Person object. At last collect the Person instances into a List.

split

String

Person

collect

Person

List

Notice that I have used LocalDateTime for the dateOfBirth field in the Person class. So the Person class looks like this.

LocalDateTime

dateOfBirth

Person

Person

public class Person
private final Integer id;
private final String name;
private final LocalDateTime dateOfBirth;
// remainder omitted for the sake of brevity.



However as you can see it is always intuitive to use some framework such as Jackson ObjectMapper to get the work done than writing all this. But in this case your Json is malformed so you won't be able to use such a framework unless you fix the json payload structure.

ObjectMapper

Update

Here's much more elegant Java9 solution.

String regexString = Pattern.quote(""") + "(.*?)" + Pattern.quote(""");
Pattern pattern = Pattern.compile(regexString);

List<Person> persons = pattern.matcher(from)
.results()
.map(mr -> mr.group(1)).map(s -> s.split(", "))
.map(a -> new Person(Integer.parseInt(a[0]), a[1], LocalDateTime.parse(a[2])))
.collect(Collectors.toList());


In Java 9, you can now use Matcher#results() to get a Stream<MatchResult>. Here's an excerpt from the documentation.

Matcher#results()

Stream<MatchResult>

Returns a stream of match results for each subsequence of the input
sequence that matches the pattern. The match results occur in the same
order as the matching subsequences in the input sequence.

I would rather recommend you to use Java9 to solve this.

If you need to do it just for a List of String representation of a Person, you may do it like so.

List

String

Person

List<String> personStr = Arrays.asList("1, Franck, 1980-01-01T00:00:00", "2, Marc, 1981-01-01T00:00:00");
List<Person> persons = personStr.stream()
.map(s -> s.replaceAll(" ", "").split(","))
.map(a -> new Person(Integer.parseInt(a[0]), a[1], LocalDateTime.parse(a[2])))
.collect(Collectors.toList());


Notice the latter is far more easier than the prior and contains just a subset of the original solution.

An Extention of your solution itself & with a bit of abstraction:

List<String> personStr = Arrays.asList("1, Franck, 1980-01-01T00:00:00", "2, Marc, 1981-01-01T00:00:00");

List<Person> persons = personStr.stream()
.map(Person::new)
.collect(Collectors.toList());


Where, Person class has a constructor which accepts a string arg to convert it to a Person object, as follows:

Person

string

Person

public Person(String from)
String data = from.split(",");
Arrays.parallelSetAll(data, i -> data[i].trim());
this.id = Integer.parseInt(data[0]);
this.name = data[1];
this.dateOfBirth = new Date(data[2]);



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