@guyheylens uploaded a file: OK Thanks, This did the trick

I think this means that you have only one customer per country

Hi Pavel, this is a silly example, indeed I would have More Than One Customer per country.

@pavel Thanks for the help, my api is ready now for production 😉

What's the best practice to implement UserDetails in kotlin for secure Spring Boot app with JPA? Is it still the following?

class User(
    private val username: String = ""
    ...
): UserDetails {
    override fun getUsername() = username
    ...
}

I've also tried

MyUser(...) : User(username, password, authorities)

but jpa won't work due to missing no-arg constructor, no-arg plugin also does not help here (probably due to User coming from the spring library, not my code).

are you using default

jdbcAuthentication()

?

@krotki yes, planning to

What I did is probably not 100% best practice ... but works for me ... I created table users on jdbc something along the lines:

@Entity
@Table(name = "users")
data class User(

        @Id
        @GeneratedValue
        var id: Long? = null,

        @NotNull
        @Column(nullable = false, unique = true)
        var username: String = "",

        @NotNull
        var password: String = "",

        @NotNull
        var enabled: Boolean = true,

        @NotNull
        @Column(nullable = false, unique = true)
        var principalId: Long = -1,

        var email: String? = null,

        var comment: String? = null,

        @Column(insertable = false, updatable = false, nullable = false, columnDefinition = "TIMESTAMP DEFAULT now()")
        var dateCreated: Instant? = null,

        @Column(insertable = false, updatable = false, nullable = false, columnDefinition = "TIMESTAMP DEFAULT now()")
        var dateUpdated: Instant? = null
)

default jdbcAuthentication looks for 2 tables users and authorities

So, what you're saying is: do not use UserDetails at all, just conform to jdbcAuthentication's database structure?

this is what I did ... I had problems with UserDetails and this seemed to be easier

Hm... figured it out. I'll use my custom user class, and just create spring security's User from my class in UserDetailsService implementation.

I had the same problem, what I did is implement

UserDetails

in java , lol

does jackson work with kotlin? I get the following error with my data class

no suitable constructor found, can not deserialize from Object value

add kotlin module https://github.com/FasterXML/jackson-module-kotlin

if you are using current spring boot you need only to add dependency

i am using

springBootVersion = '1.5.2.RELEASE'

i think will work 🙂

👍 let me give it a shot! i assume i only need the dep and no config?

yup

or maybe im being naive 😛

ok cool

hmm still get the same error

if you create

ObjectMapper

by yourselt ten you need to register module with

ObjectMapper().registerKotlinModule()

... see the github link. But spring boot on start registers by default all jackson modules on classpath if i remember correctly ...

you can also ask on #jackson-kotlin

@krotki or use no-arg gradle/maven plugin

my fault — sorry i should have updated i got it to work with what you are mentioning above. silly mistake on my end

@krotki You can also use Jackson2ObjectMapperBuilder in order to create a custom ObjectMapper that will use Spring defaults, including auto registration of the Jackson Kotlin support

yes yes, this is exactly what i'm doing 🙂