CODEWITHSUNDEEP
javagrailsgrails-orm
nightingale2k1
nightingale2k1

Reputation: 10335

Grails error validating on existing input

I got this error when I tried to modify my Controller (generated by scaffolding). What I want to do, is to add current user that I capture from SpringSecurity Service.

Payment class has User property inside. I tried to inject the user value from springSecurityService.getCurrentUser();

When I tried this, it will produce error : Property [user] of class [class kks.Payment] cannot be null

I tried to debug, it looks fine but still producing this error ....

This will work well if I put user.id inside the form as hidden value .. but I didn't want do this ...

How to solve this problem ?

Here is my code:

class PaymentController {

    def springSecurityService;

    @Transactional
    def save(Payment PaymentInstance) {
        println("1..... Payment");
        PaymentInstance.user = springSecurityService.getCurrentUser();
        if (PaymentInstance == null) {
            notFound()
            return
        }

        println("2..... Payment"+springSecurityService.getCurrentUser().class);
        println(PaymentInstance.user.id); // still okay !!! 

        if (PaymentInstance.hasErrors()) {
            println("3..... hasErrors?"); // entered here with  Property [user] of class > [class kks.Payment] cannot be null

            respond PaymentInstance.errors, view: 'create'
            return
        }


        println("3..... Payment");
        PaymentInstance.save flush: true

        request.withFormat {
            form multipartForm {
                flash.message = message(code: 'default.created.message', args: >[message(code: 'Payment.label',
                                        default: 'Payment'), PaymentInstance.id])
                redirect PaymentInstance
            }
            '*' { respond PaymentInstance, [status: CREATED] }
        }
    } 
}

Upvotes: 0

Views: 39

Answers (1)

MKB
MKB

Reputation: 7619

After inserting value in the paymentInstance you should validate the instance like:

paymentInstance.user = springSecurityService.getCurrentUser();
paymentInstance.validate()

and then check for error.

PS: Please follow convention: variable name should be started from lower case character.

Upvotes: 1

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