Spring Bean - Constructor with arguments

Question

I see an error in the constructor which says

"String queueName" cannot be autowired.

I have AmazonSQSAsync component defined in another class but not queueName. Why is the constructor trying to autowire the parameters and how can I resolve this?

@Configuration
public class SqsQueueHealthIndicator extends AbstractHealthIndicator {

    private final AmazonSQSAsync amazonSQSAsync;
    private final String queueName;

    public SqsQueueHealthIndicator(AmazonSQSAsync amazonSQSAsync, String queueName) {
        this.amazonSQSAsync = amazonSQSAsync;
        this.queueName = queueName;
    }

    @Override
    protected void doHealthCheck(Health.Builder builder) {
        try {
            amazonSQSAsync.getQueueUrl(queueName);
            builder.up();
        } catch (QueueDoesNotExistException e) {
            builder.down(e);
        }
    }

    @Bean
    SqsQueueHealthIndicator queueHealthIndicator(@Autowired AmazonSQSAsync amazonSQSAsync, @Value("${url}") String queueName) {
        return new SqsQueueHealthIndicator(amazonSQSAsync, queueName);
    }

    @Bean
    SqsQueueHealthIndicator deadLetterQueueHealthIndicator(@Autowired AmazonSQSAsync amazonSQSAsync, @Value("${dlqurl}") String deadLetterQueueName) {
        return new SqsQueueHealthIndicator(amazonSQSAsync, deadLetterQueueName);
    }

}

Every field is autowired when you use the @Configuration annotation. — Blake
– Blake, Commented May 11, 2018 at 3:18
Is it true only for constructors? I tried @Component and I saw the same error. — Punter Vicky
– Punter Vicky, Commented May 11, 2018 at 3:19

Mạnh Quyết Nguyễn · Accepted Answer · 2018-05-11 04:48:22Z

6

Because you're declaring it as final so its value must be initialized.

You did not provide default value so Spring understand its value will be injected.

So you have 2 options:

Remove the final modifier. Use @Value to inject your value from config file.

Create a bean type of String and inject it. You should name it:

@Bean("queueName")
public String getQueueName {return "xyz";}

And inject like:

    @Autowire
    @Qualifier("queueName")
    private final String queueName;

In normal circumstance, the option 1 is the go-to.

edited May 11, 2018 at 4:48

answered May 11, 2018 at 3:16

Mạnh Quyết Nguyễn

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1 Comment

Mehraj Malik Over a year ago

String is not a primitive data type

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