CODEWITHSUNDEEP
javaspringspring-boot
csfragstaa
csfragstaa

Reputation: 165

Spring boot read properties into Map

I am trying to create a Map<String, Product> from my .properties file in Spring boot 3.1. I have written my custom converter for this, but it seems that it is never executing. I haven't done custom conversions before and I am not sure if a converter is needed in this use case or if Spring boot is able to manage the conversion automagically (If I configure it correctly).

product-configurations.properties:

products.product[0].name=FirstProduct
products.product[0].productID=1234
products.product[0].availableOptions=Opt1
products.product[0].processing=Parallel

Product class:

import lombok.Getter;
import lombok.Setter;

@Getter
@Setter
public class Product {
    private String name;
    private String processing;
    private String availableOptions;
    private Integer productID;
}

Configuration class:

import java.util.List;

import org.springframework.boot.context.properties.ConfigurationProperties;
import org.springframework.context.annotation.Configuration;
import org.springframework.context.annotation.PropertySource;

import lombok.Getter;
import lombok.Setter;

@Configuration
@PropertySource("classpath:product-configurations.properties")
@ConfigurationProperties("products")
@Getter
@Setter
public class ProductProperties {

    private List<Product> product;
}

ProductConverter class:

import java.util.List;
import java.util.Map;

import org.springframework.boot.context.properties.ConfigurationPropertiesBinding;
import org.springframework.core.convert.converter.Converter;
import org.springframework.stereotype.Component;

import com.cb.clientfacingapp.model.Product;

@Component

@ConfigurationPropertiesBinding
public class ProductConverter implements Converter<List<Product>, Map<String, Product>> {

    @Override
    public Map<String, Product> convert(List<Product> source) {
        Product p = new Product();
        System.out.println("I am converting");
        return null;
    }

}

Can someone pls guide me on this?

Thanks in Advance.

Update:

Sorry, I forgot to mention how the map should be organized.

Map<String, Product> where String is basically the name property of the product, and Product is the POJO class for that specific product, So each Product is mapped to its Name, basically.

Upvotes: 3

Views: 5296

Answers (2)

Frankie Drake
Frankie Drake

Reputation: 1388

Seems like what you're trying to achieve can be done without conversion

root.products.productName1.name=FirstProduct
root.products.productName1.productID=1234
root.products.productName1.availableOptions=Opt1
root.products.productName1.processing=Parallel

root.products.productName2.name=SecondProduct
root.products.productName2.productID=5678
root.products.productName2.availableOptions=Opt2
root.products.productName2.processing=Parallel

Therefore productName1, productName2 would be your map key names.

You can use it this way:

@ConfigurationProperties("root")
@Getter
@Setter
public class ProductProperties {

    private Map<String, Product> products;
}

Upvotes: 2

Damian229
Damian229

Reputation: 472

You could do what @Daria has suggested, which is the simplier way to do it.

In case you're still wondering how to convert a list into a Map, it can be done using Streams:

import java.util.List;
import java.util.Map;

import org.springframework.boot.context.properties.ConfigurationPropertiesBinding;
import org.springframework.core.convert.converter.Converter;
import org.springframework.stereotype.Component;

import com.cb.clientfacingapp.model.Product;

@Component
@ConfigurationPropertiesBinding
public class ProductConverter implements Converter<List<Product>, Map<String, Product>> {

    @Override
    public Map<String, Product> convert(List<Product> source) {
        return source.stream().collect(
            Collectors.toMap(Product::getName,Function.identity));

    }
}

It would generate a map with the product name as key value, and the product object as content value.

Upvotes: 1

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