Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
324 views
in Technique[技术] by (71.8m points)

java - How do I create beans programmatically in Spring Boot?

I have an app that has a number of datasource settings listed in application.properties. I have a @ConfigurationProperties class that loads up these settings. Now I want to take the values from this ConfigurationProperties class and use them to create DataSource beans on-the-fly. I've tried using @PostConstruct and implementing BeanFactoryPostProcessor. However, with BeanFactoryPostProcessor, the processing seems to happen to early - before my ConfigurationProperties class has been populated. How can I read properties and create DataSource beans on the fly with Spring Boot?

Here's what my application.properties looks like:

ds.clients[0]=client1|jdbc:db2://server/client1
ds.clients[1]=client2,client3|jdbc:db2://server/client2
ds.clients[2]=client4|jdbc:db2://server/client4
ds.clients[3]=client5|jdbc:db2://server/client5

And my ConfigurationProperties class:

@Component
@ConfigurationProperties(prefix = "ds")
public class DataSourceSettings {
    public static Map<String, String> CLIENT_DATASOURCES = new LinkedHashMap<>();

    private List<String> clients = new ArrayList<>();

    public List<String> getClients() {
        return clients;
    }

    public void setClients(List<String> clients) {
        this.clients = clients;
    }

    @PostConstruct
    public void configure() {
        for (String client : clients) {
            // extract client name
            String[] parts = client.split("\|");
            String clientName = parts[0];
            String url = parts[1];
            // client to datasource mapping
            String dsName = url.substring(url.lastIndexOf("/") + 1);
            if (clientName.contains(",")) {
                // multiple clients with same datasource
                String[] clientList = clientName.split(",");
                for (String c : clientList) {
                    CLIENT_DATASOURCES.put(c, dsName);
                }
            } else {
                CLIENT_DATASOURCES.put(clientName, dsName);
            }
        }
    }

At the end of this @PostConstruct method, I'd like to create a BasicDataSource with these settings and add it to the ApplicationContext. However, if I try to do this by implement BeanFactoryPostProcessor and implementing postProcessBeanFactory, the clients property is null, as is the CLIENT_DATASOURCES that I've populated with @PostConstruct.

@Override
public void postProcessBeanFactory(ConfigurableListableBeanFactory beanFactory) throws BeansException {
    System.out.println("clients: " + CLIENT_DATASOURCES);
}

What's the best way to create datasources on-the-fly with Spring Boot?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

How about creating your beans and ask Spring Boot to inject values into it?

Something like

@Bean
@ConfigurationProperties("ds.client1")
public DataSource dataSourceClient1() {
    DataSourceBuilder.create().build();
}

@Bean
@ConfigurationProperties("ds.client2")
public DataSource dataSourceClient2() {
    DataSourceBuilder.create().build();
}

Then, any setting in the ds.client1 namespace belongs to the first data source (i.e. ds.client1.password is the data source password for that DataSource).

But maybe you don't know how much data sources you'll have? This is getting more complicated, especially if you need to inject those dynamic data sources in other objects. If you only need to lookup them by name, you could register them yourself as singletons. Here is an example that works

@ConfigurationProperties(prefix = "ds")
public class DataSourceSettings implements BeanFactoryAware {

    private List<String> clients = new ArrayList<>();

    private BeanFactory beanFactory;

    public List<String> getClients() {
        return clients;
    }

    public void setClients(List<String> clients) {
        this.clients = clients;
    }

    @Override
    public void setBeanFactory(BeanFactory beanFactory) {
        this.beanFactory = beanFactory;
    }

    @PostConstruct
    public void configure() {
        Map<String, String> clientDataSources = new HashMap<String, String>();
        for (String client : clients) {
            // extract client name
            String[] parts = client.split("\|");
            String clientName = parts[0];
            String url = parts[1];
            // client to datasource mapping
            String dsName = url.substring(url.lastIndexOf("/") + 1);
            if (clientName.contains(",")) {
                // multiple clients with same datasource
                String[] clientList = clientName.split(",");
                for (String c : clientList) {
                    clientDataSources.put(c, url);
                }
            }
            else {
                 clientDataSources.put(clientName, url);
            }
        }
        Assert.state(beanFactory instanceof ConfigurableBeanFactory, "wrong bean factory type");
        ConfigurableBeanFactory configurableBeanFactory = (ConfigurableBeanFactory) beanFactory;
        for (Map.Entry<String, String> entry : clientDataSources.entrySet()) {
            DataSource dataSource = createDataSource(entry.getValue());
            configurableBeanFactory.registerSingleton(entry.getKey(), dataSource);
        }
    }

    private DataSource createDataSource(String url) {
        return DataSourceBuilder.create().url(url).build();
    }
}

Note that those beans are only available by bean name lookup. Let me know if that works out for you.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

1.4m articles

1.4m replys

5 comments

57.0k users

...